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a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(=>\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(=>\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(=>m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(=>\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(=>\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(=>n=3\)
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m =5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
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M = 54 - \(\frac{1}{2}\). \(\frac{2.3}{2}\)- \(\frac{1}{3}\). \(\frac{3.4}{2}\)- \(\frac{1}{4}\). \(\frac{4.5}{2}\)- ... - \(\frac{1}{12}\).\(\frac{12.13}{2}\)
= 54- \(\frac{3}{2}\)- \(\frac{4}{2}\)- \(\frac{5}{2}\)- ...- \(\frac{13}{2}\)
= 54 -\(\frac{1}{2}\). ( 1+2+3+4+5+6+...+12 -1-2)
= 54 \(\frac{1}{2}\). \(\frac{13.14}{2}-3\)
=54-\(\frac{1}{2}\)(91-3)
=54-\(\frac{1}{2}\).88
= 10
Vậy M = 10
( lưu ý : \(\frac{13.14}{2}-3\)ở trong ngoặc do k bt ghi kiểu j nên để đạm vậy )
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a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m = 5
Vậy m = 5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
Vậy n = 3
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\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{49}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10< S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50.10}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{3}{20}< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{4}{15}+\frac{1}{5}\)
\(\frac{3}{5}< S< \frac{4}{5}\left(đpcm\right)\)
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Câu 1:
a)
\(M=9.\left(0,5+1\frac{1}{2}\right)-\sqrt{64}\)
\(M=9.\left(\frac{1}{2}+\frac{3}{2}\right)-8\)
\(M=9.2-8\)
\(M=18-8\)
\(M=10.\)
b) Ta có: \(5x=2y.\)
\(\Rightarrow\frac{x}{2}=\frac{y}{5}.\)
\(\Rightarrow\frac{3x}{6}=\frac{2y}{10}\) và \(3x+2y=32.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{3x}{6}=\frac{2y}{10}=\frac{3x+2y}{6+10}=\frac{32}{16}=2.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=2\Rightarrow x=2.2=4\\\frac{y}{5}=2\Rightarrow y=2.5=10\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4;10\right).\)
Chúc bạn học tốt!
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I, Tìm x biết :
1.\(\frac{x}{-15}=\frac{-60}{x}\)
\(\Leftrightarrow2x=\left(-15\right).\left(-60\right)\)
\(\Leftrightarrow2x=900\)
\(\Leftrightarrow x=450\)
2. \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right).\left(x+7\right)=\left(x-1\right).\left(x+4\right)\)
\(\Leftrightarrow x^2+7x-2x-14=x^2+4x-x-4\)
\(\Leftrightarrow5x-14=3x-4\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Vậy : \(x=5\)
3)\(\frac{37-x}{x+13}=\frac{-3}{-7}=\frac{3}{7}\)
\(\Leftrightarrow\left(37-x\right).7=\left(x+13\right).3\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow220=4x\)
\(\Leftrightarrow x=55\)
Vậy : \(x=55\)
I.
1) \(\frac{x}{-15}=\frac{-60}{x}\)
=> \(x.x=\left(-60\right).\left(-15\right)\)
=> \(x.x=900\)
=> \(x^2=900\)
=> \(\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy \(x\in\left\{30;-30\right\}.\)
Chúc bạn học tốt!
\(\frac{1}{2}^m=\frac{1}{32}\)
\(\frac{1}{2}^m=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow m=5\)
đề thế này phải k bn
\(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow m=5\)
hc tốt nhá