\(\Rightarrow2C=2^2+2^3+...+2^{101}-2^{102}\)

\(\Rightarrow C=2C-C=2^2+2^3+...+2^{101}-2^{102}-2-2^2-...-2^{100}+2^{101}\)

\(\Rightarrow C=2.2^{101}-2^{102}-2=2^{102}-2^{102}-2=-2\)

Đặt \(A=2^{x+9}-2^{x+8}-2^{x+7}-...-2^{x+1}-2^x\)

\(\Rightarrow2A=2\left(2^{x+9}-2^{x+8}-...-2^x\right)\)

\(\Rightarrow2A=2^{x+9}.2^1-2^{x+8}.2^1-...-2^x.2^1\)

\(\Rightarrow2A=2^{x+10}-2^{x+9}-...-2^{x+1}\)

\(\Rightarrow A=2A-A=2^{x+10}-2^{x+9}-...-2^{x+1}-\left(2^{x+9}-2^{x+8}-...-2^{x+1}-2^x\right)=2^{x+10}-2^{x+9}-2^{x+9}+2^x\)

\(\Rightarrow A=2^{x+10}-2.2^{x+9}+2^x=2^{x+10}-2^{x+10}+2^x=2^x\)

\(\Rightarrow2^x=1024\Rightarrow x=10\)

c) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)

e) \(\dfrac{1}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{1}{5}=\dfrac{1}{5}.1-\dfrac{1}{5}=0\)

\(C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{779.2}{780.2}\)

\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{38.41}{39.40}\)

\(=\dfrac{1.2.3...38}{2.3.4...39}.\dfrac{4.5.6...41}{3.4.5...40}=\dfrac{1}{39}.\dfrac{41}{3}=\dfrac{41}{117}\)

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Í giờ em mới để ý lớp 10 :(( Tại lớp 9 em mới học có nhiu đó à 

a) \(\Rightarrow x\left(x^2-0,25\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)

b) Thiếu đề rồi bạn

c) \(\Rightarrow\left(x-1\right)\left(x^2+x+1\right)=0\Rightarrow x-1=0\Rightarrow x=1\)

(do \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

d) \(\Rightarrow2x\left(3x-2\right)-1\left(3x-2\right)=0\Rightarrow\left(3x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)