16+3 (x - 7)= 43
28-2 (x + 5)= 14
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a.\(\left(-12\right)x-14=-2\)
\(\left(-12\right)x=-2+14\)
\(\left(-12\right)x=12\)
\(x=12:\left(-12\right)\)
\(x=-1\)
\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)
\(\left(-8\right)x=35-3\)
\(\left(-8\right)x=32\)
\(x=32:\left(-8\right)\)
\(x=-4\)
\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)
\(\left(-9\right)x+3=14+16\)
\(\left(-9\right)x+3=30\)
\(\left(-9\right)x=30-3\)
\(\left(-9\right)x=27\)
\(x=27:\left(-9\right)\)
\(x=-3\)
Bài 1:
1: =-5/24+16/27+3/4
=-5/24+18/24+16/27
=13/24+16/27
=117/216+128/216=245/216
2: =-1/3+1/3+6/7=6/7
3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)
4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)
Lời giải:
\(\frac{4}{9}\times \frac{3}{7}\times \frac{7}{4}=\frac{1}{3}\)
\(\frac{6}{5}\times \frac{4}{5}\times \frac{25}{16}=\frac{3}{2}\)
\(\frac{7}{8}\times \frac{16}{9}\times \frac{3}{14}=\frac{1}{3}\)
200-2(x+6)=14
=> 2(x+6)=200-14
=> 2(x+6)=186
=> x+6=186:2
=> x+6=93
=> x=93-6
=> x=87
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
a: =>x-5/14=6/14-1/14=5/14
=>x=10/14=5/7
b; =>2/9:x=3/6+4/6=7/6
=>x=2/9:7/6=2/9*6/7=4/21
c: =>32:x=8
=>x=4
`@` `\text {Answer}`
`\downarrow`
`a,`
`x - 5/14 = 3/7 - 1/14`
`x - 5/14 = 5/14`
`=> x = 5/14 + 5/14`
`=> x = 5/7`
Vậy, `x = 5/7`
`b,`
`2/9 \div x = 1/2 + 2/3`
`2/9 \div x = 7/6`
`x = 2/9 \div 7/6`
`x = 4/21`
Vậy, `x = 4/21`
`c,`
\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)
`6/(32 \div x) = 3/4`
`32 \div x = 6 \div 3/4`
`32 \div x = 8`
` x = 32 \div 8`
`x = 4`
Vậy, `x = 4`
16+3(x-7)=43
3(x-7)=43-16
a,3(x-7)=27
x-7=27:3
x-7=9
x=9+7
x=16
b,28-2(x+5)=14
2(x+5)=28-14
2(x+5)=14
x+5=14:2
x+5=7
x=7-5
x=2
a) <-->\(3\left(x-7\right)=43-16\)
<--> \(3x-21=27\)
<--> \(3x=27+21\)
<--> \(3x=48\)
<--> \(x=16\)
b) <--> \(-2\left(x+5\right)=14-28\)
<--> \(-2\left(x+5\right)=-14\)
<--> \(-2x-10=-14\)
<--> \(-2x=-14+10\)
<--> \(-2x=-4\)
<--> \(x=2\)