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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\dfrac{-1}{c^3}\) hay \(\dfrac{1}{a^3}+\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{b^3}=\dfrac{-1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{3}{abc}\)
\(a^2b^2c^2.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3}{abc}.a^2b^2c^2\)
\(\Leftrightarrow\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}=3abc\) hay\(M=3abc\left(đpcm\right)\)
`a,` Với `x=3`
\(B=\dfrac{x^2-x}{2x+1}\\ \Rightarrow\dfrac{3^2-3}{2\cdot3+1}\\ =\dfrac{9-3}{6+1}\\ =\dfrac{6}{7}\)
`b,` Ta có `M=A*B`
\(M=\left(\dfrac{1}{x-1}+\dfrac{x}{x^2-1}\right)\cdot\dfrac{x^2-x}{2x+1}\\ =\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+\text{ }1}\\ =\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x}{x+1}\)
`c,` Để `M=1/2`
`=> x/(x+1)=1/3`
`<=> (3x)/(3(x+1))= (x+1)/(3(x+1))`
`<=> 3x=x+1`
`<=>3x-x=1`
`<=>2x=1`
`<=>x=1/2`
Lời giải:
Ta có \(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left ( \frac{1}{a}-\frac{1}{b}-\frac{1}{c} \right )^2+\frac{2}{ab}+\frac{2}{ac}-\frac{2}{bc}\)
\(\Leftrightarrow P=6^2+\frac{2(b+c-a)}{abc}=6^2+\frac{2(-abc)}{abc}=34\)
Thanks bạn nhiều.
Mk nhờ bạn giải hộ mk bài mk cũng vừa mới đăng nha.
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)
\(\Rightarrow2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)
\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)
\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
\(\Rightarrow\dfrac{c+a+b}{abc}=1\)
\(\Rightarrow a+b+c=abc\)
Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)
M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)
= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)
= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac
= 4x\(^2\)-2x(a+b+c)+ab+bc+ca
Thay 2x=a+b+c,ta được:
M= 4x\(^2\)-2x.2c+ab+bc+ca
M= 4x\(^2\)-4x\(^2\)+ab+bc+ca
M= ab+bc+ca
M=\(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{7x-1}{x^2-1}\)
a, Rút gọn M
b, Tính giá trị của M tại x = -3
\(a,M=\dfrac{x+1-4x+4+7x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ b,x=-3\Rightarrow M=\dfrac{4}{-3-1}=-1\)
b.\(M=\dfrac{1}{-3-1}-\dfrac{4}{-3+1}+\dfrac{7\left(-3\right)-1}{\left(-3\right)^2-1}\)
\(M=\dfrac{-1}{4}-\left(-2\right)+\dfrac{11}{5}=3,95\)