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a: =>5x-2=0 hoặc 2x+1/3=0
=>x=-1/6 hoặc x=2/5
b: Đặt x/2=y/3=k
=>x=2k; y=3k
xy=54
=>6k^2=54
=>k^2=9
=>k=3 hoặc k=-3
TH1: k=3
=>x=6; y=9
TH2: k=-3
=>x=-6; y=-9
c: =>5050x=-213
=>x=-213/5050
a) -3x-2=0
=>-3x=2
=>3x=-2
=>x=\(\frac{-2}{3}\)
b)Biểu thức \(\frac{3-5x}{x+1}\)=0 \(\Leftrightarrow\)3-5x=0
=>5x=3
=>x=\(\frac{3}{5}\)
c)[2x+3] và [-3x-1] là các số \(\ge\)0
=>2x+3+(-3x-1)=0
=>2x+3-3x-1=0
-x+2=0
=>-x=-2
x=2
a, -3x-2=0
-3x=2
x=-2/3
b, (3-5x)/(x+1)=0
3-5x=0
-5x=-3
x=3/5
c,x=2
a) ( 5x + 3) - ( x -1 ) = 0
\(\Leftrightarrow\)5x + 3 - x +1 =0
\(\Leftrightarrow\)4x +4 = 0
\(\Leftrightarrow\)4x = -4 \(\Leftrightarrow\)x = \(\frac{-4}{4}\) =-1
b) (3x -2 ) - ( 5x + 4) = ( x - 3) - ( x +5 )
\(\Leftrightarrow\)3x -2 - 5x -4 = x-3 - x -5
\(\Leftrightarrow\)3x - 5x - x + x = -3 -5 +2 +4
\(\Leftrightarrow\)-2x = -2 \(\Leftrightarrow\)x =\(\frac{-2}{-2}\)= 1
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
Tìm x để các biểu thức sau có gt bằng 0:
a) -3x - 2
b) \(\frac{3-5x}{x+1}\)
c) | 2x + 3| + |-3x - 1|
a) Ta có: \(-3x-2=0\)
\(\Leftrightarrow-3x=0+2\)
\(\Leftrightarrow-3x=2\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(x=\dfrac{-2}{3}\)
b) Ta có: \(\dfrac{3-5x}{x+1}=0\)
\(\Leftrightarrow3-5x=0\)
\(\Leftrightarrow5x=3-0\)
\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{3}{5}\)
Vậy \(x=\dfrac{3}{5}\)
c) Dễ thấy: \(\left\{{}\begin{matrix}\left|2x+3\right|\ge0\\\left|-3x-1\right|\ge0\end{matrix}\right.\)
Để \(\left|2x+3\right|+\left|-3x-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|2x+3\right|=0\\\left|-3x-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=0\\-3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-3\\-3x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{-3}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-3}{2};\dfrac{1}{-3}\right\}\)
a) \(\left(x-\frac{1}{3}\right)\left(5x+2\right)>0\)
<=> \(\left[\begin{array}{nghiempt}x-\frac{1}{3}>0\\5x+3< 0\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x-\frac{1}{3}< 0\\5x+3>0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x>\frac{1}{3}\\5x< 3\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x< \frac{1}{3}\\5x>3\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x>\frac{1}{3}\\x< \frac{3}{5}\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x< \frac{1}{3}\\x>\frac{3}{5}\end{array}\right.\)
Vậy...
a) \(\left(x-\frac{1}{3}\right)\left(5x+2\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{3}>0\\5x+2>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{3}< 0\\5x+2< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{3}\\x< -\frac{2}{5}\end{array}\right.\)
b) \(\left(5x+3\right)\left(3x-2\right)< 0\)
\(\Leftrightarrow\begin{cases}5x+3>0\\3x-2< 0\end{cases}\) hoặc \(\begin{cases}5x+3< 0\\3x-2>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>-\frac{3}{5}\\x< \frac{2}{3}\end{cases}\) hoặc \(\begin{cases}x< -\frac{3}{5}\\x>\frac{2}{5}\end{cases}\) (loại)
\(\Leftrightarrow-\frac{3}{5}< x< \frac{2}{3}\)