rối quá :)

B = (-5)⁰ + 5¹ + (-5)² + 5³ + ... + (-5)²⁰¹⁶ + 5²⁰¹⁷

B = 1 + 5¹ + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷

5B = 5 + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷

5B - B = (5 + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷) - (1 + 5¹ + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷)

   4B    =    5²⁰¹⁷                                          -       1

   B      =   \(\dfrac{5^{2017}-1}{4}\)

\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)

\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)

\(-6B=\left(-5\right)^{2017}-1\)

\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)

Ta có : B = (-5)^0 + (-5)^1 + ......+ (-5)^2017

          (-5)B = (-5)^1 + (-5)^2 + .......+ (-5)^2018

              (-4)B = (-5)^0- (-5)^2018

           B = 1 - (-5)^2018 / (-4)

Nếu có sai sót gì mong các bạn bỏ qua

\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)

\(-4B=\left(-5\right)^{2018}-\left(-5\right)^0\)

\(\Rightarrow B=\frac{\left(-5\right)^{2018}-\left(-5\right)^0}{-4}\)

-5B=(-5)¹+(-5)²+(-5)³+...+(-5)²⁰¹⁸

-5B-B=[(-5)¹+(-5)²+...+(-5)²⁰¹⁸] - [(-5)⁰+(-5)¹+...+(-5)²⁰¹⁷]

-6B=(-5)²⁰¹⁸-(-5)⁰ = (-5)²⁰¹⁸-1

B= [(-5)²⁰¹⁸-1]:-6

Anh học tốt nha ( em mới lớp 6)

Cho e sửa lại dòng cuối :

B= [(-5)²⁰¹⁸-(-1)]:-6

\(B=1-5+5^2-5^3+...+5^{2016}-5^{2017}\) (1)

\(\Rightarrow5B=5-5^2+5^3-5^4+...+5^{2017}-5^{2018}\) (2)

Cộng vế với vế của (1) và (2):

\(6B=1+5-5+5^2-5^2+5^3-5^3+...+5^{2017}-5^{2017}-5^{2018}\)

\(\Rightarrow6B=1-5^{2018}\)

\(\Rightarrow B=\dfrac{1-5^{2018}}{6}\)

`A=sqrt{3-sqrt5}-sqrt{3+sqrt5}`

`<=>sqrt2A=sqrt{6-2sqrt5}-sqrt{6+2sqrt5}`

`<=>sqrt2A=sqrt{(sqrt5-1)^2}-sqrt{(sqrt5+1)^2}`

`<=>sqrt2A=sqrt5-1-sqrt5-1=-2`

`<=>A=-sqrt2`

Câu b đề sai sai kiểu gì ý `sqrt{a+1}/a` là sao ;-;?

a)B=x+5 +x +x-5/x(x-5)=3x/x(x-5)=3/x-5

        b)đkxđ   x khác 5

a)B=x+5 +x +x-5/x(x-5)=3x/x(x-5)=3/x-5

        b)đkxđ   x khác 5

1, đk x khác 0 ; -5 

\(B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

2, B = 3/(x+5) > 0 => x + 5 > 0 <=> x > -5 

3, \(\Rightarrow x+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)

\(P=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}\)

\(P=2\)

b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)

\(N=\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)

\(N=1^2-\left(\sqrt{5}\right)^2\)

\(N=-4\)

c) \(Q=\left(\dfrac{5+2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)

\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)

\(Q=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)

\(Q=\left(\sqrt{5}\right)^2-2^2\)

\(Q=1\)

\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

A=20+21+22+23+...++23+...+250250

2�=2+22+23+...+2512A=2+22+23+...+251

2�−�=�=251−202A−A=A=251−20

�=5+52+53+...+599+5100B=5+52+53+...+599+5100

5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101

5�−�=4�=5101−55B−B=4B=5101−5

�=5101−54B=45101−5​

�=3−32+33−34+...+C=3−32+33−34+...+32007−32008+32009−3201032007−32008+32009−32010

3�=32−33+34−35+...−32008+32009−32010+320113C=32−33+34−35+...−32008+32009−32010+32011

3�+�=4�=32011+33C+C=4C=32011+3

�=32011+34C=432011+3​

�100=5+5×9+5×92+5×93+...+5×999S100​=5+5×9+5×92+5×93+...+5×999

�100=5×(1+9+92+93+...+999)S100​=5×(1+9+92+93+...+999)

9�100=5×(9+92+93+...+999+9100)9S100​=5×(9+92+93+...+999+9100)

9�100−�100=8�100=5×(9100−1)9S100​−S100​=8S100​=5×(9100−1)

�100=5×(9100−1)8S100​=85×(9100−1)​

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