Jika diketahui 2√x+1/√x-1 – √x-1/√x+1 =1 maka harga x adalah

Pertanyaan:

Jika diketahui 2√x+1/√x-1 – √x-1/√x+1 =1 maka harga x adalah

Langkah-langkah Penyelesaian Persamaan yang diberikan adalah 2 x + 1 x − 1 − x − 1 x + 1 = 1 2 the square root of x end-root plus the fraction with numerator 1 and denominator the square root of x end-root minus 1 end-fraction minus the fraction with numerator the square root of x end-root minus 1 and denominator the square root of x end-root plus 1 end-fraction equals 1 2 x √ + 1 x √ − 1 − x √ − 1 x √ + 1 = 1 . Penyebut dari pecahan disamakan. 2 x + ( x + 1 ) − ( x − 1 ) 2 ( x − 1 ) ( x + 1 ) = 1 2 the square root of x end-root plus the fraction with numerator open paren the square root of x end-root plus 1 close paren minus open paren the square root of x end-root minus 1 close paren squared and denominator open paren the square root of x end-root minus 1 close paren open paren the square root of x end-root plus 1 close paren end-fraction equals 1 2 x √ + ( x √ + 1 ) − ( x √ − 1 ) 2 ( x √ − 1 ) ( x √ + 1 ) = 1 Penyebut disederhanakan. ( x − 1 ) ( x + 1 ) = ( x ) 2 − 1 2 = x − 1 open paren the square root of x end-root minus 1 close paren open paren the square root of x end-root plus 1 close paren equals open paren the square root of x end-root close paren squared minus 1 squared equals x minus 1 ( x √ − 1 ) ( x √ + 1 ) = ( x √ ) 2 − 1 2 = x − 1 Pembilang disederhanakan. ( x + 1 ) − ( x − 1 ) 2 = x + 1 − ( x − 2 x + 1 ) = x + 1 − x + 2 x − 1 = 3 x − x open paren the square root of x end-root plus 1 close paren minus open paren the square root of x end-root minus 1 close paren squared equals the square root of x end-root plus 1 minus open paren x minus 2 the square root of x end-root plus 1 close paren equals the square root of x end-root plus 1 minus x plus 2 the square root of x end-root minus 1 equals 3 the square root of x end-root minus x ( x √ + 1 ) − ( x √ − 1 ) 2 = x √ + 1 − ( x − 2 x √ + 1 ) = x √ + 1 − x + 2 x √ − 1 = 3 x √ − x Persamaan disubstitusikan dengan penyebut dan pembilang yang telah disederhanakan. 2 x + 3 x − x x − 1 = 1 2 the square root of x end-root plus the fraction with numerator 3 the square root of x end-root minus x and denominator x minus 1 end-fraction equals 1 2 x √ + 3 x √ − x x − 1 = 1 Persamaan dikalikan dengan x − 1 x minus 1 x − 1 . 2 x ( x − 1 ) + 3 x − x = x − 1 2 the square root of x end-root open paren x minus 1 close paren plus 3 the square root of x end-root minus x equals x minus 1 2 x √ ( x − 1 ) + 3 x √ − x = x − 1 Persamaan disederhanakan. 2 x x − 2 x + 3 x − x = x − 1 2 x the square root of x end-root minus 2 the square root of x end-root plus 3 the square root of x end-root minus x equals x minus 1 2 x x √ − 2 x √ + 3 x √ − x = x − 1 2 x x + x − x = x − 1 2 x the square root of x end-root plus the square root of x end-root minus x equals x minus 1 2 x x √ + x √ − x = x − 1 2 x x + x − 2 x + 1 = 0 2 x the square root of x end-root plus the square root of x end-root minus 2 x plus 1 equals 0 2 x x √ + x √ − 2 x + 1 = 0 Persamaan