Thu gọn đa thức :
\(A=2y-x-\left\{2x-y-\left[y+3x-\left(5y-x\right)\right]\right\}\)
với \(x=a^2+2ab+b^2\) , y \(=a^2-2ab+b^2\)
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24 tháng 6 2022
a: M+N-P
\(=7a^2-2a+1-a^2+4\)
\(=6a^2-2a+5\)
b: \(=2y-x-2x+y+y+3x-5y+x\)
\(=-3x+3y-4y+4x=x-y\)
\(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)
c: \(=\left[{}\begin{matrix}5x-3-2x+1=3x-2\left(x>=\dfrac{1}{2}\right)\\5x-3+2x-1=7x-4\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
12 tháng 1
a)
\(\begin{array}{l}\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\\ = \left( {2x + 5y} \right)\left( {2x - 5y + 2x + 5y} \right)\\ = \left( {2x + 5y} \right).4x\\ = 2x.4x + 5y.4x\\ = 8{x^2} + 20xy\end{array}\)
b)
\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)
28 tháng 3 2018
tách sai rồi bạn ơi
phải là
\(=\dfrac{1}{2}x^2y.\left(-4\right)x^2y^4+3x^2y^4.x^2y^2\)
=\(2x^4y^5+3x^4y^5\)
=\(5x^4y^5\)
28 tháng 3 2018
\(A=\dfrac{1}{2}x^2y.\left(-2xy^2\right)^2+2x^2y^3.\left(x^2y^2\right)\)
\(=\dfrac{1}{2}x^2y.\left(-2\right)x^2y^4+2x^4y^5\)
\(=\left(-1\right)x^4.y^5+2x^4y^5\)
\(=x^4y^5\)
Lại có : \(\left(x-2\right)^{18}+\left|y+1\right|=0\)
Mà \(\left\{{}\begin{matrix}\left(x-2\right)^{18}\ge0\\\left|y+1\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{18}=0\\\left|y+1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Mà \(A=x^4y^5\)
\(\Leftrightarrow A=2^4.\left(-1\right)^5\)
\(\Leftrightarrow A=-16\)
29 tháng 6 2022
a: \(M+N-P=2a^2-3a+1+5a^2+a-a^2+4=6a^2-2a+5\)
b: \(=2y-x-\left\{2x-y-\left[3x+y-5y+x\right]\right\}\)
\(=2y-x-\left\{2x-y-\left[4x-4y\right]\right\}\)
\(=2y-x-\left\{2x-y-4x+4y\right\}\)
\(=2y-x-\left[-2x+3y\right]\)
\(=-x+2y+2x-3y=x-y=\left(a-b\right)^2-\left(a-b\right)^2\)
=4ab
c: TH1: x>=1/2
A=5x-3-2x+1=3x-2
TH2: x<1/2
A=5x-3+2x-1=7x-4
\(A=2y-x-\left\{2x-y-\left[y+3x-\left(5y-x\right)\right]\right\}\)
\(=2y-x-\left\{2x-y-\left[y+3x-5y+x\right]\right\}\)
\(=2y-x-\left\{2x-y-y-3x+5y-x\right\}\)
\(=2y-x-2x+y+y+3x-5y+x\)
\(=\left(2y+y+y-5y\right)+\left(-x-2x+3x+x\right)\)
= \(-y+x\)
Thay \(x=a^2+2ab+b^2,y=a^2-2ab+b^2\) vào đa thức -y + x :
\(-\left(a^2-2ab+b^2\right)+\left(a^2+2ab+b^2\right)\)
\(=-a^2+2ab-b^2+a^2+2ab+b^2\)
\(=\left(-a^2+a^2\right)+\left(2ab+2ab\right)+\left(-b^2+b^2\right)\)
= 4ab
\(A=2y-x-\left\{2x-y-\left[y+3x-\left(5y-x\right)\right]\right\}\\ =2y-x-\left\{2x-y-y-3x+5y-x\right\}\\ =2y-x-2x+y+y+3x-5y+x\\ =-y+x=-\left(a^2-2ab+b^2\right)+\left(a^2+2ab+b^2\right)\\ =-a^2+2ab-b^2+a^2+2ab+b^2=4ab\)