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a: \(\Leftrightarrow3^x\left(1+3^2\right)=2430\)
\(\Leftrightarrow3^x=243\)
hay x=5
b: \(\Leftrightarrow2^x\left(2^8-1\right)=224\)
=>2x=32
hay x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x+3^x.3^2=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x.8-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x.7=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{x^3}{8}\)= \(\frac{y^3}{64}\)= \(\frac{z^3}{216}\) \(\Rightarrow\)\(\frac{x^3}{2^3}\)= \(\frac{y^3}{4^3}\)= \(\frac{z^3}{6^3}\)\(\Rightarrow\)\(\frac{x^2}{2^2}\)=\(\frac{y^2}{4^2}\)=\(\frac{z^2}{6^2}\)
và có : \(^{x^2+y^2+z^2=224}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{224}{56}=4\)
=> \(\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x\in4;-4\)
\(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y\in8:-8\)
\(\frac{z^2}{36}=4\Rightarrow z^2=144\Rightarrow z\in12:-12\)
Vì \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\)nên x,y,z cùng dấu
Vậy \(x,y,z\in\left(4;8;12\right);\left(-4;-8;-12\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x+3 + 5.2x+2 = 224
2x.23 + 5.2x.22 = 224
2x.(8 + 5.4) = 224
=> 2x . 28 = 224
=> 2x = 8 = 23
=> x = 3
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a) \(2^x+2^{x+3}=144\)
\(2^x\left(1+2^3\right)=144\)
\(2^x.9=144\)
\(2^x=144:9=16\)
=> \(2^x=2^4\Rightarrow x=4\)
b) \(2^{x-1}+5.2^{x-2}=224\)
\(2^{x-2}\left(2+5\right)=224\)
\(2^{x-2}.7=224\Rightarrow2^{x-2}=32\Rightarrow2^{x-2}=2^5\)
=> x - 2 = 5 => x = 7
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(x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
(2x - 1)3 = -64
=> (2x - 1)3 = -43
=> 2x - 1 = -4
=> 2x = -4 + 1
=> 2x = -3
=> x = -3/2
( x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(2^{x+3}+5\cdot2^{x+2}=224\)
\(2^x\cdot2^3+5\cdot2^x\cdot2^2=224\)
\(2^x\cdot8+2^x\cdot20=224\)
\(2^x\cdot\left(20+8\right)=224\)
\(2^x\cdot28=224\)
\(2^x=8\)
\(x=3\)
2x+3+5*2x+2 = 224
VT=7*2x+2
pt trở thành 7*2x+2=224
<=>7*2x+2=25*7
<=>2x+2=25
<=>x+2=5
<=>x=3