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a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Bài giải chi tiết đây em nhé:
\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+ \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)
\(2x+1\) = 19
2\(x\) = 19 - 1
2\(x\) = 18
\(x\) = 18: 2
\(x\) = 9
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)