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c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
{[(6,2:0,31-5/6.0,9).0,2+0,15]:0,2}:[(2+1/4/11.0,22:0,1).1/33]
={[(6,2:0,31-0,75).0,2+0,15]:0,2}:[(2+0,3:0,1).1/33]
={[19,25.0,2+0,15]:0,2}:[5.1/33]
={4:0,2}:5/33
=20:5/33
=132
\(A=\frac{2}{3}+\frac{3}{4}\left(\frac{-4}{9}\right)\)
\(A=\frac{2}{3}+\frac{-1}{3}\)
\(A=\frac{1}{3}\)
\(B=\frac{2}{5}+\frac{3}{5}\div\left(-2\right)\)
\(B=\frac{2}{5}+\frac{-3}{10}\)
\(B=\frac{1}{10}\)
\(C=2\frac{3}{11}\cdot1\frac{1}{12}\cdot\left(-2,2\right)\)
\(C=\frac{325}{132}\cdot\left(-2,2\right)\)
\(C=\frac{-65}{12}\)
\(D=\left(\frac{3}{4}-0,2\right)\left(0,4-\frac{4}{5}\right)\)
\(D=\frac{11}{20}\cdot\frac{-2}{5}\)
\(D=\frac{-11}{50}\)
Tính:
0,4(3) + 0,6(2) . \(2\frac{1}{2}\).[(\(\frac{1}{2}+\frac{1}{3}\)) : 0,5(8)] : \(\frac{50}{53}\)
0,4(3) + 0,6(2). \(2\frac{1}{2}\).\(\left[\left(\frac{1}{2}+\frac{1}{3}\right):0,5\left(8\right)\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\frac{75}{53}:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{7}{3}\)
\(=\frac{83}{30}\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)
Các phần kia tương tự, à bạn đăng 1 2 câu hỏi 1 lần thôi, đăng nhiều quá ko ai trả lời đâu
@-@
Thế à ! Vậy bạn hãy nhấp vào https://h.vn/hoi-dap/question/646555.html?pos=1792187 mà xem
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 \(x=\frac{1}{12}\)