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a) 6 . x - 15 = 15
6 . x = 15 + 15
6 . x = 30
x = 30 : 6
x = 5
b) 2 . x + 3 = 9
2 . x = 9 - 3
2 . x = 6
x = 6 : 2
x = 3
c) 17 - 6 : x = 15
6 : x = 17 - 15
6 : x = 2
x = 6 : 2
x = 3
d) x : 3 + 12 = 14
x : 3 = 14 - 12
x : 3 = 2
x = 2 . 3
x = 6
a) 6 . x -15 = 15
⇒ 6 .x = 15 + (-15)
⇒ 6 .x = 0
⇔ x = 0
Vậy x = 0.
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{x}{12}=\dfrac{6}{9}\)
\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)
b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{2}{7}\)
c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)
\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)
\(\Rightarrow2x+\dfrac{12}{5}=x\)
\(\Rightarrow2x-x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}\)
\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(-x-6\right).\left(x+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-6=0\\x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-9\end{cases}}}\)
\(b,-2.\left(x+3\right)-15=3.4+2x+7\)
\(-2x-6-15=12+2x+7\)
\(-2x-21=19+2x\)
\(-2x-2x=19+21\)
\(-4x=40\)
\(x=-10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) |2x – 5| –7 = 22
| 2x -5 | = 22+7
| 2x -5 | = 29
TH1: 2x-5 = -29
2x = -29+5
2x= -24
x= -24:2
x= -12
TH2: 2x -5 =29
2x = 29+5
2x= 34
x= 34:2
x= 17
Vậy...
a)-42
b)1
c-6
Dấu trừ trước mấy số mình viết là dấu âm nha!!
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