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\(\frac{\sqrt{ab}-1}{3}=\frac{\sqrt{bc}-3}{9}=\frac{\sqrt{ac}-5}{-6}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}-9}{6}=\frac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{\sqrt{ab}-1}{3}=\frac{1}{3}\\\frac{\sqrt{bc}-3}{9}=\frac{1}{3}\\\frac{\sqrt{ac}-5}{-6}=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{ab}=2\\\sqrt{bc}=6\\\sqrt{ac}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=4\\bc=36\\ac=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=9a\\ac=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=9\end{matrix}\right.\)
Lời giải:
Đặt \(\frac{\sqrt{ab}-1}{3}=\frac{\sqrt{bc}-3}{9}=\frac{\sqrt{ca}-5}{-6}=t\)
\(\Rightarrow \left\{\begin{matrix} \sqrt{ab}=3t+1\\ \sqrt{bc}=9t+3\\ \sqrt{ca}=5-6t\end{matrix}\right.\)
\(\Rightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6t+9\)
\(\Leftrightarrow 11=6t+9\Leftrightarrow t=\frac{1}{3}\)
Khi đó : \(\left\{\begin{matrix} \sqrt{ab}=2\\ \sqrt{bc}=6\\ \sqrt{ac}=3\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} ab=4\\ bc=36\\ ac=9\end{matrix}\right.\Rightarrow abc=\sqrt{4.36.9}=36\)
\(\Rightarrow \left\{\begin{matrix} c=\frac{abc}{ab}=9\\ a=\frac{abc}{bc}=1\\ b=\frac{abc}{ac}=4\end{matrix}\right.\)
Vậy....
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)
\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)
\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)
\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)
\(=3,75.\left(7,2+2,8\right)\)
\(=3,75.10=37,5\)
\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)
\(=\frac{-3}{7}+-\frac{4}{7}=-1\)
\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)
\(=9-\frac{1}{8}.8+0,2\)
\(=9-1+0,2=8+0,2=8,2\)
Lớp 7 vừa học hằng đẳng thức, chú ý hằng đẳng thức sau: (a - b)(a + b) = a2 - b2.
Bạn cần khử căn dưới mẫu và cộng tổng bên trái, muốn vậy bạn phải đánh giá từng phân số bằng cách làm trội nó
Sử dụng đánh giá sau: \(\frac{1}{\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k-1}}=\sqrt{k}-\sqrt{k-1}\)
Ta có:
\(\frac{1}{\sqrt{1}}>\frac{10}{\sqrt{100}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}};...;\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=10\)(đpcm)