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bài này k chắc đâu nha ( có thể sai :v )
Ta có:
\(\left|-\frac{3}{x-1}\right|=\frac{\left|-3\right|}{\left|x-1\right|}=\frac{3\left|x-1\right|}{\left(x-1\right)^2}\)
\(\left(-\frac{3}{x-1}\right)^2=\frac{9}{\left(x-1\right)^2}\)
\(\Rightarrow pt\Leftrightarrow\frac{9-3\left|x-1\right|}{\left(x-1\right)^2}=6\Leftrightarrow9-3\left|x-1\right|=6\left(x^2-2x+1\right)\)
\(\Leftrightarrow-3\left|x-1\right|=6x^2-12x-3\Leftrightarrow\left|x-1\right|=-2x^2+4x+1\)(1)
+) Nếu x<1 ta có \(\left(1\right)\Leftrightarrow1-x=-2x^2+4x+1\Leftrightarrow2x^2-5x=0\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\left(lo\text{ại}\right)\\x=0\left(tm\right)\end{cases}}\)
+) Nếu x>1 ta có \(\left(1\right)\Leftrightarrow x-1=-2x^2+4x+1\Leftrightarrow2x^2-3x-2=0\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{1}{2}\left(lo\text{ại}\right)\end{cases}}\)
Vậy pt có 2 nghiệm x=0 & x=2
`|1/x+3|+|1/x-3|=1+|1/x^2-9|`
`<=>|1/x+3|+|1/x-3|=|(1/x-3)(1/x+3)|+1`
`<=>|1/x+3|-1=|(1/x-3)(1/x+3)|-|1/x-3|`
`<=>|1/x+3|-1=|(1/x-3)|(|1/x+3|-1)`
`<=>(|1/x+3|-1)(|1/x-3|-1)=0`
`+)|1/x+3|=1`
`<=>` $\left[ \begin{array}{l}\dfrac1x+3=1\\\dfrac1x+3=-1\end{array} \right.$
`<=>` $\left[ \begin{array}{l}\dfrac1x+2=0\\\dfrac1x+4=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}2x+1=0\\4x+1=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=-\dfrac12\\x=-\dfrac14\end{array} \right.$
`+)|1/x-3|=1`
`<=>` $\left[ \begin{array}{l}\dfrac1x-3=1\\\dfrac1x-3=-1\end{array} \right.$
`<=>` $\left[ \begin{array}{l}\dfrac1x-4=0\\\dfrac1x-2=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}4x-1=0\\2x-1=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac14\end{array} \right.$
Vậy `S={1/2,-1/2,1/4,-1/4}`
Bài 2: Áp dụng \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\left|x^2+x+3\right|+\left|-x^2-x+6\right|\ge\left|x^2+x+3-x^2-x+6\right|=\left|9\right|=9\)
Bài 1
Ta có (a-b)2 >=0
=) a2 + b2 >= 2ab
Cộng 2 vế BĐT cho a2 + b2 ta được:
a2 + b2 + a2 + b2 >= a2 + b2 +2ab
=) 2( a2 + b2 ) >= ( a + b)2
=) a2 + b2 >= ( a + b)2/2
Nhân 2 vế BĐT cho 1/2 ta được
a2 + b2 /2 >= ( a + b)2/4
Hay a2 + b2 /2 >= (a+b/2)2
Dấu '=' XRK : a=b
\(\frac{\left(2x^3+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x^2+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x^2+1\right)\left(x-2\right)}{\left(x+2\right)\left(x+1\right)}\)
Thay x=\(\frac{1}{2}\)
\(=\frac{2\left(\frac{1}{2}^2+1\right)\left(\frac{1}{2}-2\right)}{\left(\frac{1}{2}+2\right)\left(\frac{1}{2}+1\right)}\)
\(=-1\)
