\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
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GIẢI CHI TIẾT RA GIÚP MK VS NHÉ!!!!MK RẤT CẢM ƠN
PA
17 tháng 7 2016
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
18 tháng 7 2016
a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)
=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)
=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)
=>\(\frac{2}{3}-\frac{4}{3}x=5\)
=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)
b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)
10 tháng 6 2017
\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)
\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)
\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)
\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)
\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)
\(A=1,3\left(2\right)\)
J
10 tháng 7 2017
a) \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)
<=>\(\hept{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{4}\end{cases}}}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}}\)
LA
21 tháng 6 2017
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
21 tháng 6 2017
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
22 tháng 10 2016
a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)
LY
22 tháng 10 2016
\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)
\(x+\frac{1}{2}=x+x+3\\\)
\(x+\frac{1}{2}=x+\left(x+3\right)\)
\(\Rightarrow\frac{1}{2}=x+3\)
\(\Rightarrow x=\frac{1}{2}-3\)
\(\Rightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
\(Ta\) \(có\)
\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)
\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(3x+2=4x\)
\(3x+2=3x+x\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
HT
2 tháng 6 2016
Ta có \(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5.\)
\(\Rightarrow\frac{2}{3}-\frac{1}{3}.x+\frac{1}{3}.\frac{3}{2}-\frac{1}{2}.2x-\frac{1}{2}=5\)
\(\Rightarrow\frac{2}{3}-\frac{x}{3}+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Rightarrow\frac{4}{6}-\frac{2x}{6}+\frac{3}{6}-\frac{6x}{6}-\frac{3}{6}=\frac{30}{6}\)
\(\Rightarrow4-2x+3-6x-3=30\)
\(\Rightarrow4-8x=30\)
\(\Rightarrow-8x=26\)
\(\Rightarrow x=\frac{26}{-8}=-\frac{13}{4}\)
Vậy \(x=-\frac{13}{4}\)
\(\left(x+\frac{1}{2}\right)\times\left(\frac{2}{3}-2x\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(\frac{2}{3}-2x=0\)
\(2x=\frac{2}{3}\)
\(x=\frac{2}{3}\div2\)
\(x=\frac{2}{3}\times\frac{1}{2}\)
\(x=\frac{1}{3}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{1}{3}\)