tìm x biết 2/3x-x+20%=3/4
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1) 14x-8x=10+5
x(14-8)=15
x6=15
x=15/6
2)5x-3x=30-15
2x=15
x=15/2
3)làm tương tự
a: =>10|x-2|=50
=>|x-2|=5
=>x-2=5 hoặc x-2=-5
=>x=7 hoặc x=-3
b: =>|3x-12|+|5x-20|=56
=>8|x-4|=56
=>|x-4|=7
=>x-4=7 hoặc x-4=-7
=>x=11 hoặc x=-3
Tách ?
`a, 70 -5.(x-3) =45`
`=> 5.(x-3)= 70-45`
`=> 5.(x-3)=25`
`=>x-3=25:5`
`=>x-3=5`
`=>x= 5+3`
`=>x=8`
______
`b,10 + 2.x = 4^5:4^3`
`=> 10 + 2.x = 4^(5-3)`
`=> 10 + 2.x =4^2=16`
`=> 2.x=16-10`
`=>2.x=6`
`=>x=6:2`
`=>x=3`
_____
`c,60-3.x-2=51`
`=> 60-3.x= 51+2`
`=> 60-3.x=53`
`=>3.x=60-53`
`=> 3.x= 7`
`=>x= 7/3`
____
`d, 4.x-20=2^5:2^3`
`=> 4.x-20=2^(5-3)`
`=> 4.x-20=2^2`
`=> 4.x= 4+20`
`=>4.x=24`
`=>x=24:4`
`=>x=6`
____
`2^x . 4=16`
`=> 2^x=16:4`
`=>2^x= 4`
`=>2^x=2^2`
`=>x=2`
____
`f, 3^x . 3=243`
`=>3^x=243:3`
`=> 3^x=81`
`=> 3^x=3^3`
`=>x=3`
_____
`g, 64. 4^x =16^8`
`=> 4^3 . 4^x=(4^2)^8`
`=> 4^3 . 4^x = 4^(16)`
`=> 4^x= 4^(16-3)`
`=>4^x=4^(13)`
`=>x=13`
_____
`2^x . 16^2 =1024`
`=> 2^x= 1024 : 16^2`
`=>2^x=4`
`=>2^x=2^2`
`=>x=2`
a: =>5(x-3)=25
=>x-3=5
=>x=8
b: =>2x=16-10=6
=>x=3
c: =>58-3x=51
=>3x=7
=>x=7/3
d: =>4x-20=4
=>4x=24
=>x=6
e: =>2^x=4
=>2^x=2^2
=>x=2
f: =>3^x=81
=>3^x=3^4
=>x=4
g: =>4^x*4^3=4^16
=>x+3=16
=>x=13
h: =>2^x=1024/256=4=2^2
=>x=2
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
\(2\left(x-1\right)+3\left(3x-2\right)=x-4\)
\(2x-2+9x-6=x-4\)
\(2x+9x-x-2-6=-4\)
\(10x-2-6=-4\)
\(10x-2=2\)
\(10x=4\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-3x-2x+2=x+20\)
\(12-5x+2=x+20\)
\(12-5x-x+2=20\)
\(12-6x+2=20\)
\(12-6x=18\)
\(6x=-6\)
\(x=-1\)
Vậy \(x=-1.\)
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(8x-9x+28+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
Vậy \(x=10\)
\(3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x-6=10\)
\(5x=16\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
2(x-1)+3(3x-2)=x-4
=>2x-2=9x-6-x+4=0
=>10x-4=0
=>x=\(\frac{2}{5}\)
Ta có: \(\left|3x+4\right|+\left|3x-1\right|=\left|3x+4\right|+\left|1-3x\right|\)
Theo bất đẳng thức: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\), ta có:
\(\left|3x+4\right|+\left|1-3x\right|\ge\left|3x+4+1-3x\right|=5\Rightarrow\left|3x+4\right|+\left|3x-1\right|\ge5\) (*)
Mặt khác:
Với mọi x ta có:
\(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+4\ge4\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}\le\dfrac{20}{4}\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}\le5\) (**)
Từ (*)(**) \(\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}=5\)
\(\Rightarrow3\left(x+1\right)^2+4=4\)
\(\Rightarrow3\left(x+1\right)^2=0\)
\(\Rightarrow\left(x+1\right)^2=0\)
\(\Rightarrow x=-1\)
ai giúp mik vs
\(\frac{2}{3}x-x+\frac{1}{5}=\frac{3}{4}\)
\(x\cdot\left(\frac{2}{3}-1\right)=\frac{3}{4}-\frac{1}{5}\)
\(x\cdot\frac{-1}{3}=\frac{11}{4}\)
Tự giải tiếp nha