a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

=>\(x^3-x^2-4x^2+4x+4x-4=0\)

=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

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

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

2: \(x^3+3x^2=x+6\)

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

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

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

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

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

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

=>\(2x+3\sqrt{x}-1=0\)

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

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

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

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

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

b.

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

\(2x+3\sqrt{x}+1=0\)

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

\(x^4+4x^2+1=3x^3+3x\)

\(\Leftrightarrow x^4-3x^3+4x^2-3x+1=0\)

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

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

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

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

c: Ta có: \(\sqrt{2x}=\sqrt{5}\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

d: Ta có: \(\sqrt{3x-1}=4\)

\(\Leftrightarrow3x-1=16\)

\(\Leftrightarrow3x=17\)

hay \(x=\dfrac{17}{3}\)

Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)

\(\Leftrightarrow2\left|x-1\right|=6\)

\(\Leftrightarrow\left|x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Ta có: \(\sqrt{4x^2-4x+9}=3\)

\(\Leftrightarrow4x^2-4x=0\)

\(\Leftrightarrow4x\left(x-1\right)=0\)

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

học ngu, dễ vậy

Quy đồng, khử mẫu

\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}-\dfrac{12x-1}{4-4x}\) Đkxđ : x≠1,-1

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}+\dfrac{12x-1}{4x-4}\)

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{20x-2}{4x-4}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{10x-1}{2\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{10x-1}{2\left(x-1\right)}=0\)

\(\Leftrightarrow\dfrac{1-9x}{2\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow1-9x=0\)

\(\Leftrightarrow9x=1\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy S=\(\left\{\dfrac{1}{9}\right\}\)

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

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

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

1.

$\sqrt{3x^2}-\sqrt{12}=0$

$\Leftrightarrow \sqrt{3x^2}=\sqrt{12}$

$\Leftrightarrow 3x^2=12$

$\Leftrightarrow x^2=4$

$\Leftrightarrow (x-2)(x+2)=0\Leftrightarrow x=\pm 2$

2. 

$\sqrt{(x-3)^2}=9$

$\Leftrightarrow |x-3|=9$

$\Leftrightarrow x-3=9$ hoặc $x-3=-9$

$\Leftrightarrow x=12$ hoặc $x=-6$

3.

$\sqrt{4x^2+4x+1}=6$

$\Leftrightarrow \sqrt{(2x+1)^2}=6$

$\Leftrightarrow |2x+1|=6$

$\Leftrightarrow 2x+1=6$ hoặc $2x+1=-6$

$\Leftrightarrow x=\frac{5}{2}$ hoặc $x=\frac{-7}{2}$

bạn tự kết luận nhé ! 

a, \(4x-3=2\left(x-3\right)\Leftrightarrow4x-3=2x-6\)

\(\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)

b, \(5x^2+x=0\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)

c, \(\left(3x-5\right)\left(x+7\right)=0\Leftrightarrow x=-7;x=\frac{5}{3}\)

d, \(\frac{2}{x-3}-\frac{3}{x+3}=\frac{7x-1}{x^2-9}\)ĐK : \(x\ne\pm3\)

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

\(\Rightarrow2x+6-3x+9=7x-1\Leftrightarrow-x+15=7x-1\)

\(\Leftrightarrow-8x=-16\Leftrightarrow x=2\)( tmđk )

e, \(\left(12x-1\right)\left(6x-1\right)\left(4x-1\right)\left(3x-1\right)=330\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-2\right)\left(12x-3\right)\left(12x-4\right)=330.24=7920\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-4\right)\left(12x-2\right)\left(12x-3\right)=7920\)

\(\Leftrightarrow\left(144x^2-60x+4\right)\left(144x^2-60x+6\right)=7920\)

Đặt \(144x^2-60x+4=t\)

\(t\left(t+2\right)=7920\Leftrightarrow t^2+2t-7920=0\)

\(\Leftrightarrow\left(t-88\right)\left(t+90\right)=0\Leftrightarrow t=88;t=-90\)

suy ra :TH1 :  \(144x^2-60x+4=88\Leftrightarrow12\left(12x+7\right)\left(x-1\right)=0\Leftrightarrow x=-\frac{7}{12};x=1\)

TH2 : \(144x^2-60x+4=-90\Leftrightarrow144x^2-60x+94=0\)

\(\Leftrightarrow x=\frac{5\pm3\sqrt{39}i}{24}\)

Đặt 1/x=a; 1/y=b

Hệ phương trình trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{2}{3}\\\dfrac{1}{4}a+\dfrac{1}{3}b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=2\\15a+20b=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15b+15b=30\\15b+20b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5b=18\\a+b=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{18}{5}\\a=\dfrac{64}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{18}\\y=\dfrac{15}{64}\end{matrix}\right.\)