mong mọi người giải giúp em vs 

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

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

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow x^2-6x+9+2x+3-2\sqrt{2x+3}+1+8=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{2x+3}-1\right)^2+8=0\)

Vế trái luôn dương nên pt vô nghiệm

Giải phương trình $x^2-4x+6=\sqrt{2x^2-5x+3}+\sqrt{-3x^2+9x-5}$ - Phương trình - hệ phương trình - bất phương trình - Diễn đàn Toán học

Đề bài \(ĐK\left(x\ge-\frac{3}{2}\right)\)

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

mà \(\left(x-3\right)^2+\left(\sqrt{2x+3}-3\right)^2\ge0\)

dấu = xảy ra khi x=3 (chọn )

zậy...

:V cách khác

Ta có:

\(x^2-4x+21=6\sqrt{2x+3}\left(x\ge-\frac{3}{2}\right)\)

\(\Leftrightarrow x^2-4x+21-18=6\left(\sqrt{2x+3}-3\right)\)

\(\Leftrightarrow x^2-4x+3=6\cdot\frac{2x-6}{\sqrt{2x+3}+3}\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)-\frac{12\left(x-3\right)}{\sqrt{2x+3}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left[x-1-\frac{12}{\sqrt{2x+3}+3}\right]=0\)

:V