Cho biểu thức M = (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) + x2
Tính M theo a, b, c biết x = \(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\)
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=x2-bx-ax+ab+x2-cx-bx+bc+x2-cx-ax+x2
=(x2+x2+x2+x2)-(ax+bx+cx+ax+bx+cx)+ab+bc+ca
=4x2-2(a+b+c)x+ab+bc+ca
Thay x=\(\frac{1}{2}\)(a+b+c) vào M ta đc:
M=4.\(\frac{1}{4}\)(a+b+c)2-2(a+b+c).\(\frac{1}{2}\)(a+b+c)+ab+bc+ca
=(a+b+c)2-(a+b+c)2+ab+bc+ca
=ab+bc+ca
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ca+x^2\)
\(=4x^2-2ax-2bc-2cx+ab+bc+ca\)
\(=4x^2-2\left(a+b+c\right)x+ab+bc+ca\)
với \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\Rightarrow2x=a+b+c\)
\(\Rightarrow M=\left(a+b+c\right)^2-\left(a+b+c\right)^2+ab+bc+ca\)
\(=ab+bc+ca\)