cái này thì triệt tiêu ra đó bn

là bằng 2020/6069

A = 1/3.5.7 + 1/5.7.9+ ... + 1/2019.2021.2023

2A = 2/3.5.7 + 2/5.7.9+ ... + 2/2019.2021.2023

2A = 1/3-1/5+1/7+1/5-1/7+1/9+....+1/2019-1/2021+1/2023

2A = 1/3 - 1/2023

2A = 2023/6069 - 3/6069

2A = 2023-3/6069

2A = 2020/6069

 A = 1010/6069

Vậy A = 1010/6069

\(A=\sqrt{\dfrac{3\sqrt{5}-1}{3+2\sqrt{5}}}-\sqrt{\dfrac{\sqrt{5}-1}{\sqrt{5}-2}}\)

\(=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}\)

\(=-\sqrt{2}\)

\(\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\dfrac{\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{4-3}}\)

\(=\sqrt{3}+1-\sqrt{3}+1=2\)

no bé ơi

Lời giải:

$\frac{1+14x}{5}=\frac{5-3x}{4}$

$\Rightarrow 4(1+14x)=5(5-3x)$

$4+56x=25-15x$

$56x+15x=25-4$

$71x=21$

$x=\frac{21}{71}$

c, \(\Rightarrow4+56x=25-15x\Leftrightarrow71x=21\Leftrightarrow x=\dfrac{21}{71}\)

số hs khá là: \(\dfrac{3}{4}\times\dfrac{1}{6}=\dfrac{1}{8}\left(hs\right)\)

a: số hs lớp 6A là : \(\dfrac{3}{4}+\dfrac{1}{8}+5=\dfrac{18}{24}+\dfrac{3}{24}+5=5\dfrac{21}{24}=5\dfrac{7}{8}=\dfrac{47}{8}\left(hs\right)\)

b: tỉ số là : \(\dfrac{3}{4}\times100:\dfrac{1}{8}=\dfrac{300}{4}:\dfrac{1}{8}=\dfrac{300\times8}{4\times1}=600\%\)

Gọi học sinh lớp 6a là x hs

Số học sinh giỏi là : \(\dfrac{3}{4}x\)(hs)

Số học sinh khác là : \(\dfrac{1}{6}.\dfrac{3}{4}x=\dfrac{1}{8}x\) (hs)

Theo bài ra có :

\(x-\dfrac{3}{4}x-\dfrac{1}{8}x=5\)

\(x\left(1-\dfrac{3}{4}-\dfrac{1}{8}\right)=5\)

\(\dfrac{1}{8}x=5\)

\(x=40\)

Vậy số hcj sinh lớp 6a là 40 hs

Số học sinh giỏi là : \(\dfrac{3}{4}.40=30\) (hs)

Số học sinh khá là :\(\dfrac{1}{6}.30=5\) (hs)

ĐS ...

Chắc đề đúng là \(\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...\)

- Với \(n=1\) đẳng thức đúng

- Giả sử đẳng thức cũng đúng với \(n=k>1\) hay:

\(\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\)

- Ta cần chứng minh nó cũng đúng với \(n=k+1\) hay:

\(\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}+\dfrac{2k+1}{4+\left(2k+1\right)^4}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)

Thật vậy, ta có:

\(\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}+\dfrac{2k+1}{4+\left(2k+1\right)^4}=\dfrac{k^2}{4k^2+1}+\dfrac{2k+1}{4+\left(2k+1\right)^4}\)

\(=\dfrac{k^2}{4k^2+1}+\dfrac{2k+1}{\left(2k+1\right)^4+4\left(2k+1\right)^2+4-4\left(2k+1\right)^2}=\dfrac{k^2}{4k^2+1}+\dfrac{2k+1}{\left(4k^2+4k+3\right)^2-\left(4k+2\right)^2}\)

\(=\dfrac{k^2}{4k^2+1}+\dfrac{2k+1}{\left(4k^2+1\right)\left(4k^2+8k+5\right)}=\dfrac{k^2\left(4k^2+8k+5\right)+2k+1}{\left(4k^2+1\right)\left(4k^2+8k+5\right)}\)

\(=\dfrac{\left(k+1\right)^2\left(4k^2+1\right)}{\left(4k^2+1\right)\left(4k^2+8k+5\right)}=\dfrac{\left(k+1\right)^2}{4k^2+8k+5}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\) (đpcm)

\(=\left(6\sqrt{2}-6\sqrt{2}+6\sqrt{2}\right):\sqrt{2}=6\sqrt{2}:\sqrt{2}=6\)

=(6√2−6√2+6√2):√2=6√2:√2=6

\(=\left(6\sqrt{2}-6\sqrt{2}+6\sqrt{2}\right):\sqrt{2}=6\sqrt{2}:\sqrt{2}=6\)

\(x+\dfrac{12}{25}x=74\\ \Rightarrow\left(1+\dfrac{12}{25}\right)x=74\\ \Rightarrow\dfrac{37}{25}x=74\\ \Rightarrow x=50\)

\(lim_{n\rightarrow+\infty}\dfrac{6^n+1}{6^n-2}=\)\(lim_{n\rightarrow+\infty}\dfrac{6^n\left(1+\dfrac{1}{6^n}\right)}{6^n\left(1-\dfrac{2}{6^n}\right)}=\)\(lim_{n\rightarrow+\infty}\dfrac{\left(1+\dfrac{1}{6^n}\right)}{\left(1-\dfrac{2}{6^n}\right)}=\dfrac{1}{1}=1\)

\(lim_{n\rightarrow-\infty}\dfrac{6^n+1}{6^n-2}=\)\(\dfrac{0+1}{0-2}=\dfrac{-1}{2}\)