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a)\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
b)\(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{12}\right)^2=\frac{1}{144}\)
c)\(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{\left(20\cdot5\right)^4}{\left(25\cdot4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
d)\(\left(-\frac{10}{3}\right)^5\cdot\left(-\frac{6}{5}\right)^4=\left(-\frac{10}{3}\right)^4\cdot\left(-\frac{10}{3}\right)\cdot\left(-\frac{6}{5}\right)^4=\left[-\frac{10}{3}\cdot\frac{-6}{5}\right]^4\cdot\frac{-10}{3}=4^4\cdot\frac{-10}{3}=256\cdot\frac{-10}{3}=\frac{-2560}{3}\)
Bài làm
\(a,\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
\(=\left(\frac{3}{7}\right)^2+\left(\frac{1}{2}\right)^2\)
\(=\frac{9}{49}+\frac{1}{4}\)
\(=\frac{36}{196}+\frac{49}{196}\)
\(=\frac{85}{196}\)
\(b,\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
\(=\left(-\frac{1}{12}\right)^2\)
\(=\frac{1}{144}\)
\(c,\frac{5^4.20^4}{25^5.4^5}\)
\(=\frac{5^4.\left(5.4\right)^4}{\left(5.5\right)^5.4^5}\)
\(=\frac{5^4.5^4.4^4}{5^5.5^5.4^5}\)
\(=\frac{1}{5.5.4}\)
\(=\frac{1}{100}\)
~ Check đúng cho minh nha. ~
# Học tốt #
\(a,\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
\(< =>\left(\frac{6}{14}+\frac{7}{14}\right)^2\)
\(< =>\left(\frac{13}{14}\right)^2\)
\(< =>\frac{169}{196}\)
\(b,\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
\(< =>\left(\frac{9}{12}-\frac{10}{12}\right)^2\)
\(< =>\left(\frac{-1}{12}\right)^2\)
\(< =>\frac{-1}{144}\)
\(c,\frac{5^4\cdot20^4}{25^5\cdot4^5}\)
\(< =>\frac{25^2\cdot\left(4\right)^4\cdot\left(5\right)^4}{25^5\cdot4^5}\)
\(< =>\frac{1\cdot1\cdot\left(5\right)^4}{25^3\cdot4}\)
\(< =>\frac{1\cdot25^2}{25^3\cdot4}\)
\(< =>\frac{1}{25\cdot4}\)
\(< =>\frac{1}{100}\)
a) \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
b) \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=-\frac{2560}{3}\)
e) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2=\frac{17}{12}.\left(\frac{1}{20}\right)^2=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
f) \(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3=2:\left(-\frac{1}{6}\right)^3=2:-\frac{1}{216}=-432\)
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
Bài 1:
a) \(\left(\frac{1}{2}\right)^2\) và \(\left(\frac{1}{2}\right)^5\)
Ta có: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}.\)
\(\left(\frac{1}{2}\right)^5=\frac{1}{32}.\)
Vì \(\frac{1}{4}< \frac{1}{32}.\)
=> \(\left(\frac{1}{2}\right)^2< \left(\frac{1}{2}\right)^5.\)
b) \(\left(2,4\right)^3\) và \(\left(2,4\right)^2\)
Ta có: \(\left(2,4\right)^3=13,824.\)
\(\left(2,4\right)^2=5,76.\)
Vì \(13,284>5,76.\)
=> \(\left(2,4\right)^3>\left(2,4\right)^2.\)
c) \(\left(-1\frac{1}{2}\right)^2\) và \(\left(-1\frac{1}{2}\right)^3\)
Ta có: \(\left(-1\frac{1}{2}\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}.\)
\(\left(-1\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}.\)
Vì số dương luôn lớn hơn số âm nên \(\frac{9}{4}>-\frac{27}{8}.\)
=> \(\left(-1\frac{1}{2}\right)^2>\left(-1\frac{1}{2}\right)^3.\)
Chúc bạn học tốt!
1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))
2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)
\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)
Vậy \(x=\frac{-92}{105}\)
b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)
Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)
a ) 169/196
b ) 1/36
c ) 1/100
d ) -432