=>2/6+2/12+2/20+...+2/x(x+1)=1010/1011

=>2(1/6+1/12+...+1/x(x+1))=1010/1011

=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=505/1011

=>1/x+1=1/2022

=>x+1=2022

=>x=2021

b: \(BC\cdot\sin B\cdot\sin C\)

\(=BC\cdot\dfrac{AC}{BC}\cdot\dfrac{AB}{BC}\)

\(=\dfrac{BC\cdot AH\cdot BC}{BC^2}=AH\)

Ta có:

\(\widehat{A}=180^o-60^o-45^o=75^o\)(tổng 3 góc trong tam giác)

Kẻ đường cao AH

\(sin60=\dfrac{AH}{AB}\)

\(\Rightarrow AH=sin60.4=2\sqrt{3}\)

\(sin45=\dfrac{AH}{AC}\)

\(\Rightarrow AC=\dfrac{2\sqrt{3}}{sin45}=2\sqrt{6}\)

\(BC=\sqrt{4^2-\left(2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-\left(2\sqrt{3}\right)^2}=2+2\sqrt{3}\)

`(2x - 1)^2 = 36`

`(2x - 1)^2 = 6^2` hoặc `(2x - 1)^2 = (-6)^2`

`2x - 1 = 6` hoặc `2x - 1 = -6`

`2x = 6 + 1` hoặc `2x = -6 + 1`

`2x = 7` hoặc `2x = -5`

`x = 7 : 2` hoặc `x = -5 : 2`

`x = 3,5` hoặc `x = -2,5`

\(\left(2x-1\right)^2=36\)

\(\Rightarrow\left(2x-1\right)^2=6^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

=>6/x=32/35

=>x=6:32/35=6*35/32=210/32=105/16

c b a b yên tâm đi mình học rồi

Bài nào ớ em?

IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d

XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is

C. 
Bài 1
1. C
2. B
3. C
4. B

(Nên double-check trước khi chép)
 

\(B=3^1+3^2+3^3+...+3^{100}\\3B=3^2+3^3+3^4+...+3^{101}\\3B-B=(3^2+3^3+3^4+...+3^{101})-(3^1+3^2+3^3+...+3^{100})\\2B=3^{101}-3\\\Rightarrow 2B+3=3^{101}\)

Mặt khác: \(2B+3=3^n\)

\(\Rightarrow 3^n=3^{101}\\\Rightarrow n=101(tm)\)

Vậy n = 101.

cảm ơn bạn nha :))