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minh chua hoc den cai nay. SORY nhe 

https://olm.vn/hoi-dap/question/595884.html

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}+\sqrt{x}+\sqrt{x+7}-42=0\)

Đặt \(\sqrt{x}+\sqrt{x+7}=t>0\)

\(\Rightarrow2x+7+2\sqrt{x^2+7x}=t^2\)

Pt trở thành:

\(t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\sqrt{x+7}=6\)

\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}=36\)

\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\) (\(x\le\frac{29}{2}\))

\(\Leftrightarrow4\left(x^2+7x\right)=\left(29-2x\right)^2\)

\(\Leftrightarrow144x-841=0\Rightarrow x=\frac{841}{144}\)

a: 

ĐKXĐ: x>=5/2

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)

=>\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\cdot\sqrt{2x-5}}=14\)

=>\(\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

=>\(\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)

=>\(2\sqrt{2x-5}+4=14\)

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

=>2x-5=25

=>2x=30

=>x=15

b: \(x^2-4x=\sqrt{x+2}\)

=>\(x+2=\left(x^2-4x\right)^2\) và x^2-4x>=0

=>x^4-8x^3+16x^2-x-2=0 và x^2-4x>=0

=>(x^2-5x+2)(x^2-3x-1)=0 và x^2-4x>=0

=>\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\)

Sai đề r bạn ơi !!!

https://olm.vn/hoi-dap/question/595884.html

sai đề rồi

ĐK \(x\ge0\)

\(\Leftrightarrow\sqrt{x}+\sqrt{x+7}+x+2\sqrt{x\left(x+7\right)}+x+7=42\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)+\left(\sqrt{x}+\sqrt{x+7}\right)^2=42\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)^2+\left(\sqrt{x}+\sqrt{x+7}\right)-42=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{x+7}=6\\\sqrt{x}+\sqrt{x+7}=-7\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)^2=36\)

\(\Leftrightarrow2x+7+2\sqrt{x\left(x+7\right)}=36\)

\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\)

bình phương 2 vế

\(\Leftrightarrow4\left(x^2+7x\right)=4x^2-116x+841\)

\(\Leftrightarrow4x^2+28x=4x^2-116x+841\)

\(\Leftrightarrow144x=841\Leftrightarrow x=\dfrac{841}{144}\)