có:a^3+b^3=2a^2b^2(4ab-3)
Tính a+b-2ab
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a, \(\left(a^2+b^2-2ab+2a-2b+1\right)+\left(b^2-2b+1\right)=0\)
=> \(\left(a-b+1\right)^2+\left(b-1\right)^2=0\)
Mà \(\left(a-b+1\right)^2\ge0,\left(b-1\right)^2\ge0\)
=> \(\hept{\begin{cases}a-b+1=0\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}}\)
b,Tương tự
\(\left(a-2b+1\right)^2+\left(b-1\right)^2=0\)
=>\(\hept{\begin{cases}a=1\\b=1\end{cases}}\)
a, \(\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2=\left(a+b\right)^2+2c\left(a+b\right)+c^2=a^2+b^2+c^2+2ab+2ac+2bc\)
b, \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2=2a^2+2b^2\)
c, \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab\)
\(\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2=\left(a+b\right)^2+2\cdot\left(a+b\right)\cdot c+c^2\\ =a^2+2ab+b^2+2ac+2bc+c^2\\ =a^2+b^2+c^2+2ab+2ac+2bc\)
\(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\\ 2a^2+2b^2\)
\(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)\\ =2a\cdot2b=4ab\)
\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.
a) ta có : \(M=\left(a+b\right)^3+2a^2+4ab+2b^2\)
\(M=\left(a+b\right)^3+2\left(a^2+2ab+b^2\right)=\left(a+b\right)^3+2\left(a+b\right)^2\)
\(M=\left(7\right)^3+2.\left(7\right)^2=343+98=441\) vậy \(M=441\) khi \(a+b=7\)
b) ta có : \(N=\left(a-b\right)^3-a^2+2ab-b^2\)
\(N=\left(a-b\right)^3-\left(a^2-2ab+b^2\right)=\left(a-b\right)^3-\left(a-b\right)^2\)
\(N=\left(5\right)^3-\left(5\right)^2=125-25=100\) vậy \(N=100\) khi \(a-b=5\)
c) ta có : \(\left(a-b\right)^2=a^2-2ab+b^2=a^2+2ab+b^2-4ab\)
\(=\left(a+b\right)^2-4ab=\left(5\right)^2-4.2=25-8=17\)
vậy \(\left(a-b\right)^2=17\) khi \(a+b=5\) và \(ab=2\)