\(a,A=y^2-\dfrac{1}{2}y+\dfrac{1}{16}\)

\(=y^2-2.y.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)

\(=\left(y-\dfrac{1}{4}\right)^2\)

Với \(y=100,25\), ta được:

\(A=\left(100,25-\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{401}{4}-\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{400}{4}\right)^2=100^2=10000\)

\(------\)

\(b,B=4x^2-9y^2-6y-1\)

\(=\left(2x\right)^2-\left[\left(3y\right)^2+2.3y.1+1\right]\)

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

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

Với \(x=23;y=1\), ta được:

\(B=\left(2.23-3.1-1\right)\left(2.23+3.1+1\right)\)

\(=\left(46-4\right)\left(46+4\right)\)

\(=42.50=2100\)

Ta có:

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

Thay \(x=-3\) vào A, ta có:

\(A=2.\left(-3\right)^2=18\)

Vậy A=18

\(A=x\left(x+y\right)-x\left(y-x\right)=x\left(x+y\right)+x\left(x+y\right)=\left(x+y\right).2x=\left(-3+2\right).2.\left(-3\right)=6\)

Giúp vs ạ

a) Thay \(x=0,25y\) vào M ta có:

\(M=26\cdot\left(0,25y\right)^2+y\left(2\cdot0,25y+y\right)-10\cdot0,25y\cdot\left(0,25y+y\right)\)

\(M=1,625y^2+y\cdot1,5y-2,5y\cdot1,25y\)

\(M=1,625y^2+1,5y^2-3,125y^2\)

\(M=0\)

b) Thay \(x+6y=9\Rightarrow x=9-6y\) vào N ta có:

\(N=50y^2+\left(9-6y\right)\left(9-6y-2y\right)+14y\left(9-6y-y\right)\)

\(N=50y^2+\left(9-6y\right)\left(9-8y\right)+14\left(9-7y\right)\)

\(N=50y^2+81-72y-54y+48y^2+126-98y\)

\(N=2y^2-224y+207\)

\(a,M=26x^2+y\left(2x+y\right)-10x\left(x+y\right)\\ =26x^2+2xy+y^2-10x^2-10xy\\ =16x^2-8xy+y^2\\ =16\left(x^2-\dfrac{1}{2}xy+\dfrac{1}{16}y^2\right)\\ =16\left(x^2-2.x.y.\dfrac{1}{4}+\dfrac{1}{16}y^2\right)=16\left(x-\dfrac{1}{4}y\right)^2\\ Vì:x=0,25y\Rightarrow y=4x\\ Vậy:M=16\left(x-\dfrac{1}{4}y\right)^2=16\left(x-x\right)^2=16.0^2=0\\ Vậy:tại.x=0,25y.thìM=0\)

\(N=50y^2+x\left(x-2y\right)+14y\left(x-y\right)=50y^2+x^2-2xy+14xy-14y^2\\ =36y^2+12xy+x^2\\ =36\left(y^2+\dfrac{1}{3}xy+\dfrac{1}{36}x^2\right)\\ =36.\left(y+\dfrac{1}{6}x\right)^2\\ Ta.có:x+6y=9\Leftrightarrow y+\dfrac{1}{6}x=\dfrac{9}{6}=\dfrac{3}{2}\\ Vậy:N=36.\left(y+\dfrac{1}{6}x\right)^2=36.\left(\dfrac{3}{2}\right)^2=36.\dfrac{9}{4}=81\\ Vậy:Tại.x+6y=9.thì.N=81\)

\(A=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\\ A=-2\cdot\dfrac{1}{2}\left(-100\right)=100\)

A = (-25).8.2.5.(-4).13 = [(-25).(-4)].8.2.5.13 = 100.8.10.13 = 10400

B = 13.(-7) + (-13).57 + 13.(-36) = 13.(-7) - 13.57 + 13.(-36)

                                                      = 13.[(-7) + (-57) + (-36)]

                                                      = 13. (-100)

                                                      = -1300

Ta có: \(A=\left(-25\right)\cdot8\cdot2\cdot5\cdot\left(-4\right)\cdot13\)

\(=\left[\left(-25\right)\cdot\left(-4\right)\right]\cdot\left(2\cdot5\right)\cdot\left(8\cdot13\right)\)

\(=104000\)

Ta có: \(B=13\cdot\left(-7\right)+\left(-13\right)\cdot57+13\cdot\left(-36\right)\)

\(=13\cdot\left(-7\right)-13\cdot57-13\cdot36\)

\(=13\left(-7-57-36\right)\)

\(=13\cdot\left(-100\right)=-1300\)

gúp mình với

A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

\(A=x\left(x+y\right)-y\left(x+y\right)\)

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

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

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

\(=\dfrac{1}{4}-4\)

\(=-\dfrac{15}{4}\)

a) Ta có: -7x+13>-7y+13

\(\Leftrightarrow-7x>-7y\)

hay x<y

b) Ta có: 11x-1>11y+1

mà 11x+1>11x-1

nên 11x+1>11y+1

\(\Leftrightarrow11x>11y\)

hay x>y

c) Ta có: -19x-37<-19y-37

\(\Leftrightarrow-19x< -19y\)

hay x>y

b: \(N=3x-2y+5x-y-7y+2x=10x-10y=10\cdot\left(x-y\right)=0\)

\(a,M=12-x+x-73+96+x-23=x+12\\ M=101+12=113\\ b,N=3x-2y+5x-y-7y+2x=10x-10y\\ N=10\cdot2021-10\cdot2021=0\)

tại sao lại bắt buộc bọn mình thích trả lời kiểu nào thì trả lời kiểu đấy cần j phải nghe bạn

a, Với x = 3 và y = -2 ta có:

\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|3\right|\right)+\left(-2\right)\)

\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-3\right)-2\)

\(A=\dfrac{3}{2}+\dfrac{4}{9}.3-2\)

\(A=\dfrac{3}{2}+\dfrac{4}{3}-2\)

\(A=\dfrac{5}{6}\)

 Với x = 3 và y = -3 ta có:
\(B=\left|2.3-1\right|+\left|3.\left(-3\right)+2\right|\)

\(B=\left|5\right|+\left|-7\right|\)

\(B=5+7=12\)

Hoctot ! ko hiểu chỗ nào cứ hỏi cj nhé