tìm x biết \(^{2^{x+2}\cdot3^{x+1}\cdot5^x=10800}\)
\(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)
\(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\\\Rightarrow2^x\cdot2^2\cdot3^x\cdot3\cdot5^x=10800\\\Rightarrow(2^x\cdot3^x\cdot5^x)\cdot(2^2\cdot3)=10800\\\Rightarrow(2\cdot3\cdot5)^x\cdot(4\cdot3)=10800\\\Rightarrow30^x\cdot12=10800\\\Rightarrow30^x=10800:12\\\Rightarrow30^x=900\\\Rightarrow30^x=30^2\\\Rightarrow x=2\)
\(2^{x+2}.3^{x+1}.5^x=10800\\ \Leftrightarrow2^x.2^2.3^x.3.5^x=10800\\ \Leftrightarrow12.\left(2.3.5\right)^x=10800\\ \Leftrightarrow12.30^x=10800\\ \Leftrightarrow30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
Bài 2: Tìm các số x,y,z biết \(2^{x-2}\cdot3^{y-3}\cdot5^{z-1}=144\)
\(2^{x-2}.3^{y-3}.5^{z-1}=144=>2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.5^0\)
\(\hept{\begin{cases}2^{x-2}=2^4\\3^{y-3}=3^2\\5^{z-1}=5^0\end{cases}}=>\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}}=>\hept{\begin{cases}x=4+2\\y=2+3\\z=0+1\end{cases}}=>\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}\)
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
\(\Rightarrow\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}}\)
tìm x biết : (\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{8\cdot9\cdot10}\)) x \(x=\frac{22}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\) 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
Tìm x biết: \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\cdot x=\frac{22}{45}\)Trình bày cách tính
X=2 nhé bạn.....đúng đó, vòng 12 mk 300 mà cx gặp câu này!!! Tick nha
tìm x thuộc z, biết:
(\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\left(\right)-x=\frac{-100}{99}\)
\(\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-x\)\(=\frac{-100}{99}\)
\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\frac{98}{99}-x=\frac{-100}{99}\)
\(x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(x=\frac{198}{99}=2\)
CHÚC BN HOK TỐT!
ĐÚNG THÌ K CHO MK NHA!
(2/1.3 + 2/3.5 + ... + 2/97.99) - x = -100/99
=>( 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/97 + 1/99) - x = -100/99
=>( 1/1 - 1/99) - x = -100/99
=>98/99 - x = -100/99
=>x = 98/99 - (-100/99)
=>x = 198/99 = 2
Tìm x:
1) \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
2) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
\(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=10.531441+8.531441\)
\(2.3^x=\left(10+8\right).531441\)
\(2.3^x=18.531441\)
\(2.3^x=9565938\)
\(3^x=9565938:2\)
\(3^x=4782969\)
\(3^x=3^{14}\)
\(\Rightarrow x=14\)
tìm x biết
\(5^{x+1}-2\cdot5^x=375\)
\(9^{x+1}-5\cdot3^{2x}=324\)
\(\left(1-x\right)^5=32\)
\(3\cdot5^{2x+1}-3\cdot25^x=300\)
help me !!! mình đang cần luôn giải hộ mình với
a) \(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow3^{2\left(x+1\right)}-5.3^{2x}=324\)
\(\Rightarrow3^2\left(3^{x+1}-5.3^x\right)=324\)
\(\Rightarrow9.3^x\left(3-5\right)=324\)
\(\Rightarrow3^x.\left(-2\right)=36\)
\(\Rightarrow3^x=-18=3^2.\left(-2\right)\)(vô lí vì 3x không chia hết cho 2)
c) \(\left(1-x\right)^5=32=2^5\)
\(\Rightarrow1-x=2\)
\(\Rightarrow x=-1\)
d) \(3.5^{2x+1}-3.25^x=300\)
\(\Rightarrow3\left(5^{2x}.5-5^{2x}\right)=300\)
\(\Rightarrow5^{2x}\left(5-1\right)=100\)
\(\Rightarrow5^{2x}.4=100\)
\(\Rightarrow5^{2x}=25=5^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
tìm x biết: \(5^{2x-3}-2\cdot5^x\)=\(5^2\cdot3\)
1. Tìm x
a) \(2^x+5=21\)
b) \(2^x-1+3^2=5^2+2\cdot5\)
c) \(\left(2x-1\right)^3+5=130\)
d) \(5^{2x-3}-2\cdot5^2=5^2\)
e) \(3^{2x+1}-2=3^2+\left[5^2-3\left(2^2-1\right)\right]\)
f) \(\left(7^x-11\right)^3=2^5\cdot5^2+200\)
g) \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
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