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a, (x-35)-120=0
x-35=120+0
x-35=120
x=120+35
x=155
b,7x-8=713
7x=713+8
7x=721
x=721:7
x=103
c,4x:17=0
4x=0.17
4x=0
x=0:4
x=0
d,75+(131-x)=205
131-x=205-75
131-x=130
x=131-130
x=1
e,2x-36=4^6:4^3
2x-36=4^6-3
2x-36=4^3
2x-36=48
2x=48-36
2x=12
x=12:2
x=6
f, 2011^2.2011^x=2011^6
2011^x=2011^6: 2011^2
2011^x=2011^6-2
2011^x=2011^4
x=4
bn
\(\text{a/ ( x-35 ) - 120 = 0}\)
\(\Rightarrow\left(x-35\right)=120\)
\(\Rightarrow x=120+35\)
\(x=155\)
\(\text{b/ 7x - 8 = 713}\)
\(\Rightarrow7x=713+8=721\)
\(x=721:7=103\)
\(\text{c/ 4x : 17 = 0}\)
\(4x=0\)
\(x=0\)
\(\text{d/ 75+(131-x) = 205}\)
\(131-x=205-75=130\)
\(x=131-130=1\)
\(e.2x-36=4^6:4^3\)
\(2x-36=4^3=64\)
\(2x=64+36=100\)
\(x=100:2=50\)
\(f.2011^2.2011^x=2011^6\)
\(\text{Ta có công thức }:x^n.x^m=x^{n+m}\)
\(\Rightarrow x=6-2=4\)
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a: \(=\dfrac{-3}{5}\cdot\dfrac{5}{7}+\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{6}{7}\)
\(=\dfrac{-3}{5}\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)=\dfrac{-3}{5}\cdot2=-\dfrac{6}{5}\)
b: \(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)
c: =>1/2x+1+3/8=7/16
=>1/2x=-15/16
=>x=-15/8
d: =>5/2x-1/3=1/6*(-9)/2=-9/12=-3/4
=>5/2x=-3/4+1/3=-9/12+4/12=-5/12
=>x=-1/6
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1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
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1. a) 4.415.8.25.125
= (4.25). (8.125).415
= 100.1000.415
= 100000.415
= 41500000
b) 2.31.12+4.42.6+8.27.3
= (2.31.12)+(4.42.6)+(8.27.3)
= (2.12).31+(4.6).42+(8.3).27
= 24.31+24.42+24.27
= 24 (31+42+27)
= 24.100
= 2400
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c)(x-4).(2x+6)=0
=>(x-4)=0 hoặc (2x+6)=0
với x-4 = 0
x =0+4
x =4
với 2x+6=0
2x =0-6
2x =-6
x =-6:2
x =-3
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A) 2x - 22 = 32
=> 2x - 4 = 9
=> 2x = 9 + 4
=> 2x = 13
=> x = 13/2
B) 32 - 3x = 3
=> 9 - 3x = 3
=> 3x = 6
=> x = 2
C) Mình không biết đề bài là gì?
a) (2x-2)2 = 32
=> 2x-2=3
=> 2x = 5
=> x = 5/2
b) 32 - 3x = 3
=> 3.3 - 3x = 3
=> 3.(3-x)=3
=> 3-x = 1
=> x= 1+3 = 4
c) 32 . (-5) - 5.23
= 9.(-5) - 5.8
= -45 - 40 = -85
a) 7x - 8 = 713
=> 7x = 713 + 8
=> 7x = 721
=> x = 721 : 7
=> x = 103
b) 23 + 3x= 5 6 : 53
=> 23 + 3x = 5 3
=> 23 + 3x = 125
=> 3x = 125 - 23
=> 3x = 102
=> x = 102 : 3
=> x = 34
c) ( x-3) 4 = 16
=> ( x - 3 ) 4 = 2 4
=> x - 3 = 2
=> x = 2 + 3
=> x = 5
d) - 3 < |x|<1
Do | x | luôn lớn hơn hoặc bằng 0 và nhỏ hơn 1
=> | x | = 0
=> x = 0
Tui chỉ làm kết quá nha
a) x=103
b)x=34
c)x=94
d) x\(\in\)\([\)-2,-1,0\(]\)