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\(P=\frac{a^3b^2c^2}{ab+a^2bc+abc}+\frac{ab^2c}{bc+b+abc}+\frac{abc^2}{ac+c+1}\)
\(=\frac{ }{ab\left(1+ac+c\right)}+\frac{ }{b\left(c+1+ac\right)}+\frac{ }{ac+c+1}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
đpcm
a, x^2 - 2xy + 2y^2 - 2x + 6y + 5 =0
<=> x^2 - 2x(y+1) + y^2 + 2y + 1 + y^2 + 4y + 4 = 0
<=> x^2 - 2x(y+1) + (y+1)^2 + (y+2)^2 =0
<=> (x-y-1)^2 + (y+2)^2 =0
<=> x-y-1 = 0 và y+2 =0
<=> y = -2 => x= -1
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2abc\left(a+b+c\right)}{a^2b^2c^2}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(\frac{2}{ab}-9=\frac{1}{c^2}\)\(\Rightarrow\frac{2}{ab}-\frac{1}{c^2}=9\)
Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\left(\frac{2}{ab}-\frac{1}{c^2}\right)=3^2-9\)
\(\Rightarrow\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2.\frac{1}{a}.\frac{1}{b}+2.\frac{1}{b}.\frac{1}{c}+2.\frac{1}{c}.\frac{1}{a}-\frac{2}{ab}+\frac{1}{c^2}=0\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}-\frac{2}{ab}+\frac{1}{c^2}=0\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ac}+\frac{1}{c^2}=0\)
\(\Rightarrow\left(\frac{1}{a^2}+\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}\right)=0\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{c}=0\\\frac{1}{b}+\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{-1}{c}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)\(\Rightarrow\frac{-1}{c}+\frac{-1}{c}+\frac{1}{c}=3\)\(\Rightarrow\frac{-1}{c}=3\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=3\)\(\Rightarrow c=-\frac{1}{3}\)và\(a=b=\frac{1}{3}\)
Lại có: \(P=\left(a+3b+c\right)^{2020}=\left(\frac{1}{3}+3.\frac{1}{3}+\frac{-1}{3}\right)^{2020}=1^{2020}=1\)
Đặt \(a=x;2b=y;3c=z\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow Q=\frac{\frac{y}{2}.\frac{z}{3}}{x^2}+\frac{x.\frac{c}{3}}{2y^2}+\frac{x.\frac{y}{2}}{3z^2}\)
\(=\frac{x^3y^3+y^3z^3+z^3x^3}{6x^2y^2z^2}\)
\(=\frac{x^3y^3+y^3z^3+z^3x^3-3x^2y^2z^2+3x^2y^2z^2}{6x^2y^2z^2}\)
\(=\frac{\left(xy+yz+zx\right)\left(x^2y^2+y^2z^2+z^2x^2-x^2yz-y^2zx+z^2xy\right)+3x^2y^2z^2}{6x^2y^2z^2}\)
\(=\frac{3x^2y^2z^2}{6x^2y^2z^2}=\frac{1}{2}\)
\(\frac{ab+bc+ca}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(A=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
Pham Van Hung mình ko hiểu tại sao \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(Q=\dfrac{2024a}{ab+2024a+2024}+\dfrac{b}{bc+b+2024}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{abc.a}{ab+abc.a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+1+ac}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac+c+1}{ac+c+1}=1\)