PT này bạn tham khảo công thức Cardano tìm nghiệm pt bậc 3:

https://www.mathvn.com/2021/08/cong-thuc-cardano-tim-nghiem-cua-phuong.html

Nghiệm là \(x=\frac{1}{3}(-1+\sqrt[3]{\frac{1}{2}(79-9\sqrt{77})}+\sqrt[3]{\frac{1}{2}(79+9\sqrt{77})})\)

2x²-7x+6=0

=> 2x²-3x-4x+6=0

=>x(2x-3)-2(2x-3)=0

=>(x-2)x(2x-3)=0

=>TH1 x-2=0=>x=2

=>TH2 2x-3=0=>2x=3=>x=3/2

Lời giải:
$2x^2-7x+6=0$

$\Leftrightarrow (2x^2-4x)-(3x-6)=0$

$\Leftrightarrow 2x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(2x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $2x-3=0$

$\Leftrightarrow x=2$ hoặc $x=\frac{3}{2}$

2x² - 7x + 6 = 0

\(\Leftrightarrow\) 2x² - 4x - 3x + 6 = 0

\(\Leftrightarrow\) (2x² - 4x) - (3x - 6) = 0

\(\Leftrightarrow\) 2x(x - 2) - 3(x - 2) = 0

\(\Leftrightarrow\) (x - 2)(2x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\) 

S = \(\left\{2,\dfrac{3}{2}\right\}\)

(x - 1)(2x² - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x^2-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{1;\sqrt{5}\right\}\)
(2x - 7)² - 6(2x - 7)(x - 3) = 0

\(\Leftrightarrow\left(2x-7\right)\left(2x-7-6x+18\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(11-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\4x=11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\frac{7}{2};\frac{11}{4}\right\}\)
(5x + 3)(x² + 4) = 0

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x^2+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-3\\x^2=-4\left(Loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\frac{3}{5}\)

Vậy phương trình có tập nghiệm là: \(S=\left\{-\frac{3}{5}\right\}\)

a)

\(\left(x-1\right)\cdot\left(2x^2-10\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot2\cdot\left(x^2-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{5}\end{matrix}\right.\)

b)

\(\left(2x-7\right)^2-6\cdot\left(6x-7\right)\cdot\left(x-3\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left[\left(2x-7\right)-6\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(2x-7-6x+18\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(11-4x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

c)

\(\left(5x+3\right)\cdot\left(x^2+4\right)=0\)

Vì \(\left(x^2+4\right)>0\Rightarrow\left(loại\right)\)

\(\Rightarrow5x+3=0\\ \Rightarrow x=-\frac{3}{5}\)

a, ĐKXĐ: ...

\(\sqrt{3x^2-2x+6}+3-2x=0\)

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

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

<=> (x - 3) (x - 2) (x + 1) (2 x + 1) = 0

\(x=3;x=2;x=-1;x=-\frac{1}{2}\)