1/2+1/3+1/6+1/12+1/15+1/20+1/30+1/35+1/42+1/63

=(1/2+1/6+1/12+1/20+1/30+1/42) + (1/3 +1/15+1/35+1/63)

=(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6,7)+(1/1.3+1/3.5+1/5.7+1/7.9)

=(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6 -1/7) + 1/2.(2/1.3+2/3.5+2/5.7+2/7,9)

=(1-1/7)+ 1/2.(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)

=6/7+1/2 .(1-1/9)

=6/7+1/2.(8/9)

=6/7+4/9

=82/63

học tốt

sr bạn mình làm hơi chậm nhé

82/63 nha cậu

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

KHUYÊN NHỦ BẠN HÃY THAY ĐỔI NHỮNG HÀNH VI KO TỐT CỦA MIK.

Khoảng cách có rồi thì bạn áp dụng công thức : \(\frac{a}{m.n}=\frac{1}{m}-\frac{1}{n}\)(với n-m=a) là làm được

S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{93.95}+\frac{3}{95.98}+\frac{4}{98.102}+\frac{5}{102.17}+\frac{2012}{107..2119}\)

S=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+\frac{1}{102}-\frac{1}{107}+\frac{1}{107}-\frac{1}{2119}\)

S=\(\frac{1}{5}-\frac{1}{2119}\)

S=\(\frac{2114}{10595}\)

Bạn đăng tách 2 bài ra cho mn cùng giúp nhé 

Câu 2: 

a: Thay m=-1 vào (1), ta được:

\(x^2-2x+2\cdot\left(-1\right)+3=0\)

=>x=1

b: \(\text{Δ}=\left(2m+4\right)^2-4\left(2m+3\right)=4m^2+16m+16-8m-12\)

\(=4m^2-4m+4=\left(2m-1\right)^2+3>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Theo đề, ta có: \(\left(x_1+x_2\right)^2-2x_1x_2-1< =0\)

\(\Leftrightarrow\left(2m+4\right)^2-2\left(2m-3\right)-1< =0\)

\(\Leftrightarrow4m^2+16m+16-4m+6-1< =0\)

\(\Leftrightarrow4m^2+12m+21< =0\)

\(\Leftrightarrow m\in\varnothing\)

\(\frac{5x+7}{4}+\frac{3x+5}{8}>\frac{9x+4}{5}\)

\(\frac{10\cdot\left(5x+7\right)}{40}+\frac{5\cdot\left(3x+5\right)}{40}>\frac{8\cdot\left(9x+4\right)}{40}\)

10.(5x + 7) + 5.(3x + 5) > 8.(9x + 4)

10.(5x + 7) + 5.(3x + 5) - 8.(9x + 4) > 0

50x + 70 + 15x + 25 - 72x - 32 > 0

- 7x + 63 > 0

- 7.(x - 9) > 0

\(\Rightarrow x-9