c) Ta có:

\(\sqrt{x+\frac{3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\)

\(\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\\sqrt{x^3+3x}+2x=2\left(x+1\right)\end{cases}}\)

+) \(x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

+) \(\sqrt{x^3+3x}+2x=2x+2\Rightarrow x=1\)

a/ Đặt \(\sqrt{2\left(x^2-x\right)}=a\)

\(\Rightarrow a^4-2a^2=a\)

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

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

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

\(\Leftrightarrow x=0\)

\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)

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

\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)

\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)

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

Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)

Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.

x^3-4x^2+5x-1-căn 2x-3=0

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

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

=>\(\left(x-2\right)\left[\left(x-1\right)\left(x-2\right)-\dfrac{2}{\sqrt{2x-3}+1}\right]=0\)

=>x-2=0

=>x=2

.

b) Đặt \(u=\sqrt{1-x}\); \(v=\sqrt{1+x}\)

phương trình trở thành

\(2u-v+3uv=u^2+2\)\(\Rightarrow u^2-2u+v-3uv+2=0\)

lại có \(u^2+v^2=2\)

\(\Rightarrow u^2-2u-3uv+v+u^2+v^2=0\)

\(\Rightarrow\left(u-v-1\right)\left(2u-v\right)=0\)

đến đây thì easy rồi

a)

Đặt \(\sqrt{2x+1}=t\) ;\(\sqrt{x}=k\)

Phương trình trở thành

\(\left(3k^2+t^2\right)t-\left(3t^2+k^2\right)k-1=0\)

\(\Leftrightarrow3k^2t+t^3-3t^2k-k^3-1=0\)

\(\Leftrightarrow\left(t-k\right)\left(t^2+kt+k^2\right)-3tk\left(t-k\right)-1=0\)

\(\Leftrightarrow\left(t-k\right)^3-1=0\)

\(\Leftrightarrow\left(t-k-1\right)\left(\left(t-k\right)^2+t-k+1\right)=0\)

do t > k => t - k > 0

\(\Rightarrow\left(t-k\right)^2+t-k+1>0\)

\(\Rightarrow t-k-1=0\)

\(\Leftrightarrow t=1+k\)\(\Leftrightarrow\sqrt{2x+1}=1+\sqrt{x}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

END

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

ĐK:\(x\ge 0\)

\(\Leftrightarrow\left(5x+1\right)\sqrt{2x+1}-\left(\dfrac{31}{2}x+1\right)-\left(\left(7x+3\right)\sqrt{x}-\dfrac{31}{2}x\right)=0\)

\(\Leftrightarrow\dfrac{\left(5x+1\right)^2\left(2x+1\right)-\left(\dfrac{31}{2}x+1\right)^2}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{x\left(7x+3\right)^2-\left(\dfrac{31}{2}x\right)^2}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)

\(\Leftrightarrow\dfrac{\dfrac{1}{4}x\left(200x+19\right)\left(x-4\right)}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{\dfrac{1}{4}x\left(x-4\right)\left(196x-9\right)}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)

\(\Leftrightarrow\dfrac{1}{4}x\left(x-4\right)\left(\dfrac{200x+19}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{196x-9}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Nghe t đi phần nào khó cho qua :)) b tương tự

ĐK \(x\ge-\frac{2}{3}\)

Pt

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

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

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

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

Pt (2) vô nghiệm do VT>0 với mọi \(x\ge-\frac{2}{3}\)

=> \(x=\frac{1\pm\sqrt{5}}{2}\)(tmĐKXĐ)

Vậy \(x=\frac{1\pm\sqrt{5}}{2}\)