c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

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

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

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

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

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

=>x-2=0

=>x=2(nhận)

ai làm ơn làm phước tick cho mk lên 190 với

(1)Phương trình đã cho tương đương với:
$$\sqrt{3x^2-7x+3}-\sqrt{3x^2-5x-1}=\sqrt{x^2-2}-\sqrt{x^2-3x+4}$$
$$\iff \frac{-2x+4}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x-1}}=\frac{3x-\mathrm{6<...}}{}$$

1.

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow3x^2-3x+\left(x+1-\sqrt{3x+1}\right)+\left(x+2-\sqrt{5x+4}\right)=0\)

\(\Leftrightarrow3\left(x^2-x\right)+\dfrac{x^2-x}{x+1+\sqrt{3x+1}}+\dfrac{x^2-x}{x+2+\sqrt{5x+4}}=0\)

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

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

\(\Leftrightarrow...\)

2.

Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt[3]{2-8x^3}=b\end{matrix}\right.\)

Ta được hệ:

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

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

\(\Rightarrow8\left(ab\right)^3-6\left(ab\right)^2=2\)

\(\Leftrightarrow\left(ab-1\right)\left[4\left(ab\right)^2+ab+1\right]=0\)

\(\Leftrightarrow ab=1\Rightarrow a+b=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\ab=1\end{matrix}\right.\) \(\Leftrightarrow a=b=1\)

\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

=>\(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+2\sqrt{x-4}\)

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

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

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

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

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

=>\(4\sqrt{x-1}=x+3\)

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

=>\(\sqrt{x-1}<1\)

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

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

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

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

\(\Leftrightarrow\frac{3x^2-5x-2}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-4}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-6}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

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

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

Dễ thấy: \(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

sao cái từ "dễ thấy" nó khó thấy quá v