Câu hỏi của Phan Lê Kim Chi

Question

(x-1)²-(x+2)(x-5)+(4x-1)²=(-x-1)(2-16x)

tìm x

Xyz OLM · Accepted Answer

(x - 1)² - (x + 2)(x - 5) + (4x - 1)² = (-x - 1)(2 - 16x)

=> x² - 2x + 1 - x² + 5x - 2x + 10 + 16x² - 8x + 1 = -2x + 16x² -2 + 16x

=> 16x² - 7x + 12 = 14x - 2 + 16x²

=> -21x = -14

=> 21x = 14

=> x = 2/3

Câu trả lời:

(x - 1)² - (x + 2)(x - 5) + (4x - 1)² = (-x - 1)(2 - 16x)

=> x² - 2x + 1 - x² + 5x - 2x + 10 + 16x² - 8x + 1 = -2x + 16x² -2 + 16x

=> 16x² - 7x + 12 = 14x - 2 + 16x²

=> -21x = -14

=> 21x = 14

=> x = 2/3

Khánh Ngọc · Answer

$\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)$

$\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0$

$\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0$

$\Rightarrow-21x+14=0$

$\Rightarrow-21x=-14$

$\Rightarrow x=\frac{2}{3}$

Câu trả lời:

$\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)$

$\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0$

$\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0$

$\Rightarrow-21x+14=0$

$\Rightarrow-21x=-14$

$\Rightarrow x=\frac{2}{3}$

Capheny Bản Quyền · Answer

$x^2-2x+1-\left(x^2-3x-10\right)+16x^2-8x+1=16x^2+14x-2$

$x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0$

$-21x-14=0$

$-21x=14$

$x=\frac{-2}{3}$

Câu trả lời:

$x^2-2x+1-\left(x^2-3x-10\right)+16x^2-8x+1=16x^2+14x-2$

$x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0$

$-21x-14=0$

$-21x=14$

$x=\frac{-2}{3}$