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a)
=> \(x+2=69\)
=> \(x=67\)
b)
=> \(2^{x-5}=2^{30}\)
=> \(x-5=30\)
=> \(x=35\)
c)
=> \(3^x\left(3^2+1\right)=810\)
=> \(3^x.10=810\)
=> \(3^x=81\)
=> \(x=4\)
d)
=> \(5^x\left(5-1\right)=500\)
=> \(5^x.4=500\)
=> \(5^x=125\)
=> \(x=3\)
a) 3x + 2 = 369
=> x + 2 = 69
=> x = 67
b) 2x - 5 = 810
=> 2x - 5 = (23)10
=> 2x - 5 = 230
=> x - 5 = 30
=> x =35
c) 3x + 2 + 3x = 810
=> 3x(32 + 1) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
d) 5x + 1 - 5x = 500
=> 5x(5 - 1) = 500
=> 5x.4 = 500
=> 5x = 125
=> 5x = 53
=> x = 3
Bài 1:
b) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\frac{1}{2}\\x=0+\frac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{4}\right\}.\)
c) \(\left(2x-5\right)^4=81\)
\(\Rightarrow2x-5=\pm3\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3+5=8\\2x=\left(-3\right)+5=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8:2\\x=2:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{4;1\right\}.\)
d) \(3^{x+1}+3^{x+3}=810\)
\(\Rightarrow3^x.3^1+3^x.3^3=810\)
\(\Rightarrow3^x.\left(3^1+3^3\right)=810\)
\(\Rightarrow3^x.30=810\)
\(\Rightarrow3^x=810:30\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)
\(5^3.21=5^3.3.7⋮7\) (đpcm)
b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)
\(=7^4.55=7^4.5.11⋮11\) (đpcm)
c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)
\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)
\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)
d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)
\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)
a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)
\(\)\(\RightarrowĐPCM\)
b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)
\(\Rightarrowđpcm\)
Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)
Vì 53.3 là số nguyên nên \(5^3.3.7⋮7\)
Vậy \(5^5-5^4+5^3⋮7\)
c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)
\(=\left(3^{x+3}+3^{x+1}\right)+\left(2^{x+3}+2^{x+2}\right)\)
\(=3^x\left(3^2+3\right)+2^x\left(2^2+2\right)\)
\(=3^x.12+2^x.6\)
\(=6\left(2.3^x+2^x\right)\)
Vì \(2.3^x+2^x\in Z\)
Nên : \(6\left(2.3^x+2^x\right)⋮6\)
Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)
a, \(\dfrac{13}{32}+\dfrac{8}{24}+\dfrac{19}{32}+\dfrac{2}{3}\)
\(=\left(\dfrac{13}{32}+\dfrac{19}{32}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{32}{32}+\dfrac{3}{3}=1+1=2\)
b, \(\dfrac{3}{4}.36\dfrac{1}{5}-\dfrac{3}{4}.2\dfrac{1}{5}\)
\(=\dfrac{3}{4}.\left(36\dfrac{1}{5}-2\dfrac{1}{5}\right)\)
\(=\dfrac{3}{4}.\left[\left(36-2\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\right]\)
\(=\dfrac{3}{4}.34=\dfrac{102}{4}=26\)
Bài 2:
a: x=27/10:9/5=27/10*5/9=135/90=3/2
b: =>|x|=1,75
=>x=1,75 hoặc x=-1,75
c: =>\(2-x=\sqrt[3]{25}\)
hay \(x=2-\sqrt[3]{25}\)
d: =>3^x-1*6=162
=>3^x-1=27
=>x-1=3
=>x=4
a: =>13/15x=3/4-1/2=1/4
=>x=15/52
b: =>x-3=4
=>x=7
c: =>2x+1=9
=>2x=8
=>x=4
d: =>x+3=-2
=>x=-5
e: =>(x+6)(x-4)=0
=>x=4 hoặc x=-6
f: =>(x-3)(x-7)=0
=>x=3 hoặc x=7
a/
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)\(=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)\(\Rightarrow x=20;y=12;z=42\)
b/\(3x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{3};7y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+20}=2\)
\(\Rightarrow x=20;y=30;z=42\)
tao hỏi đấy
cút đi 2007_8
mấy bn đừng chửi nó , sao cứ phải đau đầu lmj
ai hỏi
a: Ta có: \(\frac52x-\frac13=\frac43\)
=>\(\frac52x=\frac43+\frac13=\frac53\)
=>\(x=\frac53:\frac52=\frac23\)
b: \(\frac25+\frac35:x=\frac13\)
=>\(\frac35:x=\frac13-\frac25=\frac{5}{15}-\frac{6}{15}=\frac{-1}{15}\)
=>\(x=\frac35:\frac{-1}{15}=\frac35\cdot\left(-15\right)=-9\)
c: \(2\left(x+\frac12\right)^2=32\)
=>\(\left(x+\frac12\right)^2=\frac{32}{2}=16\)
=>\(\left[\begin{array}{l}x+\frac12=4\\ x+\frac12=-4\end{array}\right.=>\left[\begin{array}{l}x=4-\frac12=\frac72\\ x=-4-\frac12=-\frac92\end{array}\right.\)
d: \(3^{x+2}+3^{x}=810\)
=>\(3^{x}\cdot9+3^{x}=810\)
=>\(3^{x}\left(9+1\right)=810\)
=>\(3^{x}=\frac{810}{10}=81=3^4\)
=>x=4