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\(a,\dfrac{3}{8}=\dfrac{6}{x}\\ \Rightarrow x=6:\dfrac{3}{8}\\ \Rightarrow x=16\\ b,\dfrac{1}{9}=\dfrac{x}{27}\\ \Rightarrow x=\dfrac{1}{9}.27\\ \Rightarrow x=3\\ c,\dfrac{4}{x}=\dfrac{8}{6}\\ \Rightarrow x=4:\dfrac{4}{3}\\ \Rightarrow x=3\\ d,\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Rightarrow3x+6=-4x+20\\ \Rightarrow3x+6+4x-20=0\\ \Rightarrow7x-14=0\\ \Rightarrow7x=14\\ \Rightarrow x=2\)
a: =>6/x=3/8
hay x=16
b: =>x/27=1/9
nên x=3
c: =>4/x=4/3
nên x=3
d: =>3/x-5=-4/x+2
=>3x+2=-4x+20
=>7x=18
hay x=18/7
b, x+x-1+x-2+...+x-50=255
=>x+x+x+...+x-1-2-3-4-...-50=255
=>x.51-(1+2+3+...+50)=255
=>x.51-[(50+1).(50-1+1):2]=255
=>x.51-[51.50:2]=255
=>x.51-1275=255
=>x.51=255+1275=1530
=>x=1530:51=30
Vậy x = 30
a, x + 1/2x - 25%x = 10
=> x + 1/2x - 1/4x = 10 => x.( 1/2 - 1/4 ) =10 => x.1/4 = 10 =>x = 10 : 1/4 = 40 Vậy x = 40
`#040911`
`a)`
`6 \times x - 5 = 613`
`=> 6 \times x = 613 + 5`
`=> 6 \times x = 618`
`=> x = 618 \div 6`
`=> x = 103`
Vậy, `x = 103`
`b)`
`12 \times x + 3 \times x = 30`
`=> x \times (12 + 3) = 30`
`=> x \times 15 = 30`
`=> x = 30 \div 15`
`=> x = 2`
Vậy, `x = 2`
`c)`
`125 - 25 \times (x - 1) = 100`
`=> 25 \times (x - 1) = 125 - 100`
`=> 25 \times (x - 1) = 25`
`=> x - 1 = 25 \div 25`
`=> x - 1 = 1`
`=> x = 1 + 1`
`=> x = 2`
Vậy, `x = 2`
`d)`
`(x - 2) \times (9x - 4) = 0?`
`=>`
TH1: `x - 2 = 0`
`=> x = 0 + 2`
`=> x = 2`
TH2: `9x - 4 = 0`
`=> 9x = 4`
`=> x = 4/9`
Vậy, `x \in {2; 4/9}.`
\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
_Mấy bác cứ thik đăng nhiều :v , nhìn mak ko muốn lm . E lm bài 1 thôi :v còn các bài còn lại bác tự lm ( nó cx dễ thôi mà ) _
Bài 1 :
\(a) 2x-13=25+6x\)
\(\Rightarrow2x-6x=25+13\)
\(\Rightarrow-4x=38\)
\(\Rightarrow x=-\dfrac{19}{2}\)
Vậy .......
\(b) 12-x=3x+6\)
\(\Rightarrow-x-3x=6-12\)
\(\Rightarrow-4x=-6\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy .....
\(c) 40-(25-2x)=x\)
\(\Rightarrow40-25+2x=x\)
\(\Rightarrow15+2x=x\)
\(\Rightarrow2x-x=-15\)
\(\Rightarrow x=-15\)
Vậy ......
\(d) |x-3|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
Vậy ....
e) \(|x-3|+(x+2)+(x+1)=12\)
\(\Rightarrow\left|x-3\right|+x+2+x+1=12\)
\(\Rightarrow\left|x-3\right|+2x+3=12\)
\(\Rightarrow\left|x-13\right|+2x=9\)
\(\Rightarrow\left[{}\begin{matrix}x-3+2x=9\\-\left(x-3\right)+2x=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\) \(( x = 6 \) ko thỏa mãn điều kiện )
Vậy ....
