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1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
| 2x+1 | 1 | 3 | 7 | 21 |
| x | 0 | 1 | 3 | 10 |
| TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
a, 7\(x\).(2\(x\) + 10) = 0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\){-5; 0}
b, - 9\(x\) : (2\(x\) - 10) = 0
- 9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2023; 2024}
a)\(\left(2x-1\right)^5=32\)
\(\Rightarrow\left(2x-1\right)^5=2^5\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)
b)\(\left(2x+1\right)^2=169\)
\(\Rightarrow\left(2x+1\right)^2=13^2=\left(-13\right)^2\)
\(\Rightarrow2x+1=13\) hoặc \(2x+1=-13\)
\(\Rightarrow2x=12\) hoặc \(2x=-14\)
\(\Rightarrow x=6\) hoặc \(x=-7\)
c)\(x^{100}=x\)
\(\Rightarrow x^{100}-x=0\)
\(\Rightarrow x\left(x^{99}-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^{99}-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^{99}=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
a, (2x-1)^5=32
(2x-1)^5=2^5
2x-1=2
2x=2+1
2x=3
x=3:2
x=1,5
Vậy x=1,5
b, (2x+1)^2=169
(2x+1)^2=13^2
2x+1=13
2x=13-1
2x=12
x=12:2
x=6
Vậy x=6
c, x^100=x
=>x=0
x=1
\(a,\left(x-36\right):\left(2\cdot3^2\right)=2^3\cdot3\\ \Leftrightarrow x-36=432\\ x=468\\ b,2^x=32\\ \Leftrightarrow x=5\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x^3=3^3\\ \Leftrightarrow x=3\\ d,1579+\left(625-x\right)=2023\\ \Leftrightarrow x=1579+625-2023\\ \Leftrightarrow x=181\)
A. \(\left(x-36\right):\left(2.3^2\right)=2^3.3\)
\(\left(x-36\right):\left(2.9\right)=8.3\)
\(\left(x-36\right):18=24\)
\(x-36=24.18\)
\(x-36=432\)
\(x=432+36\)
\(x=468\)
B. \(2^x=32\)
\(2^x=2^5\)
\(x=5\)
C. \(x^3=27\)
\(x^3=3^3\)
\(x=3\)
D. \(1579+\left(625-x\right)=2023\)
\(625-x=2023-1579\)
\(625-x=444\)
\(x=625-444\)
\(x=181\)
Lời giải:
Ta có:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2023.\frac{2024}{2023}$
$\Leftrightarrow 1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=2024$
$\Leftrightarrow 3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}=2024$
$\Leftrightarrow 3+B=2024$
$\Leftrightarrow B=2021$
a) \(\left(x-2024\right)^{2023}=1\)
\(\Rightarrow\left(x-2024\right)^{2023}=1^{2023}\)
\(\Rightarrow x-2024=1\)
\(\Rightarrow x=2025\)
b) \(\left(2x-1\right)^5=32\)
\(\Rightarrow\left(2x-1\right)^5=2^5\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
c) \(5< 2^x< 100\)
\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)
\(\Rightarrow2< x< 7\)
b , x = 3/2 a và b mình ko biết