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a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)
=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)
=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)
=>2x=4
=>x=4/2=2
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)
=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)
=>\(\dfrac{7}{3}x=2,4\)
=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)
f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)
=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)
=>\(x=-\dfrac{2}{15}\cdot45=-6\)
g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)
=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)
=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)
=>x=10/2=5
h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)
=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)
=>4x=7
=>\(x=\dfrac{7}{4}\)
1,
( 2x-5) + 17=6
\(2x-5=6-17\)
\(2x-5=-11\)
\(2x=-11+5\)
\(2x=-6\)
\(x=-6:2\)
\(x=-3\)
Vậy \(x=3\)
2,
10-2(4-3x)=-4
\(-2\left(4-3x\right)=-4-10\)
\(-2\left(4-3x\right)=-14\)
\(4-3x=-14:\left(-2\right)\)
\(4-3x=7\)
\(-3x=7-4\)
\(-3x=3\)
\(x=3:\left(-3\right)\)
\(x=-1\)
Vậy \(x=-1\)
3,
-12+3(-x+7)=-18
\(3\left(-x+7\right)=-18+12\)
\(3\left(-x+7\right)=-6\)
\(-x+7=-6:3\)
\(-x+7=-2\)
\(-x=-2-7\)
\(-x=-9\)
\(x=9\)
\(\text{Vậy }x=9\)
4,
24:(3x -2)=-3
\(3x-2=24:\left(-3\right)\)
\(3x-2=-8\)
\(3x=-8+2\)
\(3x=-6\)
\(x=-6:3\)
\(x=-2\)
\(\text{Vậy }x=-2\)
5,
-45:5.(-3-2x)=3
\(5\left(-3-2x\right)=-45:3\)
\(5\left(-3-2x\right)=-15\)
\(-3-2x=-15:5\)
\(-3-2x=-3\)
\(-2x=-3+3\)
\(-2x=0\)
\(x=0\)
Vậy \(x=0\)
6,
x.(x+7)= 0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
\(\text{Vậy}\left\{{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
7,
( x+ 12 ) .( x-3)=0
\(\Rightarrow\left\{{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
8,
(-x+5).(3-x) =0
\(\Rightarrow\left\{{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-x=-5\\-x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy \(x=5\text{ hoặc }x=3\)
9, x.( 2+x).(7-x)=0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
10,
(x-1).(x+2).(-x-3)=0
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\\-x=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
(x - 2)(2x - 6) = 0
=> x - 2 = 0 hoac 2x - 6 = 0
=> x = 2 hoac x = 3
vay_
(3x + 9)(1 - 3x) = 0
=> 3x + 9 = 0 hoac 1 - 3x = 0
=> x = -3 hoac x = 1/3
vay_
|2 - x| + 2 = x
=> |2 - x| = x - 2
=> 2 - x = x - 2 hoac 2 - x = 2 - x
=> -2x = -4 hoac x thuoc tap hop rong
=> x = 2
\(a,\left(x-2\right).\left(2x-6\right)=0\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(b,\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow3\left(x+3\right).\left(1-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
\(c,\left(x^2+1\right).\left(81-x^2\right)=0\Leftrightarrow\left(x^2+1\right).\left(9-x\right).\left(9+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=0\\9+x=0\end{cases}}\) ( vì \(x^2+1\ne0\forall x\)) \(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)
x.4027=5040.2
x.4027=10080
x=10080/4027
x=2.5