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a: \(=\dfrac{\left(x^2-5\right)\left(x^2+5\right)+2x\left(x^2+5\right)}{x^2+5}=x^2+2x-5\)

b: \(=2x^2-2x+3x-3-2x-7\)

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

c: \(=\left(x-y-x-y\right)^2=\left(-2y\right)^2=4y^2\)

d: \(=18x^{2n-3}+3x^n-18x^{2n-3}+2x^n=5x^n\)

26 tháng 7 2020

a, 6xn+2-3.2xn + 3.2x1+n-1 + 2x = 2x(3n+1+1)

b,3xn-2+n-2- 3xn-2yn+2 + yn+2.3xn-2 = 3x2(n-2)

a: \(=12^{n+3}-8x^{n+2}\)

b: \(=3x^3-5x-2x^3-x^2+x^2=x^3-5x\)

c: \(=12x^4-28x^3+8x^2\)

d: \(=x^3+3x^2y+3xy^2+y^3\)

30 tháng 8 2017

Bài 1

A= (5-x)(x3-2x2+x-1)

=5x3-10x2+5x-5-x4+2x3-x2+x

=-x4+7x3-12x2+6x-5

30 tháng 8 2017

B= 2y-x - {2x-y-[y+3x-(5y-x)]}
B= 2y-x- [ 2x-y- ( y + 3x-5y+x)
B= 2y-x-( 2x-y-y-3x+5y-x)
B= 2y-x-2x+y+y+3x-5y+x
B= -y +x

17 tháng 7 2019

a) =2x^3-10x^2-2x+3x^2-x

=2x^3-7x^2-3x

17 tháng 7 2019

b) -10x^4y^2z^2+35x^3y^2z^2+4x^4y^2z^2+4x^3y^2z^2

=-6x^4y^2z^2+39x^3y^2z^2

18 tháng 12 2017

4.a) \(2x^2-10x-3x-2x^2-26=0\)

\(-13x-26=0\Rightarrow-13\left(x+2\right)=0\)

\(\Rightarrow x=-2\)

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

\(2x+10-x^2-5x=0\Leftrightarrow-x^2-3x+10=0\)

\(-\left(x^2+3x-10\right)=0\)

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

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

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

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

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

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

\(\left\{{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

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

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

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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

h) \(x^2-4x+8-2x+1=x^2-6x+9=0\)

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