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Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(\left|x-3\right|=2x+4\)
\(\left|x-3\right|=2x+2\cdot2\)
\(\left|x-3\right|=2\left(x+2\right)\)
\(\Rightarrow\orbr{\begin{cases}x-3=-\left[2\cdot\left(x+2\right)\right]\\x-3=2\left(x+2\right)\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-3=-\left[2x+4\right]\\x-3=2x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=-2x-4\\x=2x+2+3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-2x-4+3\\x=2x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2x-1\\x=2x+5\end{cases}}\) \(.....................\)
a) \(\frac{21}{52}=\frac{210}{520}=1-\frac{310}{520}\)
\(\frac{213}{523}=1-\frac{310}{523}\)
Vì \(520< 523\)\(\Rightarrow\frac{1}{520}>\frac{1}{523}\)\(\Rightarrow\frac{310}{520}>\frac{310}{523}\)
\(\Rightarrow1-\frac{310}{520}< 1-\frac{310}{523}\)
hay \(\frac{21}{52}< \frac{213}{523}\)
b) \(\frac{1515}{9797}=\frac{15.101}{97.101}=\frac{15}{97}\); \(\frac{171171}{991991}=\frac{171.1001}{991.1001}=\frac{171}{991}\)
Ta có: \(\frac{15}{97}=\frac{150}{970}=1-\frac{820}{970}\); \(\frac{171}{991}=1-\frac{820}{991}\)
Vì \(970< 991\)\(\Rightarrow\frac{1}{970}>\frac{1}{991}\)\(\Rightarrow\frac{820}{970}>\frac{820}{991}\)
\(\Rightarrow1-\frac{820}{970}< 1-\frac{920}{991}\)
hay \(\frac{1515}{9797}< \frac{171171}{991991}\)
c) \(\frac{n+2}{n+3}=1-\frac{1}{n+3}\); \(\frac{n+3}{n+4}=1-\frac{1}{n+4}\)
Vì \(n\inℕ^∗\)\(\Rightarrow n+3< n+4\)\(\Rightarrow\frac{1}{n+3}>\frac{1}{n+4}\)
\(\Rightarrow1-\frac{1}{n+3}< 1-\frac{1}{n+4}\)
hay \(\frac{n+2}{n+3}< \frac{n+3}{n+4}\)
d) \(\frac{n+7}{n+6}=1+\frac{1}{n+6}\); \(\frac{n+1}{n}=1+\frac{1}{n}\)
Vì \(n\inℕ^∗\)\(\Rightarrow n+6>n\)\(\Rightarrow\frac{1}{n+6}< \frac{1}{n}\)
\(\Rightarrow1+\frac{1}{n+6}< 1+\frac{1}{n}\)
hay \(\frac{n+7}{n+6}< \frac{n+1}{n}\)
B=3^10.11+3^10.5/3^9.2^4
= 3^10( 11+5)/3^9.16
= 3^10.16/3^9.16
= 3^10/3^9
= 3
Vậy B = 3 (1)
C = 2^10.13+2^10.65/2^8.104
= 2^10(13+65)/2^8.2^2.26
= 2^10.78/2^10.26
= 78/26
= 3
Vậy C = 3 (2)
Từ (1) v (2) suy ra B=C
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7