Tìm x biết:l 5x - 3 l -2x = 14
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a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)
\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)
\(\Rightarrow-12-56=24x-56x\)
\(\Rightarrow-12+156=24x+56x\)
\(\Rightarrow144=80x\)
\(\Rightarrow x=144:80\)
\(\Rightarrow x=\frac{9}{5}\)
b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)
\(\Rightarrow-4x=0\)
\(\Rightarrow-4.0\)
\(\Rightarrow x=0\)
\(2|5x-3|-2x=14\)
\(2|5x-3|=14+2x\)
\(|5x-3|=\frac{14+2x}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{14+2x}{2}\\5x-3=-\frac{14+2x}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(5x-3\right)2=14+2x\\\left(5x-3\right)2=-14-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}10x-6-2x=-14\\10x-6+2x=14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}8x=14+6\\12x=-14+6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}8x=20\\12x=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2,5\\x=-\frac{2}{3}\end{cases}}\)
vậy \(x=2,5\)hoặc\(x=-\frac{2}{3}\)
2.|5x-3|-2x=14 (1)
+Xét 5x-3>=0 =>5x>=3 =>x>=0.6
Từ (1) ta có:2.(5x-3)-2x=14
10x-6-2x=14
8x=20
x=2.5(thỏa mãn x>=0.6)
+Xét 5x-3<0=>5x<3=>x<0.6
Từ (1) ta có:2.(-(5x-3))-2x=14
2.(-5x+3)-2x=14
-10x+6-2x=14
-12x=8
x=-2/3(thỏa mãn x<0.6)
Vậy x=2.5 hoặc x=-2/3 thì 2.|5x-3|-2x=14
k cho toi nhe!
2.|5x - 3| - 2x = 14
2.|5x - 3| = 14 + 2x
<=> |5x - 3| = 7 + x
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}\)
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
TH1: \(x\ge\frac{3}{5}\)
Ta có \(5x-3-2x=14\)
\(3x=17\)
\(x=\frac{17}{3}\) (tmđk)
TH1: \(x< \frac{3}{5}\)
Ta có \(-5x+3-2x=14\)
\(7x=-11\)
\(x=-\frac{11}{7}\) (tmđk)
Vậy \(x\in\left\{\frac{17}{3};-\frac{11}{7}\right\}\)