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Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{z}{\dfrac{1}{2}}=\dfrac{x-z}{\dfrac{2}{3}-\dfrac{1}{2}}=\dfrac{-13}{2}:\dfrac{1}{6}=-39\)
Do đó: x=-26; y=-117/5; z=-39/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5D=1+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+...+\dfrac{1}{6.5^{99}}\)
\(6D=\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{6.5^{100}}\)
\(D=\dfrac{\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{36.5^{100}}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(A=6.7+6.7^2+6.7^3+...+6.7^{100}\)
\(\Rightarrow7A=7.\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(7A=6.7^2+6.7^3+6.7^4+...+6.7^{101}\)
\(\Rightarrow7A-A=6A=\left(6.7^2+6.7^3+6.7^4+...+6.7^{101}\right)-\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(6A=6.7^{101}-6.7=6\left(7^{101}-7\right)\Leftrightarrow A=7^{101}-7\)
vậy \(A=7^{101}-7\)
ta có : \(B=6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(\Rightarrow5B=5\left(6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\right)\)
\(5B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}\)
\(\Rightarrow5B+B=6B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}+6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(6B=6.5-6.5^{101}=6.\left(5-5^{101}\right)\Leftrightarrow B=5-5^{101}\)
vậy \(B=5-5^{101}\)
\(A=6\cdot7+6\cdot7^2+6\cdot7^3+...+6\cdot7^{100}\\ =6\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot7+\left(7-1\right)\cdot7^2+\left(7-1\right)\cdot7^3+...+\left(7-1\right)\cdot7^{100}\\ =7^2-7+7^3-7^2+7^4-7^3+...+7^{101}-7^{100}\\ =7^{101}-7=7\cdot\left(7^{100}-1\right)\)
\(B=6\cdot5-6\cdot5^2+6\cdot5^3-...+6\cdot5^{99}-6\cdot5^{100}\\ =6\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\ =\left(5+1\right)\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\=\left(5+1\right)\cdot5-\left(5+1\right)\cdot5^2+\left(5+1\right)\cdot5^3-...+\left(5+1\right)\cdot5^{99}-5^{100}\\ =5^2+5-5^3-5^2+5^4+5^3+...+5^{100}+5^{99}-5^{101}-5^{100}\\ =5-5^{101}\\ =5\cdot\left(1-5^{100}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{5^5}{5^x}=5^{18}\)
=>5-x=18
hay x=-13
b: \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)
=>4-x=50
hay x=-46
c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)
=>2x-3=49
=>2x=52
hay x=26
d: \(\dfrac{2^3}{2^x}=4^5\)
\(\Leftrightarrow2^{3-x}=2^{10}\)
=>3-x=10
hay x=-7
e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
f: \(7\cdot2^x=2^9+5\cdot2^8\)
\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)
\(\Leftrightarrow2^x=2^8\)
hay x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{4^2.25^2+32.5^3}{2^3.5^2}=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}\)
\(=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}\)
\(=2.5.7\)
\(=70\)
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1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
![](https://rs.olm.vn/images/avt/0.png?1311)
a,?????
b, Với mọi giá trị của x;y ta có:
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\)
Để \(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..........
c, \(\left|2x\right|-\left|3,5\right|=\left|-6,5\right|\)
\(\Rightarrow\left|2x\right|=6,5+3,5=10\)
\(\Rightarrow\left\{{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy..........
d, \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\left\{{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
\(=\dfrac{2^5\cdot5^3}{2^6\cdot5^2}=\dfrac{5}{2}\)