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minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
Cách khác:
Áp dụng BĐT AM-GM:
\(\frac{a}{b^2}+\frac{1}{a}\geq 2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)
\(\frac{b}{c^2}+\frac{1}{b}\geq 2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)
\(\frac{c}{a^2}+\frac{1}{c}\geq 2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)
Cộng theo vế và rút gọn:
\(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
\(\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}-3=0\)
\(\Leftrightarrow\dfrac{x-b-c}{a}-1+\dfrac{x-c-a}{b}-1+\dfrac{x-a-b}{c}+1=0\)\(\Leftrightarrow\dfrac{x-a-b-c}{a}+\dfrac{x-a-b-c}{b}+\dfrac{x-a-b-c}{c}=0\)\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
\(\) vì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\Rightarrow x-a-b-c=0\)
\(\Rightarrow x=a+b+c\)
\(PT\Leftrightarrow\left(\dfrac{x-b-c}{a}-1\right)+\left(\dfrac{x-a-c}{b}-1\right)+\left(\dfrac{x-a-b}{c}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-a-b-c}{a}+\dfrac{x-a-b-c}{b}+\dfrac{x-a-b-c}{c}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\) nên \(x-a-b-c=0\Rightarrow x=a+b+c\)
Vậy nghiệm của PT là \(x=a+b+c\)
\(\dfrac{a+b-x}{c}+\dfrac{b+c-x}{a}+\dfrac{c+a-x}{b}+\dfrac{4x}{a+b+c}=1\)
\(\Leftrightarrow\dfrac{a+b-x}{c}+\dfrac{b+c-x}{a}+\dfrac{c+a-x}{b}+\dfrac{4x}{a+b+c}-1=0\)
\(\Leftrightarrow(\dfrac{a+b-x}{c}+1)+(\dfrac{b+c-x}{a}+1)+(\dfrac{c+a-x}{b}+1)+(\dfrac{4x}{a+b+c}-4)=0\)\(\Leftrightarrow\dfrac{a+b+c-x}{c}+\dfrac{a+b+c-x}{a}+\dfrac{a+b+c-x}{b}+\dfrac{-4\left(a+b+c-x\right)}{a+b+c}=0\)\(\Leftrightarrow\left(a+b+c-x\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{4}{a+b+c}\right)=0\)
Hiển nhiên: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{4}{a+b+c}>0\left(a,b,c>0\right)\)
\(\Rightarrow x=a+b+c\)