\(P=27y^3+9y^2+y+\dfrac{1}{27}=\left(3y+3\right)^3\)

Với \(y=\dfrac{2}{3}\) ta có:

\(P=\left(3.\dfrac{2}{3}+3\right)^3=5^3=125\)

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

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

\(=\left[x^2-2x\left(1-2y\right)+\left(1-2y\right)^2\right]+4y^2-4y+10-\left(1-2y\right)^2\)\(=\left(x+2y-1\right)^2+4y^2-4y+10-1+4y-4y^2\)\(=\left(x+2y-1\right)^2+9\)

Với \(x+2y=5\) , ta có:

\(Q=\left(5-1\right)^2+9=25\)

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 : 

\(=1-2=-1\)

Ta có

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

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

\(\Rightarrow C=5^2-2.5+10\)

\(\Rightarrow C=25-10+10=25\)

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

    \(=\left[x^2+2.x.2y+\left(2y\right)^2\right]-\left(2x+4y\right)+10\)

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

    \(=5^2-2.5+10\)

    \(=5^2-10+10\)

    \(=25-10+10\)

    \(=25\)

CHÚC BẠN HỌC TỐT !!!

2.

a. Ta có: x + y = 5 ⇒ x = 5 - y

Thay vào A ta được:

\(A=3\left(5-y\right)^2+3y^2-2y+6\left(5-y\right).y-100\)

\(A=75-30y+3y^2+3y^2-2y+30y-6y^2-100\)

\(A=75-100=-25\)

b. Ta có: x - y = 7 ⇒ x = 7 + y

Thay x = 7 + y vào A ta được:

\(A=\left(7+y\right)\left(7+y+2\right)+y\left(y-2\right)-2\left(7+y\right).y+37\)

\(A=y^2+16y+63+y^2-2y-14y-2y^2+37\)

\(A=100\)

c. Ta có: x + 2y = 5 ⇒ x = 5 - 2y

Thay vào A ta có:

\(A=\left(5-2y\right)^2+4y^2-2\left(5-2y\right)+10+4\left(5-2y\right).y-4y\)

\(A=25-20y+4y^2+4y^2-19+4y+10+20y-8y^2-4y\)

\(A=16\)