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a: 3-2|4x-5|=2/6
=>2|4x-5|=3-1/3=8/3
=>|4x-5|=4/3
=>4x-5=4/3 hoặc 4x-5=-4/3
=>4x=19/3 hoặc 4x=11/3
=>x=19/12 hoặc x=11/12
c: (7-3x)(2x+1)=0
=>2x+1=0 hoặc -3x+7=0
=>x=-1/2 hoặc x=-7/3
d: 2x(5-3x)>0
=>x(3x-5)<0
=>0<x<5/3
a) * Nếu 4x - 5 \(\ge\) 0 thì x \(\ge\) \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(4x-5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(-8x=-3-10+\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{19}{12}\) (t/m)
* Nếu 4x - 5 < 0 thì x < \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(-4x+5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(3+8x-10=\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{11}{12}\) (t/m)
b) Không hiểu đề :v
c) \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d) \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow0< x< \dfrac{5}{3}\)
e) \(\left(4-2x\right)\left(5x+3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-2x< 0\\5x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-2x>0\\5x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\dfrac{3}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
Loại TH1, nhận TH2
Vậy \(-\dfrac{3}{5}< x< 2\)
g) \(\left|3x+1\right|+\left|1-3x\right|=0\) (1)
* Nếu x < \(\dfrac{-1}{3}\)
PT (1) \(\Leftrightarrow-3x-1-1+3x=0\)
0x - 2 = 0
0x = 2 \(\Rightarrow\) PT vô nghiệm
* Nếu \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1-1+3x=0\)
6x = 0
x = 0 (t/m)
* Nếu x > \(\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1+1-3x=0\)
0x + 2 = 0
0x = -2
PT vô nghiệm.
Vậy x = 0
a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)
\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)
\(\Rightarrow\left|4x-5\right|=\dfrac{4}{3}\)
+) Xét \(x\ge\dfrac{5}{4}\) có:
\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )
+) Xét \(x< \dfrac{5}{4}\) có:
\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )
Vậy...
b, tương tự
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
d, \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )
Vậy \(0< x< \dfrac{3}{5}\)
e, tương tự
g, \(\left|3x+1\right|+\left|1-3x\right|=0\)
\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)
+) Xét \(x\ge\dfrac{1}{3}\) có:
\(3x+1+3x-1=0\)
\(\Rightarrow6x=0\)
\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:
\(3x+1+1-3x=0\)
\(\Rightarrow2=0\) ( vô lí )
+) Xét \(x< \dfrac{-1}{3}\) có:
\(-3x-1+1-3x=0\)
\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )
Vậy ko có giá trị x thỏa mãn đề bài
a: =>|5/4x-7/2|=|5/8x+3/5|
=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5
=>5/8x=41/10 hoặc 15/8x=29/10
=>x=164/25 hoặc x=116/75
b: =>3:|x/4-2/3|=6-21/5=9/5
=>|1/4x-2/3|=5/3
=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3
=>1/4x=7/3 hoặc 1/4x=-1
=>x=28/3 hoặc x=-4
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)
e: =>|2x-7|=2x-7
=>2x-7>=0
=>x>=7/2
Bài 7 :
\(\frac{1}{4}-\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2=\frac{1}{4}-0\)
\(\left(2x-1\right)^2=\frac{1}{4}\)
\(\left(2x-1\right)^2=\left(\frac{1}{2}\right)^2\)
TH1:\(\Rightarrow2x-1=\frac{1}{2}\)
\(2x=\frac{1}{2}+1\)
\(2x=\frac{3}{2}\)
\(x=\frac{3}{4}\)
TH2:\(\Rightarrow2x-1=-\frac{1}{2}\)
\(2x=-\frac{1}{2}+1\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{4}\)
Vậy x \(\in\left\{\frac{1}{4};\frac{3}{4}\right\}\)
Bài 6 :
\(3^{x+1}=81\)
\(3^{x+1}=3^4\)
\(x+1=4\)
\(\Rightarrow x=3\)
Vậy x = 3
Câu 1: \(\frac{5}{12}-\left(\frac{7}{3}-x\right)=\frac{8}{9}\)
=> \(\left(\frac{7}{3}-x\right)=\frac{5}{12}-\frac{8}{9}\)
=> \(\frac{7}{3}-x=-\frac{17}{36}\)
=> \(x=\frac{7}{3}-\left(-\frac{17}{36}\right)\)
=> \(x=\frac{101}{36}\)
Vậy \(x=\frac{101}{36}.\)
Câu 3: \(\frac{5}{6}+\frac{1}{6}:x=\frac{4}{5}\)
=> \(\frac{1}{6}:x=\frac{4}{5}-\frac{5}{6}\)
=> \(\frac{1}{6}:x=-\frac{1}{30}\)
=> \(x=\frac{1}{6}:\left(-\frac{1}{30}\right)\)
=> \(x=-5\)
Vậy \(x=-5.\)
Còn mấy câu sau bạn làm tương tự nhé.
Chúc bạn học tốt!
1) \(\frac{5}{12}\)- (\(\frac{7}{3}\)- x) = \(\frac{8}{9}\)
\(\frac{7}{3}\)- x = \(\frac{-17}{36}\)
=> x = \(\frac{101}{36}\)
3) \(\frac{5}{6}\)+ \(\frac{1}{6}\): x = \(\frac{4}{5}\)
\(\frac{1}{6}\): x = \(\frac{-1}{30}\)
=> x = -5
4) \(\frac{6}{7}\)+ \(\frac{1}{7}\). x = \(\frac{3}{4}\)
\(\frac{1}{7}\). x = \(\frac{-3}{28}\)
=> x = \(\frac{-3}{4}\)
5) \(\frac{1}{3}\)+ \(\frac{2}{3}\): x = \(\frac{5}{9}\)
\(\frac{2}{3}\): x = \(\frac{2}{9}\)
=> x = 3
Bạn xem lại câu 2 đi nha. Mk ko hiểu đề bài câu 2 cho lắm