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27 tháng 10 2020

( x2 - 6x + 9 ) : ( x - 3 ) - x( x + 7 ) - 9

= ( x - 3 )2 : ( x - 3 ) - x2 - 7x - 9

= x - 3 - x2 - 7x - 9

= -x2 - 6x - 12

4 tháng 10 2021

\(a,=x^4+6x^3+8x^2\\ b,=x^2+3x-28\\ c,=x^2-3x-x^2+6x-9+9=3x\)

25 tháng 2 2021

`a,3(x-2)^2+9(x-1)=3(x^2+x-3)`

`<=>3(x^2-4x+4)+9x-9=3x^2+3x-9`

`<=>3x^2-12x+12+9x-9=3x^2+3x-9`

`<=>3x^2-3x+3=3x^2+3x-9`

`<=>6x=12`

`<=>x=12`

`b,(x+3)^2-(x-3)=6x+18`

`<=>(x+3-x+3)(x+3+x-3)+6x+18`

`<=>6.2x=6(x+3)`

`<=>2x=x+3`

`<=>x=3`

`c,(2x+7)^2=9(x+2)^2`

`<=>(2x+7)^2=(3x+6)^2`

`<=>(3x+6-2x-7)(3x+6+2x+7)=0`

`<=>(x-1)(5x+13)=0`

`<=>` $\left[ \begin{array}{l}x-1=0\\5x+13=0\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=1\\5x=-13\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{13}{5}\end{array} \right.$

a) Ta có: \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\)

\(\Leftrightarrow-6x+12=0\)

\(\Leftrightarrow-6x=-12\)

hay x=2

Vậy: x=2

10 tháng 1 2022

\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)

\(=\dfrac{4x-12+x+7}{x^2-9}\)

\(=\dfrac{5x-5}{x^2-9}\)

10 tháng 1 2022

Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\ b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\ c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)

Bài 2:
\(7a-7b=7\left(a-b\right)\\ b,x^2-8x+16=\left(x-4\right)^2\\ c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\ d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

1 tháng 9 2021

\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)

\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)

\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)

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

2 tháng 9 2021

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

 

a: \(3x^2-7x-10\)

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

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

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

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

c: \(x^2-2xy+y^2-5x+5y\)

\(=\left(x-y\right)^2-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-5\right)\)

10 tháng 9 2021

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

14 tháng 10 2021

giúp mik vs cảm ơn các bẹn iu quý

14 tháng 10 2021

\(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+2\right)^3+2\left(x+2\right)\left(x^2-2x+4\right)+6x\left(x+2\right)\)

\(=x^3-27-x^3-6x^2-12x-27+2\left(x^3+8\right)+6x^2+12x\)

\(=-54+2x^3+16\)

\(=2x^3-38\)

12 tháng 8 2023

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(3-2x\right)\left(4x^2+6x+9\right)\)

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

\(M=x^3+27-27+8x^3\)

\(M=9x^3\)

Thay x=20 vào M ta có:
\(M=9\cdot20^3=72000\)

Vậy: ...

\(N=\left(x-2y\right)\left(x^2+2xy+4y^2\right)+16y^3\)

\(N=x^3-\left(2y\right)^3+16y^3\)

\(N=x^3-8y^3+16y^3\)

\(N=x^3+8y^3\)

\(N=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

Thay \(x+2y=0\) vào N ta có:

\(N=0\cdot\left(x^2-2xy+4y^2\right)=0\)

Vậy: ...

18 tháng 8 2021

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

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

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)