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b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
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.
a) x + 3/5 = -1/15
=> x = -1/15 - 3/5
=> x = -2/3
b) x/14 = 1/7 - 3/14
=> x/14 = -1/14
=> x = -1
c) -1/2x + 1/5 = 3/10
=> -1/2x = 3/10 - 1/5
=> -1/2x = 1/5
=> x = 1/5 : (-1/2)
=> x = -2/5
d) 4/5 - (2/15 + x) = 1/15
=> 4/5 - 2/15 - x = 1/15
=> 2/3 - x = 1/15
=> x = 2/3 - 1/15
=> x = 3/5
a. Tìm x thuộc N sao cho : 2x + 1 thuộc Ư ( 2x + 10)
(2x + 10) ⋮ (2x + 1)
Ta có (2x + 10) = (2x + 1 + 9)
Mà (2x + 10) ⋮ (2x + 1)
Nên 9 ⋮ (2x + 1)
Do đó ta có (2x + 1) ∈ Ư (9) = {-1; 1; -3; 3; -9; 9}
| 2x + 1 | -1 | 1 | -3 | 3 | -9 | 9 |
| 2x | -2 | 0 | -4 | 2 | -10 | 8 |
| x | -1 | 0 | -2 | 1 | -5 | 4 |
Vậy x = {-1; 0; -2; 1; -5; 4}
b. A = 3 - 5 + 13 - 15 + 23 - 25 + ....... + 93 - 95 + 2020
A = (3 - 5) + (13 - 15) + (23 - 25) +.......+ (93 - 95) + 2020
A = (-2) + (-2) + (-2) + ......... + (-2) + 2020
Có 10 số (-2)
A = (-2) . 10 + 2020
A = (-20) + 2020
A = 2000
c. 2( x + 1) - x - 2 = (-5) - 3
2x + 2 - 1x - 2 = (-8)
2x - 1x + 2 - 2 = (-8)
2x - 1x + 0 = (-8)
2x - 1x = (-8)
(2 - 1)x = (-8)
1 . x = (-8) : 1
x = (-8)
a) x + 3/5 = -1/15
=> x = -1/15 - 3/5
=> x = -2/3
b) x/14 = 1/7 - 3/4
=> x/14 = -17/28
=> 2x/28 = -17/28
=> 2x = -17
=> x = -17/2
c) -1/2x + 1/5 = 3/10
=> -1/2x = 3/10 - 1/5
=> -1/2x = 1/10
=> x = 1/10 : (-1/2)
=> x = -1/5
d) 4/5 - (2/15 + x) = 1/15
=> 4/5 - 2/15 - x = 1/15
=> 2/3 - x = 1/15
=> x = 2/3 - 1/15
=> x = 3/5