Dùng liên hợp.

pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)

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

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

<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)

\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)

\(=3x-1\)

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

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

<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)

<=> \(x^2-3x+2=0\) phương trình bậc 2.

Em làm tiếp nhé!

đặt \(\sqrt{x+5}=a\);\(\sqrt{x+2}=b\)  => ab=\(\sqrt{x^2+7x+10}\) và \(a^2-b^2=3\)

 do đó pt trở thành \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

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

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

đến đây tự giải tiếp nhé

em chưa học , em mới lớp 5 thui

\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)=\left(\sqrt{x+5}+\sqrt{x+2}\right)\left(\sqrt{x+5}-\sqrt{x+2}\right)\)
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{\left(x+2\right)\left(x+5\right)}-\sqrt{x+5}-\sqrt{x+2}\right)=0\)
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x+5}-1\right)\left(\sqrt{x+2}-1\right)=0\)
Tự làm tiếp nhé ^_^

Viết đề mà ko ai đọc được vậy :v

a) \(3x^2+2x+3=\left(3x+1\right)\sqrt{x^2+3}\)

\(\Leftrightarrow3x^2+2x+3-3x\sqrt{x^2+3}-\sqrt{x^2+3}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\) ( thỏa mãn )

Vậy...

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

<=>\(\left(4x-1\right)\left[\sqrt{x^2+1}-\left(3-x\right)\right]=6x^2-11x+4\)

Xét \(\sqrt{x^2+1}+3-x=0\)

<=> \(x^2+1=x^2-6x+9\) <=>\(x=\frac{4}{3}\)(tm phương trình (1))

Xét \(\sqrt{x^2+1}+3-x\ne0\)

pt <=>\(\frac{\left(4x-1\right)\left(x^2+1-x^2+6x-9\right)}{\sqrt{x^2+1}+3-x}=\left(3x-4\right)\left(2x-1\right)\)

<=> \(\frac{\left(4x-1\right)\left(6x-8\right)}{\sqrt{x^2+1}+3-x}-\left(3x-4\right)\left(2x-1\right)=0\)

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

<=>\(\left[{}\begin{matrix}x=\frac{4}{3}\left(tm\right)\\\frac{8x-2}{\sqrt{x^2+1}+3-x}-2x+1=0\left(2\right)\end{matrix}\right.\)

pt (2) <=>\(8x-2=\left(2x-1\right)\sqrt{x^2+1}-2x^2+7x-3\)

<=>\(2x^2+x+1=\left(2x-1\right)\sqrt{x^2+1}\)( đk: \(x\ge\frac{1}{2}\))

=>\(4x^4+x^2+1+4x^3+2x+4x^2=\left(2x-1\right)^2\left(x^2+1\right)\)

<=>\(4x^4+4x^3+5x^2+2x+1=4x^4-4x^3+5x^2-4x+1\)

<=>\(8x^3+6x=0\) <=> \(x\left(8x^2+6\right)=0\) <=>x=0 (do 8x²+6>0) (không t/m (2))

=>(2) vô nghiệm

Vậy pt có tập nghiệm \(S=\left\{\frac{4}{3}\right\}\)

P/s: Hơi dài :)

Mấy anh chị khác god phân tích lắm nên em đành làm cách khác:(

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

Đặt \(\sqrt{x^2+1}=a\ge1\)

\(PT\Leftrightarrow-2a^2+\left(4x-1\right)a-2x+1=0\)

\(\Leftrightarrow\left(2a-1\right)\left(2x-a-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{2}\left(L\right)\\2x=a+1\left(1\right)\end{matrix}\right.\)

Xét (1): Do \(a\ge1\rightarrow a+1\ge2\Rightarrow x\ge1\)

(1) \(\Leftrightarrow2x=\sqrt{x^2+1}+1\)

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

\(\Leftrightarrow\left(x-\frac{4}{3}\right)\left[\frac{\frac{3}{16}\left(3x+4\right)}{\frac{5}{4}x+\sqrt{x^2+1}}+\frac{3}{4}\right]=0\)

\(\Leftrightarrow x=\frac{4}{3}\) (vì cái ngoặc to luôn > 0 với mọi \(x\ge1\))

Vậy...

\(9y^2+\left(2y+3\right)\left(y-x\right)\) nha mn mik ghi sai đề

mong mọi người giải giúp em vs