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- \(A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=a^3+b^3+c^3-3abc\)
- B không rút gọn được.
- \(C=\left(a+b+c\right)^2-\left(a+b\right)^2-\left(a+c\right)^2-\left(b+c\right)^2\)
\(=-a^2-b^2-c^2\)
- \(D=\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(b+c-a\right)^2\)
\(=3a^2+2ab-2ac+3b^2+2bc+3c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2-2ac+c^2\right)+a^2+b^2+c^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2+a^2+b^2+c^2\)
a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)
\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)
\(=\left(2m\cdot2n\right)+m^2-n^2\)
\(=4mn+m^2-n^2\)
b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)
\(=2ab-2a^3\)
c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2=a^2\)
xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2
\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)
\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)
Câu b xem lại đề đi nhé
\(\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2a\right)\)
\(=a^2-c^2-\left(a-b\right)^2+c^2+b^2-2ab\)
\(=a^2-c^2-a^2+2ab-b^2+c^2+b^2-2ab\)
\(=0\)
\(=\left(a^2-c^2\right)-\left(\left(a-b\right)^2-c^2\right)+b^2-2ab\)
\(=a^2-c^2-\left(a-b\right)^2+c^2+b^2-2ab\)
\(=\left(a^2-2ab+b^2\right)-\left(a-b\right)^2\)
\(=\left(a-b\right)^2-\left(a-b\right)^2=0\)