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14 tháng 7 2019
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a.\left(a+b+c\right)+bc\right]\left[b.\left(a+b+c\right)+ac\right]\left[c.\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ba+b^2+bc+ac\right)\left(ca+cb+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ba+b^2\right)+\left(bc+ac\right)\right]\left[\left(ca+c^2\right)\left(cb+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(b+a\right)\right]\left[c\left(a+c\right)b\left(b+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
F
1 tháng 10 2020
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ac\right]\left[c\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)\left(ac+bc+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ab+b^2\right)+\left(bc+ac\right)\right]\left[\left(ac+c^2\right)+\left(bc+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
HN
11 tháng 8 2021
a/ \(\left(a^2-b^2\right)\left(c^2-d^2\right)=a^2c^2-a^2d^2-b^2c^2+b^2d^2\)
\(=\left(a^2c^2+2abcd+b^2d^2\right)-\left(a^2d^2+2abcd+b^2c^2\right)\)
\(=\left(ac+bd\right)^2-\left(ad+bc\right)^2\)
HN
11 tháng 8 2021
b/ \(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
\(\Leftrightarrow x=y=z\)
11 tháng 8 2021
a: Ta có: \(\left(ac+bd\right)^2-\left(ad+bc\right)^2\)
\(=a^2c^2+b^2d^2+2abcd-a^2d^2-b^2c^2-2abcd\)
\(=a^2\left(c^2-d^2\right)-b^2\left(c^2-d^2\right)\)
\(=\left(a^2-b^2\right)\left(c^2-d^2\right)\)
12 tháng 8 2021
Bạn có làm đc câu b ko, nếu đc thì làm nốt giùm mink nha
7 tháng 7 2019
Chi tham khao tai day:
Câu hỏi của Vương Nguyễn Thanh Triều - Toán lớp 8 - Học toán với OnlineMath
AT
24 tháng 8 2018
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\text{vì }\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Leftrightarrow a=b=c=1\)
Vậy a=b=c=1