ĐK: \(x\ge-2\)

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

\(\Leftrightarrow3.\frac{1+\sqrt{x+2}.\sqrt{x+5}}{\sqrt{x+2}+\sqrt{x+5}}=3\)

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

\(\Leftrightarrow\left(\sqrt{x+2}-1\right)\left(\sqrt{x+5}-1\right)=0\)

\(\Leftrightarrow\sqrt{x+2}=1\text{ hoặc }\sqrt{x+5}=1\)

\(\Leftrightarrow x=-1\text{ (nhận) hoặc }x=-4\text{ (loại)}\)

Vậy tập nghiệm của pt là: \(S=\left\{1\right\}\)

1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

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

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

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

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

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

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

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

\(\Leftrightarrow x^2-5x+2=0\)

Câu 1:

ĐK: \(x\geq -2\)

Đặt \(\sqrt{x+5}=a; \sqrt{x+2}=b(a,b\geq 0)\)

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

PT trở thành:

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

\(\Leftrightarrow (a-b)(1+ab)=(x+5)-(x+2)=a^2-b^2\)

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

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

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

Vì \(a\neq b\Rightarrow \left[\begin{matrix} a-1=0\\ b-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=\sqrt{x+5}=1\\ b=\sqrt{x+2}=1\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=-4\\ x=-1\end{matrix}\right.\). Vì $x\geq -2$ nên chỉ có $x=-1$ là nghiệm duy nhất.

Câu 2:

ĐK: \(-4\leq x\leq 4\)

Ta có: \((\sqrt{x+4}-2)(\sqrt{4-x}+2)=2x\)

\(\Leftrightarrow \frac{(x+4)-2^2}{\sqrt{x+4}+2}.(\sqrt{4-x}+2)=2x\)

\(\Leftrightarrow x.\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2x\)

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

\(\Rightarrow \left[\begin{matrix} x=0\\ \sqrt{4-x}+2=2\sqrt{x+4}+4(*)\end{matrix}\right.\)

Xét $(*)$

Đặt \(\sqrt{4-x}=a; \sqrt{x+4}=b\) thì ta có hệ:

\(\left\{\begin{matrix} a^2+b^2=8\\ a+2=2b+4\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^2+b^2=8\\ a=2(b+1)\end{matrix}\right.\)

\(\Rightarrow 4(b+1)^2+b^2=8\)

\(\Leftrightarrow 5b^2+8b-4=0\Leftrightarrow (5b-2)(b+2)=0\)

\(\Rightarrow b=\frac{2}{5}\) (do \(b\geq 0)\)

\(\Rightarrow x+4=b^2=\frac{4}{25}\Rightarrow x=\frac{-96}{25}\) (t/m)

Vậy \(x\in \left\{ \frac{-96}{25}; 0\right\}\)

.

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

Kết hợp với phương trình ban đầu ta được hệ:

\(\left\{{}\begin{matrix}\left(a-b\right)\left(1+ab\right)=3\\a^2-b^2=3\end{matrix}\right.\)

Cứ thế giải .

Đặt: \(\left\{{}\begin{matrix}\sqrt{x+5}=m\\\sqrt{x+2}=n\end{matrix}\right.\Rightarrow m^2-n^{^2}=3\)

(Đk: \(m>n\ge0\) )

Thay vào, ta có:

\(\left(m-m\right)\left(1+mn\right)=m^2-n^2\Leftrightarrow\left(m-n\right)\left(n-1\right)\left(m-1\right)=0\)

Thử các trường hợp m, n ta được nghiệm của phương trình đã cho là \(x=-4;x=-1\)

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

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

\(ĐK:x\ge-2\)

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