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câu a: ta có:
(x+y)=(x-y)=x(x-y)+y(x-y)
=x2 - xy +yx - y2
=(-xy+yx) + x2 - y2 = x2 - y2
Vậy x2 - y2 = (x+y) (x-y)
còn câu b mình hông bik=)))))
\(^{x^2-y^2=x^2+xy-y^2-xy=x\left(x+y\right)-y\left(x+y\right)=\left(x+y\right)\left(x-y\right)..}\)
Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow a=3k;b=4k;c=5k\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(3k-4k\right)\left(4k-5k\right)\)
\(=4.\left[\left(3-4\right).k\right].\left[\left(4-5\right).k\right]\)
\(=4.\left[-k\right].\left[-k\right]=4k^2\left(1\right)\)
\(\Rightarrow\left(a-c\right)^2=\left(3k-5k\right)^2=\left[\left(3-5\right).k\right]^2=\left[-2k\right]^2=4k^2\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\left(dpcm\right)\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\Rightarrow\left(\frac{a^2}{c^2}\right)^3=\frac{\left(a^2+b^2\right)^3}{\left(a^2+d^2\right)^3}=\frac{a^6}{c^6}\left(1\right)\)
\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}\Rightarrow\left(\frac{a^3}{c^3}\right)^2=\frac{\left(a^3+b^3\right)^2}{\left(a^3+d^3\right)^2}=\frac{a^6}{c^6}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{\left(a^2+b^2\right)^3}{\left(c^2+d^2\right)^3}=\frac{\left(a^3+b^3\right)^2}{\left(c^3+d^3\right)^2}\left(đpcm\right)\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)
\(\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b^3}{d^3}\)
Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3=\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3.\)\(=3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)=3\left(a+b\right)\left(ab+ac+bc+c^2\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
#)Giải :
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b-c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca+a^2+ab+ac+a^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2ab+b^2+c^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\Rightarrowđpcm\)