a: Sửa đề: y=2/3

\(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)

b: \(Q=x^2+4xy+4y^2-2x-4y+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2\cdot5+10=25\)

a: Sửa đề: \(P=27y^3+9y^2+y+\dfrac{1}{27}\)

\(=\left(3y+\dfrac{1}{3}\right)^3\)

\(=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)

b: \(Q=x^2+4xy+4y^2-2x-4y+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=25-2\cdot5+10=25\)

a: \(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)

b: \(Q=x^2+4xy+4y^2-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

=25

\(\left(\dfrac{1}{3}.x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\left(\dfrac{1}{3}.x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)

b: \(f\left(x\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)

Thay x=-8 và y=6 cào C ta được:

\(C=\dfrac{\left(-8\right)^3}{2}+\dfrac{\left(-8\right)^2.6}{4}+\dfrac{\left(-8\right).6^2}{6}+\dfrac{6^3}{27}\)\(=\dfrac{-512}{2}+\dfrac{384}{4}-\dfrac{288}{6}+\dfrac{216}{27}\)\(=-256+96-48+8=-200\)

\(C=x^2\left(\dfrac{x}{2}+\dfrac{y}{4}\right)+y^2\left(\dfrac{x}{6}+\dfrac{y}{27}\right)=\left(-8\right)^2\left(-\dfrac{8}{2}+\dfrac{6}{4}\right)+6^2\left(-\dfrac{8}{6}+\dfrac{6}{27}\right)=-200\)

tích đúng mình làm cho

là sao ạk
giải giùm mình với ạk

1/Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay x-y=7 vào A ta đc:

\(A=7^2+2.7+37=49+14+37=100\)

2/Ta có :\(B=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x+2y=5 vào B ta đc

\(B=5^2-2.5+10=25-10+10=25\)

\(x^2+4y^2-5x+10y-4xy+20\)

\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)

\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)

\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được : 

\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)

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

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

\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được : 

\(=1-2=-1\)

a: A=y(x-4)-5(x-4)

=(x-4)(y-5)

Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5

b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)

=0,2*10=2

d: Khi x=5,75 và y=4,25 thì

D=5,75^3-5,75^2*4,25+4,25^3

=8087/64

a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)

\(=-4.\dfrac{1}{4}+10=9\)

b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)

\(=\left(-2\right).\left(32-32\right)=0\)

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

\(=4x^2-4x+1+9-4x^2\)

\(=-4x+10\)

\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)