cho x-y=3.tinh gia tri bieu thuc A=x(2x+4y-4)+2y(y-5x+2)+2xy
tinh nhanh gia tri cua bieu thuc voi x,y nhan bat ki gia tri nao:
\(\frac{1}{3}x^2y\left(3xy\right)^2y^4+\frac{1}{2}x\left(-2xy\right)^3y^4+x^4y^7+18\)
giup minh voi cam on cac ban . cho x+y=3 .tinh gia tri bieu thuc A= x^2+2xy+y^2-4x-4y+1
\(A=x^2+2xy+y^2-4x-4y+1\)
\(A=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(A=3^2-4.3+1\)
\(A=-2\)
\(x^2+2xy+y^2-4x-4y+\)\(1\)
\(=\left(x^2+2xy+y^2\right)-\left(4x+4y\right)+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
Thay x+y = 1, ta có:
\(=3^2-4.3+1=-2\)
Bạn tham khảo cách 2 nhé !
\(x^2+2xy+y^2-4x-4y+1\)
\(=\left(x^2+y^2+1+2xy-2x-2y\right)-2x-2y\)
\(=\left[1-\left(x+y\right)\right]^2-2\left(x+y\right)\)
Thay x+y=1 ta có
\(=\left(1-3\right)^2-2.3\)
\(=-2\)
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1
b,x^3+xy^2-x^2y-y^3+3 biet x-y=0
b, Ta co: \(x^3+xy^2-x^2y-y^3+3\)
\(=\left(x^3-y^3\right)+\left(xy^2-x^2y\right)+3\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\left(x-y\right)+3\)
= 3 ( vì x-y = 0)
Tim cac gia tri cua Y de bieu thuc sau nhan gia tri duong
a)2y^2-4y
b)5x(3y+1)x(4y-3)
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1
x^2=a;y^2=b( Đk:a,b không âm)
Từ giả thiết suy ra a+b=2
=>3x^4+5x^2y^2+2y^4+2y^2
=3a^2+5ab+2b^2+2b
=(3a^2+3ab)+(2ab+2b^2)+2b
=3a(a+b)+2b(a+b)+2b
=(a+b)(3a+2b)+2b
=2(3a+2b)+2b
=2(2a+2b)+2a+2b
=4.2+2.2=12
tinh gia tri bieu thuc:
a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1