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Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
\(a.\) Với \(a+b+c=0\) thì \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
\(b.\) Công thức tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)
\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)
Do đó, suy ra được: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
\(\frac{A}{x+2}+\frac{B}{\left(x+2\right)^2}+\frac{C}{\left(x+2\right)^3}=\frac{x^2+x+4}{\left(x+2\right)^3}\)
\(\Leftrightarrow\frac{A\left(x+2\right)^2}{\left(x+2\right)^3}+\frac{B\left(x+2\right)}{\left(x+2\right)^3}+\frac{C}{\left(x+2\right)^3}=\frac{x^2+x+4}{\left(x+2\right)^3}\)
\(\Leftrightarrow\frac{A\left(x+2\right)^2+B\left(x+2\right)+C}{\left(x+2\right)^3}=\frac{x^2+x+4}{\left(x+2\right)^3}\)
\(\Leftrightarrow A\left(x+2\right)^2+B\left(x+2\right)+C=x^2+x+4\)
\(\Leftrightarrow A\left(x^2+4x+4\right)+Bx+2B+C=x^2+x+4\)
\(\Leftrightarrow Ax^2+4Ax+4A+Bx+2B+C=x^2+x+4\)
\(\Leftrightarrow Ax^2+\left(4A+B\right)x+\left(4A+2B+C\right)=x^2+x+4\)
Áp dụng: \(ax^2+bx+c=1.x^2+1.x+4\Rightarrow\hept{\begin{cases}a=1\\b=1\\c=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}A=1\\4A+B=1\\4A+2B+C=4\end{cases}\Leftrightarrow\hept{\begin{cases}A=1\\B=1-4B=1-4.1=-3\\C=4-4A-2B=4-4.1-2.\left(-3\right)=6\end{cases}}}\)
Vậy A = 1; B = -3 ; C = 6 thì thỏa mãn
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