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Câu 1: \(\dfrac{\left(a+b\right)^2}{a-b}\)
Câu 2:
\(H+\left(3x-2y^2+5x^2-4y-3\right)=\left(2xy\right)^2+2x+2y-x^2-2y^2\)
\(\Rightarrow H=\left(4x^2y^2+2x+2y-x^2-2y^2\right)-\left(3x-2y^2+5x^2-4y-3\right)\)
\(\Rightarrow H=4x^2y^2+2x+2y-x^2-2y^2-3x+2y^2-5x^2+4y-3\)
\(\Rightarrow H=4x^2y^2+\left(2x-3x\right)+\left(2y+4y\right)+\left(-x^2-5x^2\right)+\left(-2y^2+2y^2\right)-3\)
\(\Rightarrow H=4x^2y^2-x+6y-6x^2-3\)
Giải:
Có: \(2^x=8^{y+1}\) và \(9^y=3^{x-9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2^x=2^{3y+3}\\3^{2y}=3^{x-9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y+3\\2y=x-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y+3\\y=\dfrac{x-9}{2}\end{matrix}\right.\)
\(\Leftrightarrow x+y=3y+3+\dfrac{x-9}{2}\)
Chúc bạn học tốt!
Ta có: \(A=4^{n+3}+4^{n+2}-4^{n+1}-4^n\)
\(A=4^{n-1}.4^4+4^{n-1}.4^3-4^{n-1}.4^2-4^{n-1}.4\)
\(A=4^{n-1}\left(4^4+4^3-4^2-4\right)=4^{n-1}.300\).
Vậy .......... (dpcm)
\(A=4^{n+3}+4^{n+2}-4^{n+1}-4^n\)
\(=4^{n-1}.4^4+4^{n-1}.4^3-4^{n-1}.4^2-4^{n-1}.4\)
\(=4^{n-1}\left(4^4+4^3-4^2-4\right)\)
\(=4^{n-1}.300⋮300\)
\(\Rightarrow A⋮300\left(đpcm\right)\)
Vậy...
Bài 2:
\(H+3x+5x^2-2y^2-4y-3=4x^2y^2+2x+2y-x^2-2y^2\)
\(\Leftrightarrow H+3x+5x^2-4y-3=4x^2y^2+2x+2y-x^2\)
\(\Leftrightarrow H=4x^2y^2-x+6y-6x^2+3\)
a) \(x^n.x^{2\left(n+1\right)}\)
= \(x^{n+2.\left(n+1\right)}=x^{n+2n+2}=x^{3n+2}\)
b) \(x^{n+3}.x^{2-n}=x^{n+3+2-n}=x^5\)
c) \(\left(-\dfrac{1}{3}x^{n+2}\right).\left(-3x^{n-1}\right)\)
= \(-x^{n+2+n-1}=-x^{2n+1}\)
d) \(\left(-\dfrac{1}{\dfrac{1}{2x^2y^3}}\right)^2\)
= \(\left(-1.\dfrac{2x^2y^3}{1}\right)^2=\left(-2x^2y^3\right)^2=4x^4y^6\)
e) \(\left(-0,1x^3y\right)^3=-0,001x^9y^3\)