Câu hỏi của Hồng Chan

Question

Sử dụng hằng đẳng thức $\sqrt{A^2}$= IAI để giải pt:

a) $\sqrt{9-12x+4x^2}$= 4 + x

b) $\sqrt{4-4x+x^2}$= ( x - 1 )2 + x - 6

Mysterious Person · Accepted Answer

a) $\sqrt{9-12x+4x^2}=4+x\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x$ $\Leftrightarrow\left|3-2x\right|=4+x$ th1: $3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow\Leftrightarrow x\le\dfrac{3}{2}$ $\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow3-2x=4+x\Leftrightarrow3x=-1\Leftrightarrow x=\dfrac{-1}{3}\left(tmđk\right)$ th2: $3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}$ $\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow2x-3=4+x\Leftrightarrow x=7\left(tmđk\right)$ vậy $x=\dfrac{-1}{3};x=7$ b) $\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6$ $\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6$ $\Leftrightarrow\left|2-x\right|=x^2-x-5$ th1: $2-x\ge0\Leftrightarrow x\le2$ $\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow2-x=x^2-x-5$ $\Leftrightarrow x^2=7\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{7}\left(loại\right)\x=-\sqrt{7}\left(tmđk\right)\end{matrix}\right.$ th2: $2-x< 0\Leftrightarrow x>2$ $\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow x-2=x^2-x-5$ $\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2+x-3x-3=0$ $\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0$ $\Leftrightarrow\left\{{}\begin{matrix}x-3=0\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(tmđk\right)\x=-1\left(loại\right)\end{matrix}\right.$ vậy $x=-\sqrt{7};x=3$Câu trả lời: a) $\sqrt{9-12x+4x^2}=4+x\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x$ $\Leftrightarrow\left|3-2x\right|=4+x$ th1: $3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow\Leftrightarrow x\le\dfrac{3}{2}$ $\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow3-2x=4+x\Leftrightarrow3x=-1\Leftrightarrow x=\dfrac{-1}{3}\left(tmđk\right)$ th2: $3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}$ $\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow2x-3=4+x\Leftrightarrow x=7\left(tmđk\right)$ vậy $x=\dfrac{-1}{3};x=7$ b) $\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6$ $\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6$ $\Leftrightarrow\left|2-x\right|=x^2-x-5$ th1: $2-x\ge0\Leftrightarrow x\le2$ $\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow2-x=x^2-x-5$ $\Leftrightarrow x^2=7\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{7}\left(loại\right)\x=-\sqrt{7}\left(tmđk\right)\end{matrix}\right.$ th2: $2-x< 0\Leftrightarrow x>2$ $\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow x-2=x^2-x-5$ $\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2+x-3x-3=0$ $\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0$ $\Leftrightarrow\left\{{}\begin{matrix}x-3=0\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(tmđk\right)\x=-1\left(loại\right)\end{matrix}\right.$ vậy $x=-\sqrt{7};x=3$

Huy Thắng Nguyễn · Answer

a) $\sqrt{9-12x+4x^2}=4+x$

$\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x$

$\Leftrightarrow\left|3-2x\right|=4+x$

$\Leftrightarrow\left[{}\begin{matrix}3-2x=4+x\3-2x=-4-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\x=7\end{matrix}\right.$

Vậy $x_1=-\dfrac{1}{3};x_2=7$.

b) $\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6$

$\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6$

$\Leftrightarrow\left|2-x\right|=x^2-x-5$

$\Leftrightarrow\left[{}\begin{matrix}2-x=x^2-x-5\2-x=-x^2+x+5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^2=7\x^2=2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\left(l\right)\x=-\sqrt{7}\x=3\x=-1\left(l\right)\end{matrix}\right.$

Vậy $x_1=-\sqrt{7};x_2=3$.Câu trả lời: a) $\sqrt{9-12x+4x^2}=4+x$