DKXD: x\(\ge1\)

Ta có: \(2x^2+5x-1=7\sqrt{x^3-1}\)\(\Leftrightarrow\left(2x^2+2x+2\right)+\left(3x-3\right)=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
Do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}>0\forall x\)

Nen ta chia hai ve cua phuong trinh cho \(x^2+x+1,\)ta duoc

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

Dat \(\sqrt{\frac{x-1}{x^2+x+1}}=t\)\(\left(t\ge0\right)\)ta có

\(2+3t^2=7t\Leftrightarrow3t^2-7t+2=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=2\\t=\frac{1}{3}\end{cases}}\)

+) \(t=2\Rightarrow\frac{x-1}{x^2+x+1}=4\Rightarrow4x^2+3x+5=0\)

\(\left(ptvn\right)\)

+) \(t=\frac{1}{3}\Rightarrow\frac{x-1}{x^2+x+1}=\frac{1}{9}\)

TT bạn tu tinh nhé

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 \(\sqrt{2\left(x^2-x\right)}=a\)

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

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

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

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

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

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

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

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

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 \(x=\frac{1\pm\sqrt{5}}{2}\)

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: ...

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

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

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

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

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

\(\Leftrightarrow...\)

GIÚP MK NHA CÁC BN

âm; dương  sao bình phương dễ dc.

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

\(\Leftrightarrow\left(2x^2-1\right)\left(36x^2+24x+4\right)=100x^4+9x^2+36+60x^3-36x-120x^2\)