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a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=-\frac{21}{3}\)
\(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)
\(2x=-\frac{1}{21}\)
\(x=\frac{-1}{42}\)
b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)
c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)
a) 1/4 + 1/3 : 2x = -5
=> 1/3 : 2x = -5 - 1/4
=> 1/3 : 2x = -21/4
=> 2x = 1/3 : (-21/4) = -4/63
=> x = -4/63 : 2 = -2/63
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{1}{3}:\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{-4}{63}:2\)
\(x=\frac{-2}{63}\)
\(\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)
\(\Rightarrow2x=\frac{-4}{63}\)
\(\Rightarrow x=\frac{-2}{63}\)
\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)
\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)
Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)
Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)
Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)
\(\left(3x-\frac{2}{4}\right)\cdot\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}\)
còn lại tương tự bài trên
=> 2x - 5 = 0 => 2x = 5 => x = 5/2
hoặc 3/2x + 9 = 0 => 3/2x = -9 => x = -6
hoặc 0,3x - 12 = 0 => 0,3x = 12 => x = 40
Vậy x = 5/2 ; x = -6 ; x = 40
\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)
\(\Leftrightarrow2x-5=0\left(h\right)\frac{3}{2}x+9=0\left(h\right)0,3x-12=0\)
\(\Leftrightarrow x=\frac{5}{2}\left(h\right)x=-6\left(h\right)x=40\)
Vậy ......
P/s (h) là hoặc
\(\Leftrightarrow2x-5=0\Rightarrow x=\frac{5}{2}\)
họcc\(\frac{3}{2}x+9=0\Rightarrow x=-6\)
hoăc\(0,3x-12=0\Rightarrow x=40\)
a: =>x-2,7=0,3 hoặc x-2,7=-0,3
=>x=3 hoặc x=2,4
b: =>|x+1,5|=2,4
=>x+1,5=2,4 hoặc x+1,5=-2,4
=>x=-3,9 hoặc x=0,9
c: =>|2x-3|=1/6
=>2x-3=1/6 hoặc 2x-3=-1/6
=>2x=19/6 hoặc 2x=17/6
=>x=17/12 hoặc x=19/12
d: =>3|2x-5|=7,5+0,8=8,3
=>|2x-5|=83/30
=>2x-5=83/30 hoặc 2x-5=-83/30
=>2x=233/30 hoặc 2x=67/30
=>x=233/60 hoặc x=67/60
e: =>x-y=0 và y+9/25=0
=>x=y=-9/25
a/ \(\left(3x-\dfrac{2}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ................
b/ \(\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)
Vậy ..
\(a)\left(3x-\dfrac{2}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
\(b)\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)
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