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12 + ( 5 + x ) = 20 5.22 + ( x + 3 ) = 52 23 + ( x + 3 ) = 52 43 - ( x - 2 ) = 52
17 + x = 20 5.4 + x + 3 = 25 8 + x + 3 = 25 64 - x + 2 = 25
x = 20 - 17 20 + 3 + x = 25 11 + x = 25 66 - x = 25
x = 3 23 + x = 25 x = 25 - 11 x = 66 - 25
x = 25 - 23 x = 14 x = 41
x = 2
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a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
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(x+1)=2+4+6+8+...+2500
x(x+1)=(2500+2).1250:2
x(x+1)=1563750
Ta thấy: x và x+1 là 2 số liên tiếp
Mà x(x+1)=1563750=>x=1250;x+1=1251
Vậy x=1250
b) x-3=12:{390:[500-(5^3+7^2.5)]}
x-3=12:[390:(500-254.64)]
x-3=12:(390:245.35)
x-3=12:1.58
x-3=7.54
=>x=10.5
Vậy x=10.5
c) x+3^3.12=3^3.19
=>x=2.46
Vậy x=2.46
\(7\cdot4^x=112\)
\(\Rightarrow4^x=\dfrac{112}{7}\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
_____
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=\dfrac{250}{2}\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
____
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+25\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left[400:\left(500-300\right)\right]+10\)
\(=12:\left(400:200\right)+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
\(7\cdot4^x=112\)
\(\Rightarrow4^x=112:7\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
\(---\)
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=250:2\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
Vậy \(x=6.\)
\(---\)
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-300\right]\right\}+10\)
\(=12:\left\{400:200\right\}+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
#\(Toru\)