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B1. 2x + 3 + 22 = 72
=> 2x + 3 + 4 = 72
=> 2x + 3 = 72 - 4
=> 2x + 3 = 68
=> ko có gtri x
B2 : Ta có : A = 1 + 2 + 22 + 23 + 24 + 25 + 26 + ... + 22001 + 22002
= (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (22000 + 22001 + 22002)
= 3 + 22.(1 + 2 + 22) + 25.(1 + 2 + 22 ) + ... + 22000 . (1 + 2 + 22)
= 3 + 22.7 + 25.7 + ... + 22000 . 7
= 3 + (22 + 25 + .... + 22000) . 7
=> Số dư của 7 là 3
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\(5^x+5^{x+2}=650;5^x.26=650;5^x=25;x=2\)
\(2^x+2^{x+3}=144;2^x.9=144;2^x=16;x=4\)
\(3^{x-1}+5.3^{x-1}=162;3^{x-1}.6=162;3^{x-1}=27;x=4\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\rightarrow x-5=0\&x-5=1\) hoặc x - 5 = - 1
\(x-5=1;x=6;x-5=0;x=5;x-5=-1;x=4\)
\(\left(2^2:4\right).2^n=4;2^n=2^2;n=2\)
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\(3^{2x+2}=9^{x+3}\)
\(\Rightarrow3^{2x+2}=3^{2x+6}\)
\(\Rightarrow2x+6=2x+2\)
\(\Rightarrow\left(2x-2x\right)+\left(6-2\right)=0\)
\(\Rightarrow0x=-4\left(loại\right)\)
\(b,\left(x-3\right)^4=\left(x-3\right)^6\)
\(\Rightarrow\left(x-3\right)^4-\left(x-3\right)^4.\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)^4.\left[1-\left(x-3\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^4=0\\1-\left(x-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x-3\in\left\{\pm1\right\}\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x\in\left\{4;2\right\}\end{cases}\Rightarrow}x\in\left\{2;3;4\right\}}\)
a, => 32x+2 =32.(x+3)
2x+2=2.(x+3)
2(x+1)=2(x+3)
x+1=x+3
=> x= rỗng
vậy............
c, x15-x2=0
x2(x13-1)=0
\(\orbr{\begin{cases}x^2=0\\x^{13}=0\end{cases}}< =>\orbr{\begin{cases}x=0\\x^{13}=1^{13}\end{cases}}< =>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy.........
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bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon