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10 tháng 12 2021

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

15 tháng 7 2021

B1

a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)

b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)

c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

\(=\)\(\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^3+1^3\right)\left(x^3-1^3\right)=x^6-1\)

d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

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

\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)

\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)

\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)

\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)

B2:

\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)

\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)

\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)

Bài 1: 

a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=x^2+2xy+y^2-x^2+2xy+y^2\)

=4xy

b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

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

\(=\left(2y\right)^2=4y^2\)

c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

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

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

\(=x^6-1\)

d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)

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

\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)

\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)

\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)

\(=2a^2-4bc\)

2 tháng 12 2023

Ta có \(x^2+y^2+xy+x=y-1\)

\(\Leftrightarrow2x^2+2y^2+2xy+2x-2y+2=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(\Rightarrow B=\left(-1+1-1\right)^{2023}\) \(=\left(-1\right)^{2023}\) \(=-1\)

2 tháng 12 2023

bvbbbvvbvv

28 tháng 9 2021

\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)

\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)

7 tháng 12 2019

25 tháng 7 2018

Ai giúp mik vs

25 tháng 7 2018

Huhu ai giúp vs

a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)