\(b,x^2+3x-2=0\\ \Delta=3^2-4.1.\left(-2\right)=17\\ =>\left[{}\begin{matrix}x_1=\dfrac{-3+\sqrt{17}}{2}\\x_2=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)

Mấy câu còn lại mình giải rồi 

a: =>(x+1)(x+3)=0

=>x=-1 hoặc x=-3

b: Δ=3^2-4*1*(-2)=9+8=17>0

=>Phương trình có hai nghiệm pb là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{17}}{2}\\x_2=\dfrac{-3+\sqrt{17}}{2}\end{matrix}\right.\)

c: =>3x^2-5x-8=0

=>3x^2-8x+3x-8=0

=>(3x-8)(x+1)=0

=>x=8/3 hoặc x=-1

d: =>(3x-1)^2=0

=>3x-1=0

=>x=1/3

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

a. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 1=2\sqrt{x-2}$

$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

a) \(x^3-7x+6=x^3+3x^2-x^2-3x-2x^2-6x+2x+6\)

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

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

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

=\(\left\{\begin{matrix}x+3=0=>x=-3\\x-2=0=x=2\\x-1=0=>x=1\end{matrix}\right.\)

\(b...x^3-19x+30=0\)

\(=>x^3+5x^2-2x^2-10x-3x^2-15x+6x+30=0\)

=>\(x^2\left(x+5\right)-2x\left(x+5\right)-3x\left(x+5\right)+6\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(x^2-2x-3x+6\right)=0\)

=>\(\left(x+5\right)\left(x-3\right)\left(x-2\right)=0\)

=>\(\left\{\begin{matrix}x-3=0=>x=3\\x-2=0=>x=2\\x+5=0=>x=-5\end{matrix}\right.\)

Vậy x=-5;2;3

d, \(\Delta'=225-25.9=0\)pt có nghiệm kép 

\(x_1=x_2=\dfrac{-15}{9}=-\dfrac{5}{3}\)

e, \(\Delta'=4.5-4=16>0\)pt có 2 nghiệm pb 

\(x_1=2\sqrt{5}-4;x_2=2\sqrt{5}+4\)

d: \(\Leftrightarrow\left(3x+5\right)^2=0\)

=>3x+5=0

hay x=-5/3

e: \(\text{Δ}=\left(4\sqrt{5}\right)^2-4\cdot1\cdot4=80-16=64>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{4\sqrt{5}-8}{2}=2\sqrt{5}-4\\x_2=2\sqrt{5}+4\end{matrix}\right.\)

d, \(\Delta=30^2-9.4.25=0\)

Vậy pt có nghiệm kép:\(x_{1,2}=\dfrac{-b}{2a}=\dfrac{-30}{2.9}=\dfrac{-30}{18}=\dfrac{-5}{3}\)

e, \(\Delta=\left(-4\sqrt{5}\right)^2-4.1.4=80-16=64\)

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}+\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}+8}{2}=4+2\sqrt{5}\)

\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}-\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}-8}{2}=-4+2\sqrt{5}\)

b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)

\(\Leftrightarrow x^2-6x+9=3\)

\(\Leftrightarrow x^2-6x+6=0\)

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)

\(x^4-9x^2+24x-16=\)\(0\)

\(\Leftrightarrow x^4-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow x^4-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(x^2+3x-4\right)\left(x^2-3x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]=0\)

Vì \(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)nên:

\(\left(x+4\right)\left(x-1\right)=0:\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

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

Vậy phương trình có tập nghiệm \(S=\left\{1;-4\right\}\)

\(x^4=6x^2+12x+\)\(8\)

\(\Leftrightarrow x^4-2x^2+1=4x^2+12x+9\)

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

\(\Leftrightarrow|x^2-1|=|2x+3|\)\(|\)

xét các trường hợp:

- Trường hợp 1:

\(x^2-1=2x+3\)

\(\Leftrightarrow x^2-1-2x-3=0\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}}\)

-Trường hợp 2:

\(x^2-1=-2x-3\)

\(\Leftrightarrow x^2-1+2x+3=0\)

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

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

\(\Leftrightarrow\left(x+1\right)^2=-1\left(vn\right)\)(vô nghiệm)

Vậy phương trình có tập nghiệm: \(S=\left\{1\pm\sqrt{5}\right\}\)

\(x^4=4x+1\)

\(\Leftrightarrow x^4+2x^2+1=2x^2+4x+2\)

\(\Leftrightarrow\left(x^2+1\right)^2=2\left(x+1\right)^2\)

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

Xét các trường hợp sau:

-Trường hợp 1:

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

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

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

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

Vì \(\frac{2\sqrt{2}-1}{2}>0\)nên:

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

Lại xét các trường hợp:

+Trường hợp 1.1:

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

+Trường hợp 1.2:

\(x-\frac{1}{\sqrt{2}}=\frac{\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\Leftrightarrow x=\frac{1-\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\)

-Trường hợp 2:

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

\(\Leftrightarrow x^2+1+x\sqrt{2}+\sqrt{2}=0\)

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

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

Do đó phương trình (2) vô nghiệm.

Vậy phương trình có tập nghiệm : \(S=\left\{\frac{1\pm\sqrt{2\sqrt{2}-1}}{\sqrt{2}}\right\}\)