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ĐKXĐ: ....
\(\Leftrightarrow x^2-8x+16+3x+4-8\sqrt{3x+4}+16=0\)
\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{3x+4}-4\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{3x+4}-4=0\end{matrix}\right.\) \(\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a) ĐKXĐ: $x\ge \frac{-4}{3}$
Ta có:
PT \(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)
\(\Leftrightarrow (x^2-8x+16)+(3x+4-8\sqrt{3x+4}+16)=0\)
\(\Leftrightarrow (x-4)^2+(\sqrt{3x+4}-4)^2=0\)
Dễ thấy \((x-4)^2\geq 0; (\sqrt{3x+4}-4)^2\geq 0, \forall x\geq \frac{-4}{3}\)
Do đó để tổng của chúng bằng $0$ thì \((x-4)^2=(\sqrt{3x+4}-4)^2=0\Leftrightarrow x=4\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq \frac{2}{3}$
\(\sqrt{3x-2}=2-\sqrt{3}\)
\(\Rightarrow 3x-2=(2-\sqrt{3})^2=7-4\sqrt{3}\)
\(\Rightarrow x=\frac{7-4\sqrt{3}+2}{3}=\frac{9-4\sqrt{3}}{3}\) (thỏa mãn)
Vậy.......
Lời giải:
a) ĐKXĐ: $x\ge \frac{-4}{3}$
Ta có:
PT \(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)
\(\Leftrightarrow (x^2-8x+16)+(3x+4-8\sqrt{3x+4}+16)=0\)
\(\Leftrightarrow (x-4)^2+(\sqrt{3x+4}-4)^2=0\)
Dễ thấy \((x-4)^2\geq 0; (\sqrt{3x+4}-4)^2\geq 0, \forall x\geq \frac{-4}{3}\)
Do đó để tổng của chúng bằng $0$ thì \((x-4)^2=(\sqrt{3x+4}-4)^2=0\Leftrightarrow x=4\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq \frac{2}{3}$
\(\sqrt{3x-2}=2-\sqrt{3}\)
\(\Rightarrow 3x-2=(2-\sqrt{3})^2=7-4\sqrt{3}\)
\(\Rightarrow x=\frac{7-4\sqrt{3}+2}{3}=\frac{9-4\sqrt{3}}{3}\) (thỏa mãn)
Vậy.......
![](https://rs.olm.vn/images/avt/0.png?1311)
dk \(x\ge-\frac{4}{3}\)
\(x^2-5x+4=8\sqrt{3x+4}-32\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=8\left(\sqrt{3x+4}-4\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)-8\frac{\left(\sqrt{3x+4}-4\right)\left(\sqrt{3x+4}+4\right)}{\sqrt{3x+4}+4}=0\)
\(\left(x-1\right)\left(x-4\right)-8.\frac{3\left(x-4\right)}{\sqrt{3x+4}+4}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1-\frac{24}{\sqrt{3x+4}+4}=0\right)\)
đến đây để rồi tự làm nhé ^^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-5x+36=8\sqrt{3x+4}\)
\(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)
\(\Leftrightarrow\left(-8\sqrt{3x+4}+32\right)+\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow-8\left(\sqrt{3x+4}-4\right)+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x+4-16}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x-12}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\left(x-4\right)\left(\frac{-24}{\sqrt{3x+4}+4}+x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\frac{-24}{\sqrt{3x+4}+4}+x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-\frac{24}{\sqrt{3x+4}+4}+3+x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-3.\frac{16-3x-4}{\left(\sqrt{3x+4}+4\right)^2}+\left(x-4\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(x-4\right)\left[\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1\right]=0\end{cases}}\)
Mà \(\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1>0\forall x\) nên \(x-4=0\Rightarrow x=4\)
Vật PT có nghiệm duy nhất là \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)