a/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)

\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)

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

\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)

b/ ĐKXĐ: ....

Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)

\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)

\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)

a/ ĐK: \(x\ge0\)

\(\Leftrightarrow\sqrt{3+x}=x^2-3\)

Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:

\(a=x^2-\left(a^2-x\right)\)

\(\Leftrightarrow x^2-a^2+x-a=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)

\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))

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

d/ ĐKXĐ: ...

\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)

\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)

\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))

e/ ĐKXĐ: \(x\ge0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)

Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)

\(\Rightarrow5c\left(a-b\right)=5c^2\)

\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)

f/ ĐKXĐ: \(x>0\)

\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)

Khó quá, tớ mới học lớp 5 thôi.

đặt x-2016=a

y-2017=b

z-2018=c

ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)

=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)

=>\(a=b=c=4\)

còn lại tự lm nốt

oke cao van duc

thank nhiều nha

hok tốt

Đặt \(\hept{\begin{cases}a=\sqrt{x-2009}\\b=\sqrt{y-2010}\\c=\sqrt{z-2011}\end{cases}}\)(với a,b,c>0). Khi đó phương trình đã cho trở thành

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2\)

\(\Leftrightarrow a=b=c=2\)\(\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)