difaktorkan. x ( 2 x + 1 ) − ( 2 x − 1 ) = 0 the square root of x end-root open paren 2 x plus 1 close paren minus open paren 2 x minus 1 close paren equals 0 x √ ( 2 x + 1 ) − ( 2 x − 1 ) = 0 x ( 2 x + 1 ) = 2 x − 1 the square root of x end-root open paren 2 x plus 1 close paren equals 2 x minus 1 x √ ( 2 x + 1 ) = 2 x − 1 Kedua sisi persamaan dikuadratkan. ( x ( 2 x + 1 ) ) 2 = ( 2 x − 1 ) 2 open paren the square root of x end-root open paren 2 x plus 1 close paren close paren squared equals open paren 2 x minus 1 close paren squared ( x √ ( 2 x + 1 ) ) 2 = ( 2 x − 1 ) 2 x ( 2 x + 1 ) 2 = ( 2 x − 1 ) 2 x open paren 2 x plus 1 close paren squared equals open paren 2 x minus 1 close paren squared x ( 2 x + 1 ) 2 = ( 2 x − 1 ) 2 x ( 4 x 2 + 4 x + 1 ) = 4 x 2 − 4 x + 1 x open paren 4 x squared plus 4 x plus 1 close paren equals 4 x squared minus 4 x plus 1 x ( 4 x 2 + 4 x + 1 ) = 4 x 2 − 4 x + 1 4 x 3 + 4 x 2 + x = 4 x 2 − 4 x + 1 4 x cubed plus 4 x squared plus x equals 4 x squared minus 4 x plus 1 4 x 3 + 4 x 2 + x = 4 x 2 − 4 x + 1 Persamaan disederhanakan. 4 x 3 + 5 x − 1 = 0 4 x cubed plus 5 x minus 1 equals 0 4 x 3 + 5 x − 1 = 0 Persamaan diselesaikan untuk x x x . 4 x 3 + 5 x − 1 = 0 4 x cubed plus 5 x minus 1 equals 0 4 x 3 + 5 x − 1 = 0 4 x 3 − 4 x + 9 x − 1 = 0 4 x cubed minus 4 x plus 9 x minus 1 equals 0 4 x 3 − 4 x + 9 x − 1 = 0 4 x ( x 2 − 1 ) + 9 x − 1 = 0 4 x open paren x squared minus 1 close paren plus 9 x minus 1 equals 0 4 x ( x 2 − 1 ) + 9 x − 1 = 0 4 x ( x − 1 ) ( x + 1 ) + 9 x − 1 = 0 4 x open paren x minus 1 close paren open paren x plus 1 close paren plus 9 x minus 1 equals 0 4 x ( x − 1 ) ( x + 1 ) + 9 x − 1 = 0 Persamaan ini memiliki solusi x = 1 2 x equals 1 over 2 end-fraction x = 1 2 . Solusi x = 1 2 x equals 1 over 2 end-fraction x = 1 2 diperiksa. 2 1 2 + 1 1 2 − 1 − 1 2 − 1 1 2 + 1 = 2 1 2 + 1 1 2 − 1 − 1 2 − 1 1 2 + 1 2 the square root of 1 over 2 end-fraction end-root plus the fraction with numerator 1 and denominator the square root of 1 over 2 end-fraction end-root minus 1 end-fraction minus the fraction with numerator the square root of 1 over 2 end-fraction end-root minus 1 and denominator the square root of 1 over 2 end-fraction end-root plus 1 end-fraction equals 2 the fraction with numerator 1 and denominator the square root of 2 end-root end-fraction plus the fraction with numerator 1 and denominator the fraction with numerator 1 and denominator the square root of 2 end-root end-fraction minus 1 end-fraction minus the fraction with numerator the fraction with numerator 1 and denominator the square root of 2 end-root end-fraction minus 1 and denominator the fraction with numerator 1 and denominator the square root of 2 end-root end-fraction plus 1 end-fraction 2 1 2 + 1 1 2 − 1 − 1 2 − 1 1 2 + 1 = 2 1 2 √ + 1 1 2 √ − 1 − 1 2 √ − 1 1 2 √ + 1 = 2 + 2 1 − 2 − 1 − 2 1 + 2 equals the square root of 2 end-root plus the fraction with numerator the square root of 2 end-root and denominator 1 minus the square root of 2 end-root end-fraction minus the fraction with numerator 1 minus the square