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`a)C=-|x+2/5|`
`|x+2/5|>=0`
`<=>-|x+2/5|<=0`
`<=>C<=0`
Dấu "=" `<=>x=-2/5`
`b)D=5/17-|3x-2|`
`|3x-2|>=0`
`<=>5/17-|3x-2|<=5/17`
`<=>D<=5/17`
Dấu "=" `<=>x=2/3`.
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
Giải:
a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{11}\)
\(2x=\dfrac{16}{11}-\dfrac{7}{2}\)
\(2x=\dfrac{-45}{22}\)
\(x=\dfrac{-45}{22}:2\)
\(x=\dfrac{-45}{44}\)
b) \(3-\left(17-x\right)=-12\)
\(3-17+x=-12\)
\(x=-12-3+17\)
\(x=2\)
c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}:\dfrac{2}{3}\)
\(x=\dfrac{-3}{5}\)
d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)
Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\)
e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\)
\(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\)
\(-0,6x=\dfrac{-7}{18}\)
\(x=\dfrac{-7}{18}:-0.6\)
\(x=\dfrac{35}{54}\)
f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\)
\(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\)
\(\dfrac{3}{5}.x=\dfrac{13}{9}\)
\(x=\dfrac{13}{9}:\dfrac{3}{5}\)
\(x=\dfrac{65}{27}\)
Chúc bạn học tốt!
\(C=-\left|x+\dfrac{2}{5}\right|\le0\forall x\)
\("="\Leftrightarrow x=\dfrac{-2}{5}\\ C_{max}=0\)
\(D=\dfrac{5}{17}-\left|3x-2\right|\le\dfrac{5}{17}\)
\("="\Leftrightarrow x=\dfrac{2}{3}\\ D_{max}=\dfrac{5}{17}\)