\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)

Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)

\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)

PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)

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

\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)

+ Với a=1

\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)

+ Với b=1

\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)

Vậy \(S=\left\{-1\right\}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)

Thì được:

\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)

Làm tiếp

\(ĐK:x\ge-2\)

\(PT\Leftrightarrow\dfrac{x+5-x-2}{\sqrt{x+5}+\sqrt{x+2}}\left(1+\sqrt{x^2+7x+10}\right)=3\\ \Leftrightarrow\dfrac{3\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)}{\sqrt{x+5}+\sqrt{x+2}}=3\\ \Leftrightarrow1+\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x+5}+\sqrt{x+2}\\ \Leftrightarrow\left(\sqrt{x+5}-1\right)\left(1-\sqrt{x+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x+2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x=-1\)

ĐK: \(x\ge1\)

Đặt \(\sqrt{3x-2}+2\sqrt{x-1}=t\left(t\ge1\right)\)

\(pt\Leftrightarrow3t=t^2-4\)

\(\Leftrightarrow t^2-3t-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-1\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow7x-6+4\sqrt{\left(3x-2\right)\left(x-1\right)}=16\)

\(\Leftrightarrow4\sqrt{3x^2-5x+2}=22-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}48x^2-80x+32=484+49x^2-308x\\22-7x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}452+x^2-228x=0\\x\le\dfrac{22}{7}\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3.\)

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

Đặt : \(\sqrt{x+5}=a\Rightarrow x+5=a^2\)

\(\sqrt{x+2}=b\Rightarrow x+2=b^2\)\(\left(đk:a,b\ge0\right)\)

\(\Rightarrow a^2-b^2=x+5-x-2=3\left(1\right)\)

Mà theo phương trình, ta có :

\(\left(a-b\right)\left(1+ab\right)=3\)

\(\Rightarrow a+a^2b-b-ab^2=3\)\(\left(2\right)\)

Tự giải hệ 

\(\Leftrightarrow1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

\(\Leftrightarrow\sqrt{x^2+7x+10}-2-\sqrt{x+5}+2-\sqrt{x+2}+1=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+10}+2}+\frac{x+1}{2+\sqrt{x+5}}+\frac{x+1}{1+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{x+6}{\sqrt{x^2+7x+10}+2}+\frac{1}{2+\sqrt{x+5}}+\frac{1}{1+\sqrt{x+2}}\right)=0\)

Giải nốt nhá ^.^

a/ ĐKXĐ: \(x\ge-\frac{5}{2}\)

\(\sqrt{2x+5}=5\Rightarrow2x+5=25\Rightarrow x=10\)

b/ \(\sqrt{x-7}+3=0\)

Do \(\sqrt{x-7}\ge0\Rightarrow\sqrt{x-7}+3>0\Rightarrow ptvn\)

c/ ĐKXĐ: \(x\ge0\)

\(\sqrt{3x}=\sqrt{10}-1\Rightarrow3x=11-2\sqrt{10}\Rightarrow x=\frac{11-2\sqrt{10}}{3}\)

d/ \(4-7x=11\Rightarrow-7x=7\Rightarrow x=-1\)

Đk x>= -2 

Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\Rightarrow\sqrt{x^2+7x+10}=a+b;a^2-b^2=x+5-x-2=3\)

pt <=> \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

<=> \(\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)

<=> \(\left(a-b\right)\left(ab+1\right)-\left(a-b\right)\left(a+b\right)=0\)

<=> \(\left(a-b\right)\left(ab+1-a-b\right)=0\)

<=> \(\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)

=> a = b hoặc b = 1 hoặc a = 1 

(+) a = b => x + 5 = x +2 => 0x = -3 (loại )

(+) a = 1 => x + 5 = 1 => x = -4 (loại )

(+) b = 1 => x + 2 = 1=> x = -1 ( TM)

Vậy x = -1 là nghiệm của pt