Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}.1\)
=\(\frac{3+39}{7+91}\)
=\(\frac{42}{98}\)
=\(\frac{3}{7}\)
b)\(\sqrt{\left(2,5-0,7\right)^2}\)
=\(|2,5-0,7|\)
=2,5-0,7
=1,8
*\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
Vậy phương trình đã cho có \(S=\left\{0;\dfrac{1}{7}\right\}\)
* \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4x}=\dfrac{2}{5}\)
\(\Leftrightarrow15x+1=8x\)
\(\Leftrightarrow7x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{7}\left(tmđk\right)\)
Vậy phương trình đã cho có \(S=\left\{\dfrac{-1}{7}\right\}\)
a, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\Rightarrow x=\dfrac{-5}{7}\)
Vậy \(x=\dfrac{-5}{7}\)
Chúc bạn học tốt!!!
A=\(\sqrt{\frac{9}{16}}\)+\(2016^0+\left|-0,25\right|\)
A=\(\frac{3}{4}\)+1+0,25
A=2
Ta có:1+2+3+..+(n-1)
=>số số hạng của tổng trên là:\(\frac{\left(n-1\right)-1}{1}\) +1=n-2+1=n-1
vậy:1+2+3+..+(n-1)=[(n-1)+1].(n-1):2=n(n-1):2
=>\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+..+3+2+1}\)
\(\sqrt{n\left(n-1\right):2.2+n}\)
\(\sqrt{n\left(n-1\right)+n}\)
\(\sqrt{n.n-n+n}\)
\(\sqrt{\sqrt{n}}\)=n
vậy\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+..+3+2+1}\)
=n(dpcm)
Ta có:
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(...............\)
\(\dfrac{1}{\sqrt{98}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
Cộng theo vế ta có:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{99}{10}\)
Lại có \(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\) suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=\dfrac{100}{10}=10\)
Ta có:
1/√1>1/√100=1/10
1/√2>1/√100=1/10
........
1/√100=1/√100=1/10
Nên:
1/√1+1/√2+...+1/√100>1/10+1/10+...+1/10(100 phân số 1/10)
=1/√1+1/√2+..+1/√100>100/10
1/√1+1/√2+..+1/√100>10(đpcm)
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+....+\dfrac{1}{\sqrt{100}+\sqrt{99}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{100}}\)
\(=\dfrac{\sqrt{100}-1}{\sqrt{100}}\)