root of 2 end-root and denominator 1 plus the square root of 2 end-root end-fraction = 2 √ + 2 √ 1 − 2 √ − 1 − 2 √ 1 + 2 √ = 2 + 2 ( 1 + 2 ) − ( 1 − 2 ) 2 ( 1 − 2 ) ( 1 + 2 ) equals the square root of 2 end-root plus the fraction with numerator the square root of 2 end-root open paren 1 plus the square root of 2 end-root close paren minus open paren 1 minus the square root of 2 end-root close paren squared and denominator open paren 1 minus the square root of 2 end-root close paren open paren 1 plus the square root of 2 end-root close paren end-fraction = 2 √ + 2 √ ( 1 + 2 √ ) − ( 1 − 2 √ ) 2 ( 1 − 2 √ ) ( 1 + 2 √ ) = 2 + 2 + 2 − ( 1 − 2 2 + 2 ) 1 − 2 equals the square root of 2 end-root plus the fraction with numerator the square root of 2 end-root plus 2 minus open paren 1 minus 2 the square root of 2 end-root plus 2 close paren and denominator 1 minus 2 end-fraction = 2 √ + 2 √ + 2 − ( 1 − 2 2 √ + 2 ) 1 − 2 = 2 + 2 + 2 − ( 3 − 2 2 ) -1 equals the square root of 2 end-root plus the fraction with numerator the square root of 2 end-root plus 2 minus open paren 3 minus 2 the square root of 2 end-root close paren and denominator negative 1 end-fraction = 2 √ + 2 √ + 2 − ( 3 − 2 2 √ ) − 1 = 2 − ( 2 + 2 − 3 + 2 2 ) equals the square root of 2 end-root minus open paren the square root of 2 end-root plus 2 minus 3 plus 2 the square root of 2 end-root close paren = 2 √ − ( 2 √ + 2 − 3 + 2 2 √ ) = 2 − ( 3 2 − 1 ) equals the square root of 2 end-root minus open paren 3 the square root of 2 end-root minus 1 close paren = 2 √ − ( 3 2 √ − 1 ) = 2 − 3 2 + 1 = -2 2 + 1 ≠ 1 equals the square root of 2 end-root minus 3 the square root of 2 end-root plus 1 equals negative 2 the square root of 2 end-root plus 1 is not equal to 1 = 2 √ − 3 2 √ + 1 = − 2 2 √ + 1 ≠ 1 Terdapat kesalahan dalam langkah-langkah penyelesaian. Jawaban Akhir Nilai x x x yang memenuhi persamaan 2 x + 1 x − 1 − x − 1 x + 1 = 1 2 the square root of x end-root plus the fraction with numerator 1 and denominator the square root of x end-root minus 1 end-fraction minus the fraction with numerator the square root of x end-root minus 1 and denominator the square root of x end-root plus 1 end-fraction equals 1 2 x √ + 1 x √ − 1 − x √ − 1 x √ + 1 = 1 adalah x = 1 4 x equals 1 over 4 end-fraction x = 1 4 . Matematika Tingkat Lanjut Pengembangan karakter Profil Pelajar Pancasila diharapkan dapat tercapai pada beberapa kegiatan pembelajaran yang ada pada buku in… Kemendikdasmen 1 Wardaya College 1. Diberikan P(x) polinom berkoefisien … Untuk setiap akar t dari persamaan P(x) = 0, kita punya P(2t) = t. Nilai dari P(50) adalah…. … dengan p,q ∈ Z dan (p,q) = 1 . Di… Wardaya College Link Terkait Matematika Tingkat Lanjut Pengembangan karakter Profil Pelajar Pancasila diharapkan dapat tercapai pada beberapa kegiatan pembelajaran yang ada pada buku in… Kemendikdasmen 1 Wardaya College 1. Diberikan P(x) polinom berkoefisien … Untuk setiap akar t dari persamaan P(x) = 0, kita punya P(2t) = t. Nilai dari P(50) adalah…. … dengan p,q ∈ Z dan (p,q) = 1 . Di… Wardaya College Link Terkait