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a) \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
\(=1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
=> 7S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}\)
Lấy 7S trừ S ta có :
7S - S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}-\left[1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\right]\)
6S = \(7-1-1+\left(\frac{1}{7}\right)^{2007}=5+\left(\frac{1}{7}\right)^{2007}\Rightarrow S=\frac{5+\left(\frac{1}{7}\right)^{2007}}{6}\)
b)
(x-7)x+1 - (x-7)x+11 = 0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0 hoặc 1-(x-7)10=0
=>x-7=0 hoặc (x-7)10=1
=>x=7 hoặc x-7=1 hoặc x-7=-1
=>x=7 hoặc x=8 hoặc x=6
a)
(x-1)x+2=(x-1)x+6
(x-1)x+2-(x-1)x+6=0
(x-1)x+2 . [1-(x-1)4]=0
=> (x-1)x+2=0 hoặc 1-(x-1)4=0
=>x-1=0 =>(x-1)4=1
=>x=1 =>x-1=1 hoặc x-1=-1
=> x=2 hoặc x=0
vậy x \(\in\) {0;1;2}
Câu 1
4 p/s cộng thêm 1,p/s cuối trừ 4 rồi nhóm vs nhau
d/s la x= - 329
Câu 2
NHân vs 7 thành 7S rồi rút gọn là đc
Câu 1 :
a) \(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\) \(\Rightarrow x+329=0\Rightarrow x=-329\)
\(S=1+2+2^2+.....+2^{2017}\)
\(\Leftrightarrow2A=2+2^2+.....+2^{2018}\)
\(\Leftrightarrow2A-A=\left(2+2^2+....+2^{2018}\right)-\left(1+2+.....+2^{2017}\right)\)
\(\Leftrightarrow A=2^{2018}-1\)
a/ \(7^{2x}+7^{2x+2}=2450\)
\(\Leftrightarrow7^{2x}+2^{2x}.7^2=2450\)
\(\Leftrightarrow7^{2x}\left(1+49\right)=2450\)
\(\Leftrightarrow7^{2x}.50=2450\)
\(\Leftrightarrow7^{2x}=79\)
\(\Leftrightarrow7^{2x}=7^2\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ....
b/ Ta có :
\(A=1+2+2^2+.......+2^{2016}\)
\(\Leftrightarrow2A=2+2^2+......+2^{2017}\)
\(\Leftrightarrow2A-A=\left(2+2^2+.......+2^{2017}\right)-\left(1+2+....+2^{2016}\right)\)
\(\Leftrightarrow A=2^{2017}-1\)
Mà \(B=2^{2017}-1\)
\(\Leftrightarrow A=B\)