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\(\frac{1}{15}\)chứ ko phải \(\frac{1}{5}\)nha các bạn
\(\frac{1}{5}+\frac{9}{10}+\frac{14}{15}-\frac{1}{9}-\frac{20}{10}+\frac{1}{157}\)
\(=\frac{3}{15}+\frac{9}{10}+\frac{14}{15}-\frac{1}{9}-\frac{20}{10}+\frac{1}{157}\)
\(=\left(\frac{3}{15}+\frac{14}{15}\right)+\left(\frac{9}{10}-\frac{20}{10}\right)-\frac{1}{9}+\frac{1}{157}\)
\(=\frac{17}{15}+\frac{-11}{10}-\frac{1}{9}+\frac{1}{157}\)
\(=\left(\frac{51}{45}-\frac{5}{45}\right)+\frac{-11}{10}+\frac{1}{157}\)
\(=\frac{46}{45}+\frac{-11}{10}+\frac{1}{157}\)
\(=\left(\frac{92}{90}+\frac{-99}{90}\right)+\frac{1}{157}\)
\(=\frac{-1099}{14130}+\frac{90}{14130}\)
\(=\frac{-1099}{14130}\)

\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\\ \Leftrightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\\ \Rightarrow x+2=0\Leftrightarrow x=-2\)

Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)

\(\frac{7^5}{14^2}=\frac{7^5}{7^2.2^2}=\frac{7^3}{2^2}=\frac{343}{4}\)
\(\frac{14^4}{7^3}=\frac{7^4.2^4}{7^3}=7.2^4=7.16=112\)
\(\frac{15^3}{\left(-5\right)^4}=\frac{5^3.3^3}{\left(-5\right)^4}=\frac{3^3}{-5}=\frac{27}{-5}=-\frac{27}{5}\)

2) so sánh
Ta có \(\sqrt{17}\)>\(\sqrt{16}\)=4
\(\sqrt{26}\)>\(\sqrt{25}\)=5
=> \(\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}\)
=>\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1\)
=>\(\sqrt{17}+\sqrt{25}+1>5+4+1=10\)
Mà \(\sqrt{99}< \sqrt{100}=10\)
Vậy \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)
mk giúp bạn được câu 2 thôi
Xin lỗi nhá

\(\frac{-11}{14}-\frac{-4}{...}=\frac{-3}{14}\)
\(\frac{-11}{14}+\frac{4}{...}=\frac{-3}{14}\)
\(\frac{4}{...}=\frac{-3}{14}-\frac{-11}{14}\)
\(\frac{4}{...}=\frac{-3}{14}+\frac{11}{14}\)
\(\frac{4}{...}=\frac{8}{14}\)
\(\frac{4}{...}=\frac{4}{7}\)
\(\frac{1}{...}-\frac{-2}{15}=\frac{7}{15}\)
\(\frac{1}{...}+\frac{2}{15}=\frac{7}{15}\)
\(\frac{1}{...}=\frac{7}{15}-\frac{2}{15}\)
\(\frac{1}{...}=\frac{5}{15}\)
\(\frac{1}{...}=\frac{1}{3}\)
Hok tốt !!!!!!!!!
**Trả lời:
\(\frac{-15}{14}-\left(-\frac47\right)\)
\(=\frac{-15}{14}-\left(-\frac{8}{14}\right)\)
\(=-\frac{7}{14}\).