e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

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

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

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

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

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

=(x-1)(x+1)(x-2)(x+2)

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`

\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)

\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)

\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)

\(=-165+22-108=22-273=-251\)

\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)

\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)

\(=3-7.0+5.\left(-3\right)=3-15=-12\)

\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)

\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)

\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)

\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)

\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)

\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)

\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)

\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)

\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)

\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)

\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)

\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)

\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)

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

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

=>x=1 hoặc x=4/7

e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

=>10x-4=15-9x

=>19x=19

hay x=1

f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

=>2x-1+x-1=1

=>3x-2=1

hay x=1(loại)

g: =>1+3x-6=3-x

=>3x-5-3+x=0

=>4x-8=0

=>x=2(loại)

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)

Bài 2: 

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

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