\(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)
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\(2^{x-2}.3^{y-3}.5^{z-1}=144=>2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.5^0\)
\(\hept{\begin{cases}2^{x-2}=2^4\\3^{y-3}=3^2\\5^{z-1}=5^0\end{cases}}=>\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}}=>\hept{\begin{cases}x=4+2\\y=2+3\\z=0+1\end{cases}}=>\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}\)
vậy \(\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}\)
Tách số 144 ra ta có :
\(144=2^4.3^2.1=2^4.3^2.5^0\)
Theo đề bài
\(\Rightarrow\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}}\)
\(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=10.531441+8.531441\)
\(2.3^x=\left(10+8\right).531441\)
\(2.3^x=18.531441\)
\(2.3^x=9565938\)
\(3^x=9565938:2\)
\(3^x=4782969\)
\(3^x=3^{14}\)
\(\Rightarrow x=14\)
\(\frac{2}{2.3}\)+ \(\frac{2}{3.4}\)+ \(\frac{2}{4.5}\)+........+ \(\frac{2}{x+\left(x+1\right)}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+..........+ \(\frac{1}{x+\left(x+1\right)}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+.........+ \(\frac{1}{x}\)- \(\frac{1}{x+1}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\)
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\): 2
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\). \(\frac{1}{2}\)
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{502}{1005}\)
= \(\frac{1}{x+1}\)= \(\frac{1}{2}\)- \(\frac{502}{1005}\)
= \(\frac{1}{x+1}\)= \(\frac{1}{2010}\)
\(\Rightarrow\)\(x+1\)= 2010
\(\Leftrightarrow\) \(x\) = 2010 - 1
\(\Rightarrow\) \(x\)= 2009
Vậy \(x\)= 2009
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x\left(x+1\right)}=\frac{2008}{2010}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1004}{1005}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
\(=>x+1=2010\)
\(=>x=2009\)
Vậy \(x=2009\)
\(15-\left\{2.\left[\left(2x-4\right).5\right].3.\left(x+1\right)\right\}=12-x\)
\(15-\left\{\left[10x-20\right].6.\left(x+1\right)\right\}=12-x\)
\(15-\left\{10x-20.6x+1\right\}=12-x\)
\(15-\left\{10x-120x+1\right\}=12-x\)
\(15-\left(-110x\right)-1=12-x\)
\(15+110x-1=12-x\)
\(110x+x=12-15+1\)
\(111x=-2\)
\(x=\dfrac{-2}{111}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\) vậy
\(\frac{11}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}\div\frac{11}{45}=2\)
vậy suy ra x =2
mình chắc chắn 100% luôn đó, cái này ở trong violympic toán 7 vòng 12 phải ko
a) \(2^x+5=21\)
\(\Rightarrow2^x=21-5=16\Rightarrow2^x=2^4\)
Vậy x = 4
b) \(2^x-1+3^2=5^2+2.5\)
\(\Rightarrow2^x-1+9=35\)
\(\Rightarrow2^x=35-9+1=27\)
Vậy x không có giá trị
c;d;e;f làm tương tự
X=2 nhé bạn.....đúng đó, vòng 12 mk 300 mà cx gặp câu này!!! Tick nha
\(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\\\Rightarrow2^x\cdot2^2\cdot3^x\cdot3\cdot5^x=10800\\\Rightarrow(2^x\cdot3^x\cdot5^x)\cdot(2^2\cdot3)=10800\\\Rightarrow(2\cdot3\cdot5)^x\cdot(4\cdot3)=10800\\\Rightarrow30^x\cdot12=10800\\\Rightarrow30^x=10800:12\\\Rightarrow30^x=900\\\Rightarrow30^x=30^2\\\Rightarrow x=2\)
\(2^{x+2}.3^{x+1}.5^x=10800\\ \Leftrightarrow2^x.2^2.3^x.3.5^x=10800\\ \Leftrightarrow12.\left(2.3.5\right)^x=10800\\ \Leftrightarrow12.30^x=10800\\ \Leftrightarrow30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)