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ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]
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`@` `\text {Ans}`
`\downarrow`
`(2^2+1) \times (x+14) = 5^2 \times 4 + (2^5 + 3^2 + 7^2) \div 2`
` \Rightarrow (4+1) \times (x+14) = 5^2\times 2^2 + ( 32 + 9 + 49) \div 2`
`\Rightarrow 5 \times (x+14) = (5*2)^2 + (32+58) \div 2`
`\Rightarrow 5 \times (x+14) = 10^2+90 \div 2`
`\Rightarrow 5 \times (x+14) = 100 + 45`
`\Rightarrow 5 \times (x+14) = 145`
`\Rightarrow x+14 = 145 \div 5`
`\Rightarrow x+14=29`
`\Rightarrow x=29-14`
`\Rightarrow x=15`
Vậy, `x=15.`
\(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
\(\Rightarrow\left(4+1\right)\cdot\left(x+14\right)=25\cdot4+\left(32+9+49\right):2\)
\(\Rightarrow5\cdot\left(x+14\right)=100+45\)
\(\Rightarrow5x+70=145\)
\(\Rightarrow5x=75\)
\(\Rightarrow x=\dfrac{75}{5}=15\)
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\(2^2\cdot3\left(x+5\right)-6^2=\left(2^3+2^2\right)\cdot2^2\)
\(\Rightarrow4\cdot\left(3x+15\right)-36=12\cdot4\)
\(\Rightarrow12x+60-36=48\)
\(\Rightarrow12x+24=48\)
\(\Rightarrow12x=24\)
\(\Rightarrow x=24:12=2\)
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`(2^x+1)^2 =25`
`=> (2^x+1)^2 = (+-5)^2`
\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)
\(\left(x+6\right)\left(5^x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
\(\left(x-3\right)^{2023}=x-3\)
\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
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A, 2^5 x 8^4 = 2^5 x (2^3)^4 B,25^6 x 125^3=(5^2)^6 x (5^3)^3
= 2^5 x 2^12 =5^12 x 5^9
=2^17 =5^21
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a)\(3^2.9^3=9.9^3=9^{1+3}=9^4\)
b)\(2^2.5^2=4.25=100=10^2\)
c)\(8^5.2^3=8^5.8=8^{5+1}=8^6\)
d)\(9^8:3^2=9^8:9=9^{8-1}=9^7\)
\(2^{x+1}-2^x=3^2\)
\(\Rightarrow2^x\cdot\left(2-1\right)=9\)
\(\Rightarrow2^x=9\)
\(\Rightarrow x\in\varnothing\)
\(2^{x+1}-2^x=3^2\)
=>2^x*2-2^x=9
=>2^x=9
=>\(x\in\varnothing\)