Biết a-b=7,Tính giá trị củ biểu thức sau
\(a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(a-b+1\right)\)
K
Khách
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Y
1 tháng 7 2019
\(A=a^3-b^3+a^2+b^2-3ab\left(a-b\right)-3ab+ab\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2-3ab\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2-2ab+b^2+3ab\right)+49-3ab\left(a-b\right)\)
\(=\left(a-b\right)^2+3ab\left(a-b\right)+49-3ab\left(a-b\right)\)
\(=49+49=98\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 9 2015
1/
\(A=a^3+a^2-b^3+b^2+ab-3a^2b+3ab^2-3ab\)
\(A=\left(a^3-3a^2b+3ab^2-b^3\right)+\left(a^2-2ab+b^2\right)=\left(a-b\right)^3+\left(a-b\right)^2=7^3+7^2=392\)
29 tháng 12 2018
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2+2ab-2ab\right)+6a^2b^2\left(a+b\right)\)
\(M=a^2+2ab+b^2-3ab+3ab-6a^2b^2+6a^2b^2\)
\(M=\left(a+b\right)^2=1\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 1
Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
24 tháng 11 2018
\(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Rightarrow\left(2x^2+4xy+2y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
Khi đó: \(A=\left(-1+1\right)^{2014}+\left(-1+2\right)^{2015}+\left(1-1\right)^{2016}\)
\(=0+1+0=1\)
26 tháng 5 2022
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
Ta có : \(a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(a-b+1\right)\)
= \(a^3+a^2-b^3+b^2+ab-3ab\left(7+1\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+a^2+b^2+ab-24ab\)
\(=7\left(a^2+b^2+ab\right)+a^2+b^2-23ab\)
\(=7a^2+7b^2+7ab+a^2+b^2-23ab\)
\(=8a^2-16ab+8b^2\)
\(=8\left(a^2-2ab+b^2\right)\)
\(=8\left(a-b\right)^2=8.7^2=392\)