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\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)
\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)
b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)
c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)
d)\(-3\left(x-2\right)=2x+1\)
\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)
\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)
e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều
\(d,-3\left(x-2\right)=2x+1\)
\(< =>-3x+6=2x+1\)
\(< =>-3x-2x+6-1=0\)
\(< =>5-5x=0\)
\(< =>5\left(1-x\right)=0< =>x=1\)
\(e,\left(x-1\right)^2-4=0\)
\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
a)Vì \(-|x-3,5|\le0;\forall x\)
\(\Rightarrow0,5-|x-3,5|\le0,5-0;\forall x\)
Hay \(A\le0,5-0;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-3,5=0\)
\(\Leftrightarrow x=3,5\)
Vậy MAX A=0,5 \(\Leftrightarrow x=3,5\)
b) Vì \(-|1,4-x|\le0;\forall x\)
\(\Rightarrow-|1,4-x|-2\le0-2;\forall x\)
Hay \(B\le-2;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow1,4-x=0\)
\(\Leftrightarrow x=1,4\)
Vậy MAX B=-2 \(\Leftrightarrow x=1,4\)
A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
x | \(\frac{-2}{3}\) 3 |
3x+2 | - 0 + | + |
x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
1: \(\Leftrightarrow3\left|x-7\right|-48\left|x+2\right|=3x-8\)
TH1: x<-2
Pt sẽ là 3(7-x)-48(-x-2)=3x-8
=>21-3x+48x+96=3x-8
=>45x+117=3x-8
=>42x=-125
=>x=-125/42(nhận)
TH2: -2<=x<7
Pt sẽ là 3(7-x)-48(x+2)=3x-8
=>21-3x-48x-96=3x-8
=>-51x-75=3x-8
=>-48x=67
=>x=-67/48(nhận)
TH3: x>=7
Pt sẽ là 3(x-7)-48(x+2)=3x-8
=>3x-21-48(x+2)=3x-8
=>48(x+2)=-21+8=-13
=>x+2=-13/48
=>x=-109/48(loại)
3: =>|3x-5|=|x-7|
=>3x-5=x-7 hoặc 3x-5=-x+7
=>2x=-2 hoặc 4x=12
=>x=3 hoặc x=-1
\(\left(\dfrac{12}{7}:x-0.5\right)\left(\dfrac{2}{5}x-1\dfrac{1}{2}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{12}{7}:x-0,5=0\\\dfrac{2}{5}x-1\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{12}{7}:x=0,5\\\dfrac{2}{5}x=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{12}{7}:\dfrac{1}{2}=\dfrac{12}{7}\cdot2=\dfrac{24}{7}\\x=\dfrac{3}{2}:\dfrac{2}{5}=\dfrac{3}{2}\cdot\dfrac{5}{2}=\dfrac{15}{4}\end{matrix}\right.\)
=>12/7:x=1/5 hoặc 2/5x=5/2
=>x=12/7:1/5=60/7 hoặc x=5/2:2/5=25/4