So sánh: \(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)\) và \(\sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\).
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3 tháng 11 2017
Ta có :
\(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)=15-\dfrac{1}{\sqrt{13}}+1=16-\dfrac{1}{\sqrt{13}}\)
\(\sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)=17-\dfrac{1}{\sqrt{14}}-1=16-\dfrac{1}{\sqrt{14}}\)
Vì 13 < 14 \(\Rightarrow\sqrt{13}< \sqrt{14}\)
\(\Rightarrow\dfrac{1}{\sqrt{13}}>\dfrac{1}{\sqrt{14}}\)
\(\Rightarrow16-\dfrac{1}{\sqrt{13}}< 16-\dfrac{1}{\sqrt{14}}\)
\(\Rightarrow\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)
3 tháng 11 2017
Ta có: \(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)\)
\(=15-\dfrac{1}{\sqrt{13}}+1\)
\(=\left(15+1\right)-\dfrac{1}{\sqrt{13}}\)
\(=16-\dfrac{1}{\sqrt{13}}\)
Và: \(\sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)
\(=17-\dfrac{1}{\sqrt{14}}-1\)
\(=\left(17-1\right)-\dfrac{1}{\sqrt{14}}\)
\(=16-\dfrac{1}{\sqrt{14}}\)
Vì \(13< 14\Rightarrow\sqrt{13}< \sqrt{14}\Rightarrow\dfrac{1}{\sqrt{13}}>\dfrac{1}{\sqrt{14}}\Rightarrow-\dfrac{1}{\sqrt{13}}< -\dfrac{1}{\sqrt{14}}\Rightarrow16-\dfrac{1}{\sqrt{13}}< 16-\dfrac{1}{\sqrt{14}}\)
Hay \(\sqrt{225}-\left(\dfrac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\dfrac{1}{\sqrt{14}}+1\right)\)
Chúc bn học tốt 
24 tháng 4 2018
Xét \(\frac{1}{\sqrt{13}}>\frac{1}{\sqrt{14}}\Rightarrow\frac{1}{\sqrt{13}}-1< \frac{1}{\sqrt{14}}+1\)
Mà \(\sqrt{225}< \sqrt{289}\)
\(\Rightarrow\sqrt{225}-\left(\frac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\frac{1}{\sqrt{14}}+1\right)\)
Vậy....................
31 tháng 12 2023
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
26 tháng 11 2021
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
26 tháng 11 2021
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
27 tháng 8 2021
a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\dfrac{1-x}{2\sqrt{x}}\right)^2=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}=\dfrac{1-x}{\sqrt{x}}\)
b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)
\(\Rightarrow P>-2\sqrt{x}\)
HP
27 tháng 8 2021
a, ĐK: \(x\ge0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
a.
\(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-1,2}=\left(5^{-\dfrac{1}{2}}\right)^{-1,2}=5^{\left(-\dfrac{1}{2}\right).\left(-1,2\right)}=5^{0,6}>1\) do \(\left\{{}\begin{matrix}5>1\\0,6>0\end{matrix}\right.\)
b.
\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}=\left(5^{-1}\right)^{\sqrt{2}}=5^{-\sqrt{2}}< 1\) do \(\left\{{}\begin{matrix}5>1\\-\sqrt{2}< 0\end{matrix}\right.\)
12 tháng 1
a: \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{6}{5}}=\left(1:\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{5}{6}}=\left(\sqrt{5}\right)^{-\dfrac{5}{6}}\)
\(1=\left(\sqrt{5}\right)^0\)
mà -5/6<0 và \(\sqrt{5}>1\)
nên \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}>1\)
b: \(0< \dfrac{1}{5}< 1\)
=>\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}< \left(\dfrac{1}{5}\right)^0=1\)
3 tháng 8 2021
a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)
\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)
\(=\dfrac{5}{2}\sqrt{5}+7\)
b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)
\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)
\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)
\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)
\(=\dfrac{1}{\sin25^0}-1\)
\(=\dfrac{1-\sin25^0}{\sin25^0}\)
22 tháng 10 2021
a: \(Q=\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{a+2\sqrt{a}+1}{a-\sqrt{a}}\)
√225−(1√13 −1) < √289−(1√14 +1).
√225−(1√13 −1) < √289−(1√14 +1).