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1: Tìm x
a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)
\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)
\(\Leftrightarrow3^x=3^{12}\cdot3^3\)
\(\Leftrightarrow3^x=3^{15}\)
hay x=15
Vậy: x=15
b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)
\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)
Vậy: x=2
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
11: Ta có: \(\left(x+3\right)^3=125\)
\(\Leftrightarrow x+3=5\)
hay x=2
12: Ta có: \(\left(2x\right)^4=16\)
\(\Leftrightarrow x^4=1\)
hay \(x\in\left\{1;-1\right\}\)
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: \(x\in\left\{1;2;3;4;6;9;12;18;36\right\}\)
mà 2<x<6
nên \(x\in\left\{3;4\right\}\)
Vậy: Có 2 cách chia nhóm
a) 25 - x = 12 + 6 =18
x=25-18=7 Vậy x=7
b) 7 + 2 x ( x -3 ) = 11
2.(x-3)=11-7=4
x-3=4:2=2
x=3+2=5
c) 102 : ( 2.x + 13) : 4) = 6
(2.x+13):4=102:6=17
2.x+13=17.4=68
2.x=68-13=55
x=27,5 Vậy x=27,5
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: x∈{1;2;3;4;6;9;12;18;36}x∈{1;2;3;4;6;9;12;18;36}
mà 2<x<6
nên x∈{3;4}x∈{3;4}
Vậy: Có 2 cách chia nhóm
còn bài 1 chắc bn làm đc nha tick mk nha
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)