Câu hỏi của Trần Thị Khiêm

Question

Tim x (x+3)^2-(x-4)(x+8)=1

nguyen thi vang · Accepted Answer

$\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\left(x^2+6x+9\right)-\left(x^2+8x-4x-32\right)=1$

$x^2+6x+9-x^2-8x+4x+32=1$

$2x+41=1$

$\Rightarrow x=\dfrac{40}{2}=20$Câu trả lời: $\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\left(x^2+6x+9\right)-\left(x^2+8x-4x-32\right)=1$

$x^2+6x+9-x^2-8x+4x+32=1$

$2x+41=1$

$\Rightarrow x=\dfrac{40}{2}=20$

Hồng Quang · Answer

$\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\Leftrightarrow x^2+6x+9-x^2+8x-4x-32=1$

$\Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1$

$\Leftrightarrow x^2+6x+9-x^2-4x+32=1$

$\Leftrightarrow2x+41=1$

$\Leftrightarrow2x=-40\Leftrightarrow x=-20$Câu trả lời: $\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\Leftrightarrow x^2+6x+9-x^2+8x-4x-32=1$

$\Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1$

$\Leftrightarrow x^2+6x+9-x^2-4x+32=1$

$\Leftrightarrow2x+41=1$

$\Leftrightarrow2x=-40\Leftrightarrow x=-20$

nguyen thi vang · Answer

$\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\left(x^2+6x+9\right)-\left(x^2+8x-4x-32\right)=1$

$x^2+6x+9-x^2-8x+4x+32=1$

$2x+41=1$

$x=\dfrac{-40}{2}=-20$Câu trả lời: $\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1$

$\left(x^2+6x+9\right)-\left(x^2+8x-4x-32\right)=1$

$x^2+6x+9-x^2-8x+4x+32=1$

$2x+41=1$

$x=\dfrac{-40}{2}=-20$

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