Bài 3: Liên hệ giữa phép nhân và phép khai phương

ĐK : \(x\ge0\) và \(x\ne1\)

\(A=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-x\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}\)

Ta có : \(\dfrac{2}{A}+\sqrt{x}=\dfrac{-2x-2\sqrt{x}-2}{\sqrt{x}}+\sqrt{x}\)

\(=\dfrac{-x-2\sqrt{x}-2}{\sqrt{x}}=-\sqrt{x}-2-\dfrac{2}{\sqrt{x}}=-\left(\sqrt{x}+\dfrac{2}{\sqrt{x}}+2\right)\)

Theo BĐT Cô - si ta có : \(\sqrt{x}+\dfrac{2}{\sqrt{x}}\ge2\sqrt{2}\)

\(\Leftrightarrow\sqrt{x}+\dfrac{2}{\sqrt{x}}+2\ge2\sqrt{2}+2\)

\(\Leftrightarrow-\left(\sqrt{x}+\dfrac{2}{\sqrt{x}}+2\right)\le-2\sqrt{2}-2\)

Vậy GTLN của Q là \(-2\sqrt{2}-2\) . Dấu \("="\) xảy ra khi \(x=2\)

1)

a) \(\sqrt{0,49.169.25}=\sqrt{0,49}.\sqrt{169}.\sqrt{25}=0,7.13.5=45,5\)

b) \(\dfrac{\sqrt{63}}{\sqrt{112}}=\sqrt{\dfrac{63}{112}}=\sqrt{\dfrac{9}{16}}=\dfrac{\sqrt{9}}{\sqrt{16}}=\dfrac{3}{4}\)

c) \(\sqrt{\dfrac{27}{17}}.\sqrt{11.\dfrac{5}{17}}=\sqrt{\dfrac{27}{17}.11.\dfrac{5}{17}}=\sqrt{\dfrac{1485}{17^2}}=\dfrac{\sqrt{1485}}{\sqrt{17^2}}=\dfrac{\sqrt{9.165}}{17}=\dfrac{\sqrt{9}.\sqrt{165}}{17}=\dfrac{3\sqrt{165}}{17}\)

d) \(\sqrt{\left(-5\right)^2.16.225}=\sqrt{\left(-5\right)^2}.\sqrt{16}.\sqrt{225}=\left|-5\right|.4.15=5.60=300\)

e) \(\dfrac{\sqrt{2,5}}{\sqrt{1,6}}=\sqrt{\dfrac{2,5}{1,6}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{5}{4}\)

Ta có : \(\dfrac{1}{\sqrt{2019}-\sqrt{2018}}=\dfrac{\sqrt{2019}+\sqrt{2018}}{\left(\sqrt{2019}-\sqrt{2018}\right)\left(\sqrt{2019}+\sqrt{2018}\right)}=\dfrac{\sqrt{2019}+\sqrt{2018}}{2019-2018}=\sqrt{2019}+\sqrt{2018}< \sqrt{2020}+\sqrt{2019}\)

c)Ta có: \(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2017+2\sqrt{\left(2016-1\right)\left(2016+1\right)}+2015\\ =2.2016+2\sqrt{2016^2-1}\)

\(\left(2\sqrt{2016}\right)^2=4.2016=2.2016+2.\sqrt{2016^2}\)

Vì \(2\sqrt{2016^2-1}< 2\sqrt{2016^2}\) nên \(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\)

\(\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)

a) Ta có \(\sqrt{5}+\sqrt{7}>\sqrt{4}+\sqrt{4}=2+2=4=\sqrt{16}>\sqrt{13}\)

Vậy \(\sqrt{5}+\sqrt{7}>\sqrt{13}\)

b) Ta có \(\sqrt{15}.\sqrt{17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)

\(16=\sqrt{16^2}\)

Vì \(16^2>16^2-1\) nên \(\sqrt{16^2}>\sqrt{16^2-1}\Leftrightarrow16>\sqrt{15}.\sqrt{17}\)

\(\text{c) Ta có }:\left(\sqrt{2015}+\sqrt{2017}\right)^2=2017+2\sqrt{2015\cdot2017}+2015\\ =2016+1+2\sqrt{\left(2016-1\right)\left(2016+1\right)}+2016-1\\ =2\cdot2016+2\sqrt{2016^2-1}\\ >2\cdot2016+2\sqrt{2016^2}\\ =2\cdot2016+2\cdot2016\\ =4\cdot2016=\left(2\sqrt{2016}\right)^2\)

\(\Rightarrow\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)

Vậy........

