\(a,=6x^2+23x+21-\left(6x^2+23x-55\right)\\ =76\left(đpcm\right)\\ b,=3x^4+6x^3+9x^2-2x^3-4x^2-6x+x^2+2x+3-4x^3+4x-3x^4-6x^2\\ =3\left(đpcm\right)\)

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

a)3x+10-2x>-11
3x - 2x > -10-11
1x > -21
x > -21
b) 3x2 - 6x + 3x2 < 36
-6x < 36
x < -6

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

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

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

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

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

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

\(=\dfrac{2}{2x-1}\)

\(---\)

b) \(\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\) (sửa đề)

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

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

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

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

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

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

\(---\)

c) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

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

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

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

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

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

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

\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}\)

\(=\dfrac{8}{1-x^8}\)

#\(Toru\)

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

__

`x^3+1` chứ cậu nhỉ?

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

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

__

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)

\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)

\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)

\(\Leftrightarrow-22x-60=-75x-30\)

\(\Leftrightarrow-22x+75x=-30+60\)

\(\Leftrightarrow53x=30\)

\(\Leftrightarrow x=\dfrac{30}{53}\)

Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)

5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)

\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)

\(\Leftrightarrow10x-6-21x+3=60\)

\(\Leftrightarrow-11x-3=60\)

\(\Leftrightarrow-11x=63\)

\(\Leftrightarrow x=-\dfrac{63}{11}\)

Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)

`9,x^3+x^2-2=0`

`x^3-x^2+2x^2-2=0`

`<=>x^2(x-1)+2(x-1)(x+1)=0`

`<=>(x-1)(x^2+2x+2)=0`

`<=>x=1`

`14,x^2-2x+1=0`

`<=>(x-1)^2=0`

`<=>x-1=0`

`<=>x=1`

`15,x^3+3x^2+3x+1=0`

`<=>(x+1)^3=0`

`<=>x+1=0`

`<=>x=-1`

Bài 6: 

1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)

\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)

\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)

\(\Leftrightarrow-14x-9=0\)

\(\Leftrightarrow-14x=9\)

\(\Leftrightarrow x=-\dfrac{9}{14}\)

Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)

`1)2x(x-5)-(x+3)^2=3x-x(5-x)`

`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`

`<=>x^2-16x-9=x^2-2x`

`<=>14x=-9`

`<=>x=-9/14`