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

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

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

=> giải tiếp nhé , mình biết lớp 10

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

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

\(ĐK:x\ge-2\)

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

đơn giản như đan rổ

1. đk: pt luôn xác định với mọi x

\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)

\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)

Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!

2.  đk: \(x\geq 1\)

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

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

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

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

Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!

em ms hok lớp 1

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

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

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 ^.^

đk -3 =< x =< 10

\(\sqrt{x+3}-2+\sqrt{10-x}-3=x^2-7x+6\)

\(\Leftrightarrow\dfrac{x+3-4}{\sqrt{x+3}+2}+\dfrac{10-x-9}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x+3}+2}+\dfrac{1-x}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x+3}+2}-\dfrac{1}{\sqrt{10-x}+3}-x+6\ne0\right)=0\Leftrightarrow x=1\)(tm)

ĐKXĐ: \(\dfrac{74}{9}\le x\le10\)

Đặt \(\sqrt{10-x}=t\Rightarrow0\le t\le\dfrac{4}{3}\) \(\Rightarrow x=10-t^2\)

Ta được:

\(2+\sqrt{4-3t}=\dfrac{10-t^2}{3}\)

\(\Leftrightarrow\sqrt{4-3t}-1=\dfrac{10-t^2}{3}-3\)

\(\Leftrightarrow\dfrac{3\left(1-t\right)}{\sqrt{4-3t}+1}=\dfrac{\left(1-t\right)\left(1+t\right)}{3}\)

\(\Rightarrow\left[{}\begin{matrix}t=1\Rightarrow x=9\\\dfrac{3}{\sqrt{4-3t}+1}=\dfrac{t+1}{3}\left(1\right)\end{matrix}\right.\)

Xét (1), do \(0\le t\le\dfrac{4}{3}\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{\sqrt{4-3t}+1}\ge1\\\dfrac{t+1}{3}\le\dfrac{\dfrac{4}{3}+1}{3}=\dfrac{7}{9}< 1\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy pt có nghiệm duy nhất \(x=9\)

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