ĐK:....

\(x^2-3x+8=4\sqrt{3x-5}\)

\(\Leftrightarrow x^2-3x+8-4\sqrt{3x-5}=0\)

\(\Leftrightarrow3x-5-4\sqrt{3x-5}+4+x^2-6x+9=0\)

\(\Leftrightarrow\left(\sqrt{3x-5}-2\right)^2+\left(x-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x-5}-2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=4\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

xem ra anh ghê nhỉ:)

Lời giải:

ĐKXĐ: \(x\geq \frac{5}{3}\)

Ta có: \(4\sqrt{3x-5}-8=0\Leftrightarrow 4\sqrt{3x-5}=8\Leftrightarrow \sqrt{3x-5}=2\)

\(\Rightarrow 3x-5=4\Rightarrow x=\frac{4+5}{3}=3\) (thỏa mãn)

Vậy $x=3$

\(dkxd:x\ge-1;\sqrt{x-4\sqrt{x+1}+3}=5\Leftrightarrow x-4\sqrt{x+1}+3=25\Leftrightarrow x+1-4\sqrt{x+1}+2=25\Leftrightarrow\left(x+1\right)-4\sqrt{x+1}+4=27\Leftrightarrow\left(\sqrt{x+1}-2\right)^2=27\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=-\sqrt{27}+2\left(< 0loai\right)\\\sqrt{x+1}=\sqrt{27}+2\left(tm\right)\end{matrix}\right.\Leftrightarrow x+1=31+4\sqrt{27}\Leftrightarrow x=30+4\sqrt{27}\)

\(\sqrt{x-4\sqrt{x+1}+3}=5\)

\(\Leftrightarrow x-4\sqrt{x+1}+3=25\)

\(\Leftrightarrow x-4\sqrt{x+1}-22=0\)

\(\Leftrightarrow x+1-4\sqrt{x+1}+4-27=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2=27=\left(\pm\sqrt{27}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}-2=\sqrt{27}\\\sqrt{x+1}-2=-\sqrt{27}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{27}+2\left(chon\right)\\\sqrt{x+1}=-\sqrt{27}-2\left(loai\right)\end{matrix}\right.\)

Xét \(\sqrt{x+1}=\sqrt{27}+2\)

\(\Leftrightarrow x+1=31+12\sqrt{3}\)

\(\Leftrightarrow x=30+12\sqrt{3}\)

Vậy...

ĐKXĐ: \(-2\le x\le2\)

Đặt \(\sqrt{2-x}+\sqrt{2+x}=a>0\Rightarrow a^2=4+2\sqrt{4-x^2}\)

Phương trình trở thành:

\(a+\frac{a^2-4}{2}=2\)

\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2-x}+\sqrt{2+x}=2\)

Mà \(\sqrt{2-x}+\sqrt{2+x}\ge\sqrt{2-x+2+x}=2\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left[{}\begin{matrix}2-x=0\\2+x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)(ĐK: \(\sqrt{2x-5}\ge0\Leftrightarrow x\ge\frac{5}{2}\)

\(\Leftrightarrow\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)+2\sqrt{2x-5}.3+9}+\sqrt{\left(2x-5\right)-2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|=4\)

\(\Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\)(vì \(\sqrt{2x-5}\ge0\) nên \(\sqrt{2x-5}+3\ge3>0\))

-TH: \(\sqrt{2x-5}-1\ge0\Leftrightarrow\sqrt{2x-5}\ge1\Leftrightarrow2x-5\ge1\Leftrightarrow x\ge3\) thì ta được phương trình:

\(\sqrt{2x-5}+3+\sqrt{2x-5}-1=4\)

\(\Leftrightarrow2\sqrt{2x-5}=2\)

\(\Leftrightarrow\sqrt{2x-5}=1\)

\(\Leftrightarrow2x-5=1\)

\(\Leftrightarrow x=3\left(chọn\right)\)

-TH: \(\sqrt{2x-5}-1< 0\Leftrightarrow x< 3\) thì ta được phương trình:

\(\sqrt{2x-5}+3+1-\sqrt{2x-5}=4\)

\(\Leftrightarrow4=4\)(luôn đúng với mọi \(\frac{5}{2}\le x< 3\))

Vậy nghiệm của phương trình là \(\frac{5}{2}\le x\le3\)

2,7612

\(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x+2}\ge0\end{matrix}\right.\Rightarrow\sqrt{x^2-4}+\sqrt{x+2}\ge0mà:\sqrt{x^2-4}+\sqrt{x+2}=0\Rightarrow\left\{{}\begin{matrix}x^2-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=-2\)

Em ko chắc đâu nhất là cái đk ý.

Nhận xét x = -2 là một nghiệm do đó xét x khác -2:

ĐK: \(x\ge2\). Đặt \(\sqrt{x+2}=a\ge2;\sqrt{x-2}=b\ge0\) . Theo đề bài thì:

ab + a = 0 <=> a(b+1) = 0 <=> a = 0 (loại) hoặc b = - 1( loại)

Vậy 1 nghiệm x = - 2???