\(|x+6.4|+\left|x+2.5\right|+\left|x+8.1\right|=4x\)
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BL
12 tháng 3 2020
\(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=-2,5\)
\(\Rightarrow\frac{4x+1}{\left(x+1\right)\left(x-2\right)}+\frac{2,5\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Rightarrow4x+1+2,5x^2-5x+2,5x-5=0\)
\(\Rightarrow2,5x^2+1,5x-4=0\)
\(\Rightarrow2,5x^2-2,5x+4x-4=0\)
\(\Rightarrow2,5x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2,5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2,5x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1,6\end{matrix}\right.\)
Vậy....
12 tháng 3 2020
Ta có: \(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=-2,5\)
⇔\(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=\frac{-5}{2}\)
⇔\(2\left(4x+1\right)=-5\left(x+1\right)\left(x-2\right)\)
⇔\(8x+2=\left(-5x-5\right)\left(x-2\right)\)
⇔\(8x+2=-5x^2+5x+10\)
⇔\(8x+2+5x^2-5x-10=0\)
⇔\(3x+5x^2-8=0\)
⇔\(5x^2+3x-8=0\)
⇔\(5x^2-5x+8x-8=0\)
⇔\(5x\left(x-1\right)+8\left(x-1\right)=0\)
⇔\(\left(x-1\right)\left(5x+8\right)=0\)
⇔\(\left[{}\begin{matrix}x-1=0\\5x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-8}{5}=-1,6\end{matrix}\right.\)
Vậy: x∈{1;-1,6}
5 tháng 1 2021
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
31 tháng 7 2016
a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)
\(\Leftrightarrow40x=46+9+25=80\)
\(\Leftrightarrow x=2\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)
\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)
\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)
\(\Leftrightarrow2x=-2+6-5=-1\)
\(\Leftrightarrow x=\frac{1}{2}\)
7 tháng 4 2017
a) 3x^3 -10x+3 =(3x-1)(x-3)
| x | -vc | 1/3 | 5/4 | 3 | +vc | |||||||||
| 3x-1 | - | 0 | + | + | + | + | + | |||||||
| x-3 | - | - | - | - | - | 0 | + | |||||||
| 4x-5 | - | - | - | 0 | + | + | + | |||||||
| VT | - | 0 | + | 0 | - | 0 | + |
Kết luận
VT< 0 {dấu "-"} khi x <1/3 hoắc 5/4<x<3
VT>0 {dấu "+"} khi x 1/3<5/4 hoặc x> 3
VT=0 {không có dấu} khi x={1/3;5/4;3}
Bỏ | | đi và giải bình thường (vì x>0)
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