1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

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

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

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

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

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)

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

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

a) 3(x - 1)² - 3x(x - 5) = 2

=> 3(x² - 2x + 1) - 3x² + 15x = 2

=> 3x² - 6x + 3 - 3x² + 15x = 2

=> 9x = 2 - 3

=> 9x = -1

=> x = -1/9

b) 4x² - 12x = -9

=> 4x² - 12x + 9 = 0

=> (2x - 3)² = 0

=> 2x - 3 = 0

=> 2x = 3

=>  x = 3/2

c) (2x - 3)² = (x  + 5)²

=> (2x - 3)² - (x + 5)² = 0

=> (2x - 3 - x - 5)(2x - 3 + x + 5) = 0

=> (x - 8)(3x + 2) = 0

=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

d) \(\left(x^4-2x^3+4x^2-8x\right):\left(x^2+4\right)-2x=-4\)

=> \(\left[x^3\left(x-2\right)+4x\left(x-2\right)\right]:\left(x^2+4\right)-2x=-4\)

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

=> \(x^2-2x-2x+4=0\)

=> \(\left(x-2\right)^2=0\)

=> x - 2 = 0

=> x = 2

e) khđ

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

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

1/\(4x^4+12x^3-47x^2+12x+4=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)

Vậy ....

1,

-_- bài này hôm qua lm rùi

\(12\left(x+5\right)+2x=130\\\Leftrightarrow 12x+60+2x=130\\ \Leftrightarrow14x=70\\ \Leftrightarrow x=5\\ ----\\ 23\left(x-5\right)-12x=138\\ \Leftrightarrow23x-115-12x=138\\ \Leftrightarrow23x-12x=138+115\\ \Leftrightarrow11x=253\\ \Leftrightarrow x=\dfrac{253}{11}=23\\ ----\\ 360-12x+23\left(x-5\right)=278\\ \Leftrightarrow360-12x+23x-115=278\\ \Leftrightarrow-12x+23x=278+115-360\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=\dfrac{33}{11}=3\)

\(6\left(x+3\right)+3\left(x-5\right)=278\\ \Leftrightarrow6x+18-3x-15=278\\ \Leftrightarrow6x-3x=278+15-18\\ \Leftrightarrow3x=275\\ \Leftrightarrow x=\dfrac{275}{3}\\ ---\\ \left(7-x\right)\left(3x-90\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-90=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=30\end{matrix}\right.\)

\(\left(x-2\right)\left(4x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\\ ---\\ \left(2x-18\right)\left(3x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-18=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\3x=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)

Câu 9:

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

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

\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(9,x^3-2x^2-x+2=0\\ \Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

\(10,\) giống 9

\(11,x^2+4x-5=0\\ \Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(12,2x^2+4x+2=72\\ \Rightarrow2x^2+4x-70=0\\ \Rightarrow x^2+2x-35=0\\ \Rightarrow\left(x^2-5x\right)+\left(7x-35\right)=0\\ \Rightarrow x\left(x-5\right)+7\left(x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

\(13,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ \Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\\ \Rightarrow x^3-25x-x^3-8=17\\ \Rightarrow-25x=25\\ \Rightarrow x=-1\)

\(14,2x^3+5x^2-12x=0\\ \Rightarrow x\left(2x^2+5x-12\right)=0\\ \Rightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Rightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ \Rightarrow x\left(2x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

\(\left(2x-4\right)\left(1-3x\right)=0\)

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

<=>  \(\orbr{\begin{cases}x-2=0\\1-3x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}\)

Vậy....

\(\left(2x-4\right)\left(1-3x\right)=0\)

\(\Rightarrow2x-4=0\)hoặc\(1-3x=0\)

\(TH1:2x-4=0\)

                     \(2x=0+4\)

                     \(2x=4\)

                       \(x=4:2\)

                       \(x=2\)

\(TH2:1-3x=0\)

                    \(3x=1-0\)

                   \(3x=1\)

                      \(x=\frac{1}{3}\)

Vậy:\(x=2\)hoặc \(x=\frac{1}{3}\)

a. (2x-4)(1-3x)=0

<=> 2x-4=0 hoặc 1-3x=0

<=> x=2 hoặc x=1/3

b. (x-3)(x²-3x+2)=0

<=> x-3=0 hoặc x²-3x+2=0

<=> x=3 hoặc x=2 hoặc x=1

c. (x-1)(5x+3)=(2x-7)(x-1)

<=> (x-1)(3x+10)=0

<=> x-1=0 hoặc 3x+10=0

<=> x=1 hoặc x=-10/3