a. Thay a = 14, b = 15, c = 10, ta có:

\(a=a\times b+200\)

\(=>a=14\times15+200\)

\(=>a=210+200=410\)

___

\(b=a\times b\times c\)

\(=>b=14\times15\times10=2100\)

b. Vì 410 < 2100 nên a < b.

\(#NqHahh\)

a: Khi a=14 và b=15 thì \(A=14\cdot15+200=210+200=410\)

Khi a=14 và b=15 và c=10 thì \(B=14\cdot15\cdot10=210\cdot10=2100\)

b: A=410

B=2100

=>A<B

a: \(17628+3547\cdot6\)

\(=17628+21282\)

\(=38910\)

b: \(57924-15760:5\)

\(=57924-3152\)

=54772

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: =6450+50+2/17=6500+2/17=110502/17

b: =100+26x901=23526

c: =13600:68=200

\(A=x^4+y^4+\left(x+y\right)^4\)

\(=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(=2y^4+4y^2\left(x^2+xy\right)+2\left(x^4+2x^3y+x^2y^2\right)\)

\(=2y^4+4y^2\left(x^2+xy\right)+2\left(x^2+xy\right)^2\)

\(=2\left(y^2+xy+x^2\right)^2=2.5^2=50\)

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:

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

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

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

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

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

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

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:

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

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

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