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:
\(N=50y^2+\left(9-6y\right)\left(9-6y-2y\right)+14y\left(9-6y-y\right)\)
\(N=50y^2+\left(9-6y\right)\left(9-8y\right)+14\left(9-7y\right)\)
\(N=50y^2+81-72y-54y+48y^2+126-98y\)
\(N=2y^2-224y+207\)
\(a,M=26x^2+y\left(2x+y\right)-10x\left(x+y\right)\\ =26x^2+2xy+y^2-10x^2-10xy\\ =16x^2-8xy+y^2\\ =16\left(x^2-\dfrac{1}{2}xy+\dfrac{1}{16}y^2\right)\\ =16\left(x^2-2.x.y.\dfrac{1}{4}+\dfrac{1}{16}y^2\right)=16\left(x-\dfrac{1}{4}y\right)^2\\ Vì:x=0,25y\Rightarrow y=4x\\ Vậy:M=16\left(x-\dfrac{1}{4}y\right)^2=16\left(x-x\right)^2=16.0^2=0\\ Vậy:tại.x=0,25y.thìM=0\)
\(N=50y^2+x\left(x-2y\right)+14y\left(x-y\right)=50y^2+x^2-2xy+14xy-14y^2\\ =36y^2+12xy+x^2\\ =36\left(y^2+\dfrac{1}{3}xy+\dfrac{1}{36}x^2\right)\\ =36.\left(y+\dfrac{1}{6}x\right)^2\\ Ta.có:x+6y=9\Leftrightarrow y+\dfrac{1}{6}x=\dfrac{9}{6}=\dfrac{3}{2}\\ Vậy:N=36.\left(y+\dfrac{1}{6}x\right)^2=36.\left(\dfrac{3}{2}\right)^2=36.\dfrac{9}{4}=81\\ Vậy:Tại.x+6y=9.thì.N=81\)
