tinh gia tri cua bieu thuc A=\(x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
Những câu hỏi liên quan
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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tinh nhanh gia tri cua bieu thuc voi x,y nhan bat ki gia tri nao:
\(\frac{1}{3}x^2y\left(3xy\right)^2y^4+\frac{1}{2}x\left(-2xy\right)^3y^4+x^4y^7+18\)
cau 1: tinh gia tri cua x thoa manleft(x-3right)left(x^2+3x+9right)+xleft(x+2sqrt{2}right)left(2sqrt{2}-xright)-3cau 2.tinh GTLN cua bieu thuc2x-2x^2+13cau 3. tinh gia tri cua bieu thuc frac{3^{left(x+yright)^2}}{3^{left(x-yright)^2}}voi xyfrac{1}{2}cau 4. tim GTLN cua-3x^2-6x-4cau 5. cho ham so : f(x)frac{1}{5x+9}tinh gia tri cua fleft(frac{40}{25}right)cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
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cau 1: tinh gia tri cua x thoa man
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\sqrt{2}\right)\left(2\sqrt{2}-x\right)=-3\)
cau 2.tinh GTLN cua bieu thuc
\(2x-2x^2+13\)
cau 3. tinh gia tri cua bieu thuc
\(\frac{3^{\left(x+y\right)^2}}{3^{\left(x-y\right)^2}}\)voi xy=\(\frac{1}{2}\)
cau 4. tim GTLN cua
\(-3x^2-6x-4\)
cau 5. cho ham so : f(x)=\(\frac{1}{5x+9}\)
tinh gia tri cua \(f\left(\frac{40}{25}\right)\)
cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
a, biet x+y=0
tinh gia tri bieu thuc : M=\(x^4-xy^3+x^3y-y^4-1\)
b, biet xyz=2 va x+y+z=0
tinh gia tri bieu thuc : M= \(\left(x+y\right)\left(y+2\right)\left(x+2\right)\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
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Tinh gia tri cua bieu thuc A, biet A= \(\frac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2\)voi \(^{x^2-y^2=4}\)
\(A=\frac{1}{2}x^4+x^2y^2+\frac{1}{2}y^4-2x^2y^2\)
\(=\frac{1}{2}\left(x^4-2x^2y^2+y^4\right)=\frac{1}{2}\left(x^2-y^2\right)^2=\frac{1}{2}.4^2=8\)
1. biết x2-2y2=xy,y\(\ne\)0,x+y\(\ne\)0. thì gia tri cua bieu thuc Q=\(\frac{x+y}{x-y}\)=
2.cho x\(\ne\)0,y\(\ne\)0 thoa man x+y=4 ;xy=2 .gia tri cua bieu thuc A=\(\frac{1}{x^3}+\frac{1}{y^3}\)la
3.gia tri cua bieu thuc A=\(\frac{81^8-1}{\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}\)la
Bài 3:
Ta có:
\(81^8-1=\left(9^2\right)^8-1=\left[\left(3^2\right)^2\right]^8-1=3^{32}-1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Do đó:
\(A=3^4-1=80\)
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A=\(\frac{\left(1^3+2^3+3^3+...+10^3\right)\cdot\left(x^2+y^2\right)\cdot\left(x^3+y^3\right)\cdot\left(x^4+y^4\right)}{1^2+2^2+3^2+...+10^2}\)
Tinh gia tri bieu thuc
tinh gia tri bieu thuc
C=x^3+y^3+3xy(x^2+y^2)+6x^2y^2(x+y) với x+y=1
nhanh nha mik tick cho
Cho x - y = 11. Tinh gia tri cua bieu thuc : M = x^3-3xy(x- y) - y^3 -x^2 + 2xy - y^2
Ta có: x^3 -3xy(x-y) -y^3 -x^2 + 2xy-y^2
= x^3 -y^3 - 3xy(x-y) -( x^2 -2xy+y^2)
= (x-y)(x^2+xy +y^2) - 3xy(x-y) -(x-y)^2
= (x-y)(x^2+xy+y^2 -3xy-x+y)
=11( x^2 -2xy+y^2 -x+y)
= 11[ (x-y)^2 -(x-y)]
= 11[ 11^2 -11]
= 11^3 -11^2=...
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