Đặt \(A=1+3^2+3^4+...+3^{100}\)

\(9A=3^2+3^4+3^6+...+3^{102}\)

\(9A-A=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)

\(8A=3^{102}-1\)

\(A=\frac{3^{102}-1}{8}\)

Vậy \(A=\frac{3^{102}-1}{8}\)

Chúc bạn học tốt ~ 

Đặt A = 1 + 3^2 + 3^4 + 3^6 + ....+ 3^100

3^2A = 3^2 + 3^4 + 3^6 + ..+3^102

8A=3^2A - A = 3¹⁰² - 1

A = 3¹⁰² - 1/8

=. A = 3¹⁰² - 1 /8

Cam on nha

áp dụng công thức tính tổng

\(1^3+2^3+......+n^3=\frac{n^2.\left(n+1\right)^2}{4}\)

=>  tổng là

\(\frac{100^2.\left(100+1\right)^2}{4}=....\)

bạn tự tính tiếp nhé

giúp mình đi

3 + 1,3 = a,b = 4,3

Vậy số a,b là 4,3

4/5

4/5

\(\dfrac{6}{7}x\dfrac{2}{3}x\dfrac{7}{5}=\dfrac{4}{5}\)

\(1-\left(x-1\right):3=\dfrac{2}{3}\)

\(\Rightarrow1-\left(x-1\right)=\dfrac{2}{3}.3\)

\(\Rightarrow1-\left(x-1\right)=2\)

\(\Rightarrow x-1=1-2\)

\(\Rightarrow x-1=\left(-1\right)\)

\(\Rightarrow x=\left(-1\right)+1\)

\(\Rightarrow x=0\)

câu này bị sai nhé câu đúng mình làm ở trên roi

1+1= mấy giúp mik vs

\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)

\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)

\(=\frac{1}{2}-\frac{1}{23}\)

\(=\frac{21}{46}\)

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

`14/9:8/3=14/9xx3/8=[2xx7xx3]/[3xx3xx4xx2]=7/12`

`5/4+9/5xx5/4=5/4+9/4=14/4=7/2`

14/9:8/3=7/12
5/4+9/5x5/4=5/4+9/4=14/4=7/2

14/9x3/8=7/12

5/4+9/4=7/2