Rút gọn : a . P = 3+2√3 / √3 + 2+√2 / √2+1 - ( √2 + √3 ) ; b. N = ( 1 - 5 + √5 / 1 + √5 ) ( 5 - √5 / 1- √5 - 1 ) ; c. Q = ( 5 - 2√5 / 2 - √5 - 2 ) ( 3+3 √5 / 3 + √5 - 2 ). Giúp mik vs ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
I2x+3I = 5
=> 2x+3 = 5 hoặc 2x+3 = -5
=> 2x = 5 - 3 hoặc 2x = -5 - 3
=> 2x = 2 hoặc 2x = -8
=> x = 2 hoặc x = -4
2:
B = 1/2.2/3.3/4.4/5.....27/28
= 1.2.3.4.5.6...27/2.3.4.5.6...28
= 1/28
3:
2A = 2(1+1/2+1/2^2+1/2^3+1/2^4+...+1/2^2015) = 2+1+1/2+1/2^2+1/2^3+...+1/2^2014
=> 2A-A = ( 2+1+1/2+1/2^2+1/2^3+...+1/2^2014)-(1+1/2+1/2^2+1/2^3+...+1/2^2015)
=> A = 2-1/2^2015
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+...+2^{25}\)
\(2A=2+2^2+2^3+...+2^{26}\)
\(2A-A=A=2^{26}-1\)
\(B=1+3+3^2+...+3^{19}\)
\(3B=3+3^2+3^3+...+3^{20}\)
\(3B-B=3^{20}-1\)
\(B=\dfrac{3^{20}-1}{2}\)
a: A=1+2+2^2+...+2^25
=>2A=2+2^2+...+2^26
=>A=2^26-1
b: B=1+3+3^2+...+3^19
=>3B=3+3^2+3^3+...+3^20
=>2B=3^20-1
=>B=(3^20-1)/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(15/10-4/10+1/10)/(18/12-8/12+1/12)
A=(6/5)/(11/12)
A=(6/5)*(12/11)
A= 72/55
minh chi biet cach tinh ra the thoi
chuc ban hoc gioi
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) C = 1 + 3 + 32 + 33 + ....... + 320
<=> 3C = 3.( 1 + 3 + 32 + 33 + ...... + 320 )
<=> 3C = 3 + 32 + 33 + 34 + ....... + 321
<=> 3C - C = ( 3 + 32 + 33 + 34 + ....... + 321 ) - ( 1 + 3 + 32 + 33 + ...... + 320 )
<=> 2C = 321 - 1
=> C = ( 321 - 1 ) : 2
b ) B = 2 + 22 + 23 + ...... + 250
<=> 2B = 2.( 2 + 22 + 23 + ...... + 250 )
<=> 2B = 22 + 23 + 24 + ....... + 251
<=> 2B - B = ( 22 + 23 + 24 + ...... + 251 ) - ( 2 + 22 + 23 + ...... + 250 )
=> B = 251 - 2
a, Ta có: 3C=3+3^2+3^3+3^4+...+3^21
3C-C=(3+3^2+3^3+...+3^20+3^21)-(1+3+3^2+...+3^19+3^20)
<=>2C = 3^21 - 1 - 3^20 =3^20. (3-1) -1=3^20 .2 -1
=>C\(=\frac{3^{20}.2-1}{2}=3^{20}-\frac{1}{2}=3^{20}-0,5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=5-3=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3^2+3^4+...+3^{2022}\)
\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)
\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)
d, không đáp án nào đúng
Lời giải:
$S=1+3^2+3^4+....+3^{2022}$
$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$
$\Rightarrow 9S-S=3^{2024}-1$
$\Rightarrow S=\frac{3^{2024}-1}{8}$
Đáp án D.
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: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
Sai rồi đấy ạ câu P