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Đặt \(a-b=x\) , \(b-c=y\) và \(c-a=z\)
\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)
Chắc bạn cùng biết \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
Vậy \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Chúc bạn học tốt.
Bài 1.
\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)
\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)
a) \(A=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^2+c^3-3c^2a+3ca^2-a^3\)
\(=\left(a^3-a^3\right)+\left(-b^3+b^3\right)+\left(-c^3+c^3\right)-3\left(a^2b+ac^2-ab^2-bc^2+b^2c-a^2c\right)\)
\(=3[\left(a^2b-ab^2\right)+\left(ac^2-b^2c\right)-\left(a^2c-b^2c\right)]\)
\(=3[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)]\)
\(=3[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)]\)
\(=3\left(a-b\right)[\left(a+b+c^2-c\left(a+b\right)\right)]\)
\(=3\left(a-b\right)\left(ab+c^2-ca-cb\right)\)
\(=3\left(a-b\right)[\left(ab-ac\right)+\left(c^2-cb\right)]\)
\(=3\left(a-b\right)[a\left(b-c\right)+c\left(c-b\right)]\)
\(=3\left(a-b\right)[a\left(b-c\right)-c\left(b-c\right)]\)
\(=3\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
b) \(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
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Đặt a+b-c=x;c+a-b=y;b+c-a=z
=>x+y+z=a+b-c+a+b-c+b+c-a=a+b+c
Ta có hăng đẳng thức:(x+y+z)3-x3-y3-z3=3(x+y)(y+z)(x+z)
=>(a+b+c)3-(a+b-c)3-(c+a-b)3-(b+c-a)3
=(x+y+z)3-x3-y3-z3
=3(x+y)(y+z)(z+x)
=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)
=3.2a.2c.2b
=24abc
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=\left(ab^3-a^3b\right)+\left(bc^3-ac^3\right)+\left(a^3c-b^3c\right)\)
\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)
\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)
Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)
Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)