Với 2 số tự nhiên a;b khác nhau, ta có: \(2\sqrt{ab}< a+b\Leftrightarrow\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\).( Tự cm)

Đặt biểu thức vế trái là A thì:

\(A=2\left(\dfrac{1}{\sqrt{1.2005}}+\dfrac{1}{\sqrt{2.2004}}+...+\dfrac{1}{\sqrt{1002.1004}}\right)+\dfrac{1}{1003}\)

Áp dung bđt vừa cm, ta đc:

\(A>2\left(\dfrac{1}{1003}+\dfrac{1}{1003}+...+\dfrac{1}{1003}\right)+\dfrac{1}{1003}=\dfrac{2005}{2003}\)=> ĐPCM

Gỉa sử : \(3+\sqrt{5}>2\sqrt{2}+\sqrt{6}\)

\(\Leftrightarrow14+6\sqrt{5}>14+8\sqrt{3}\)

\(\Leftrightarrow6\sqrt{5}>8\sqrt{3}\)

\(\Leftrightarrow\sqrt{180}>\sqrt{192}\)( vô lý )

\(\Rightarrow3+\sqrt{5}< 2\sqrt{2}+\sqrt{6}\)

3.a.Ta có :\(\sqrt{40^2-24^2}=\sqrt{\left(40-24\right)\left(40+24\right)}=\sqrt{16.64}=4.8=32\)

b.Ta có :\(\sqrt{52^2-48^2}=\sqrt{\left(52-48\right)\left(52+48\right)}=\sqrt{4.100}=2.10=20\)

4.a)Ta có :

\(\sqrt{4x}=8\Leftrightarrow4x=8^2\Leftrightarrow4x=64\Leftrightarrow x=16\left(tm\right)\)

Vậy x=16

b)Ta có :

\(\sqrt{0,7x}=6\Leftrightarrow0,7x=36\Leftrightarrow x=\dfrac{36}{0.7}\left(tm\right)\)

Vậy x=\(\dfrac{36}{0.7}\)

c)Ta có:

\(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

Vậy x=4

d)Ta có :

\(\sqrt{5x}< 6\Leftrightarrow5x< 36\Leftrightarrow x< \dfrac{36}{5}\)

vậy 0≤x<\(\dfrac{36}{5}\)

Bài 3

a) \(\sqrt{40^2-24^2}\)

\(=\sqrt{\left(40+24\right)\left(40-24\right)}\)

=\(\sqrt{64.16}=\sqrt{64}.\sqrt{16}\)

\(=8.4=24\)

b)\(\sqrt{52^2-48^2}\)

\(=\sqrt{\left(52+48\right)\left(52-48\right)}\)

\(=\sqrt{100.4}=\sqrt{100}.\sqrt{4}\)

=10.2=20

Bài 4

a)\(\sqrt{4x}=8\)

\(\Leftrightarrow2\sqrt{x}=8\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\)(TM)

b)\(\sqrt{0,7x}=6\)

\(\Leftrightarrow\sqrt{\left(0,7x\right)^2}=6^2\)

\(\Leftrightarrow\left|0,7x\right|=36\)

\(\Leftrightarrow\left[{}\begin{matrix}0,7x=36\\0,7x=-36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{360}{7}\left(TM\right)\\x=-\dfrac{360}{7}\left(KTM\right)\end{matrix}\right.\)

c)\(9-4\sqrt{x}=1\)

\(\Leftrightarrow4\sqrt{x}=8\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\)(TM)

d)\(\sqrt{5x}< 6\)