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x + 12 =(-5)-x
x + x = -5 - 12
2x = -17
\(x=-\frac{17}{2}\)
x + 5 = 10 - x
x + x = 10 - 5
2x = 5
\(x=\frac{5}{2}\)
12-x=x+1
12 - 1 = x+x
11=2x
x=\(\frac{11}{2}\)
14+4x=3x-20
4x-3x=-20-14
x=-34
a, \(\frac{x}{5}=\frac{-12}{20}\)
\(x=\frac{5.\left(-20\right)}{20}\)
\(x=-5\)
b, 3x - 35 = -45 - 2x
3x + 2x = -45 + 35
5x = -10
x = (-10) : 5
x = -2
c, \(\frac{-1}{6}-x=\frac{5}{12}+\frac{-2}{3}\)
\(\frac{-1}{6}-x=\frac{-1}{4}\)
\(-x=\frac{-1}{6}-\frac{-1}{4}\)
\(-x=\frac{1}{12}\)
\(x=\frac{-1}{12}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a) 3x - 10 = 2x + 13
<=> 3x - 2x = 13+10
<=> x = 23
Vậy x = 23
b) x + 12 = -5 - x
<=> x + x = -5 - 12
<=> 2x = -17
<=> x = -17 : 2
<=> x = -8,5
Vậy x = -8,5
c) x + 5 = 10 - x
<=> x + x = 10 - 5
<=> 2x = 5
<=> x = 5 : 2
<=> x = 2,5
Vậy x = 2,5
d) 6x + 2 ^ 3 = 2x - 12
<=> 6x + 8 = 2x - 12
<=> 6x - 2x = -12 - 8
<=> 4x = -20
<=> x = -20 : 4
<=> x = -5
Vậy x = -5
e) 12 - x = x + 1
<=> -x - x = 1 - 12
<=> -2x = -11
<=> x = -11 : (-2)
<=> x = 5,5
Vậy x= 5,5
f) 14 - 4x = 3x + 20
<=> -4x - 3x = 20 - 14
<=> -7x = 6
<=> x = 6 : (-7)
<=> x = \(-\frac{6}{7}\)
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.
1) \(-12+3-\left|x+2\right|=7-12\)
\(-12+3+12-7=\left|x+2\right|\)
\(\left|x+2\right|=-4\)( vô lý )
Vậy \(x\in\varnothing\)
2) \(\left|x-20\right|-\left(8-7\right)=5-\left|-7\right|\)
\(\left|x-20\right|-1=5-7\)
\(\left|x-20\right|-1=-2\)
\(\left|x-20\right|=-1\)( vô lý )
Vậy \(x\in\varnothing\)
a ) \(15:\left(x+2\right)=3\)
\(\left(x+2\right)=15:3\)
\(x+2=5\)
\(x=5-2\)
\(x=3\)
Vậy \(x=3\)
b) \(20:\left(1+x\right)=2\)
\(\left(1+x\right)=20:2\)
\(1+x=10\)
\(x=10-1\)
\(x=9\)
Vậy \(x=9\)
c) \(240:\left(x-5\right)=2^2.5^2=20\)
\(240:\left(x-5\right)=4.25=20\)
\(240:\left(x-5\right)=100=20\)
\(\Rightarrow\orbr{\begin{cases}240:\left(x-5\right)=100\\240:\left(x-5\right)=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)=240:100\\\left(x-5\right)=240:20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-5=2,4\\x-5=12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2,4+5\\x=12+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7,4\\x=17\end{cases}}\)
Vậy \(x=7,4\)hoặc \(x=17\)
d) \(96-3\left(x+1\right)=12\)
\(3\left(x+1\right)=96-12\)
\(3\left(x+1\right)=84\)
\(\left(x+1\right)=84:3\)
\(x+1=28\)
\(x=28-1\)
\(x=27\)
Vậy \(x=27\)
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\(\dfrac{5}{x+1}=\dfrac{20}{-12}\\ \Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\\ \Rightarrow x+1=-3\\ \Rightarrow x=-4\)
\(\dfrac{5}{x+1}=\dfrac{20}{-12}\)
\(\Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\)
\(\Rightarrow x+1=-3\)
\(\Leftrightarrow x=-4\)