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2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
c) pt<=> 2x+1+3x-4 =5 <=> x = 8/5 hoặc 2x+1+3x-4 = -5 <=> x = -2/5 hoặc -2x-1 +3x-4 = 5 <=> x = 10 hoặc 2x+1 -3x +4 = 5 <=> x = 0
vậy x = { -2/5 ; 0 ;8/5 ;10 }
d) pt <=> 2x + 4/5 = x-3/2 <=> x = -23/10 hoặc 2x +4/5 = -x+3/2 <=> x = 7/30
vậy x = { -23/10 ; 7/30}
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
a) \(4:\left(x-1\right)=\left(x-1\right):9\)
\(\frac{4}{x-1}=\frac{x-1}{9}\)
\(\left(x-1\right)^2=36\)
\(\left(x-1\right)^2=6^2\)
\(\Rightarrow x-1=6\)
\(\Rightarrow x=7\)
vậy \(x=7\)
c) \(3\frac{1}{2}:x\frac{1}{2}=5\frac{1}{3}:\frac{1}{2}.1\frac{1}{5}\)
\(\frac{7}{2}:\frac{1}{2}x=\frac{16}{3}:\frac{1}{2}.\frac{6}{5}\)
\(\frac{7}{2}:\frac{1}{2}x=\frac{64}{5}\)
\(\frac{1}{2}x=\frac{7}{2}:\frac{64}{5}\)
\(\frac{1}{2}x=\frac{35}{128}\)
\(x=\frac{35}{128}:\frac{1}{2}\)
\(x=\frac{35}{64}\)
d) \(\left|2x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
f) \(\left(2x-\frac{1}{2}\right)^2=\left(1-3x\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=1-3x\)
\(\Rightarrow2x+3x=1+\frac{1}{2}\)
\(\Rightarrow5x=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{10}\)
\(\left(3-\frac{1}{2}:x\right)^2=14\)
\(\left(3-\frac{1}{2x}\right)^2=14\)
\(\frac{1}{4x^2}-2.\frac{1}{2x}.3+9=14\)
\(\frac{1}{4x^2}-\frac{3}{x}=5\)
\(\left(\frac{1}{4x}-3\right):x=5\)
a) \(\frac{3}{4}-\left|2x+1\right|=\frac{7}{8}\)
\(\left|2x+1\right|=\frac{3}{4}-\frac{7}{8}\)
\(\left|2x+1\right|=-\frac{1}{8}\)
\(\Rightarrow x\in\varnothing\)
b) \(2.\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{4}\)
TH1: 2x - 3 = 1/4
...
TH2: 2x -3 = -1/4
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rùi bn tự lm típ nhé! câu c dựa vào phần a;b là lm đk
d)\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=1,6\)
...
c) TH1 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(3x-4\right)=5x-3=5\)
\(\Rightarrow x=\frac{8}{5}\)
TH2 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(-3x+4\right)=-x+5=5\)
\(\Rightarrow x=0\)
d) \(\Rightarrow\left|2x+\frac{4}{5}\right|-\left|x-\frac{3}{2}\right|=0\)
Tương tự xét 2 trường hợp như câu c. Ta sẽ tìm được x
c ) \(\left|2x+1\right|+\left|3x-4\right|=5\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left|2x+1\right|=5\\\left|3x-4\right|=5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\\3x-4=5\\3x-4=-5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2\\-3\\3\\-\frac{1}{3}\end{array}\right.\)