Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)

a: \(=\left(\sqrt{13}-2y\right)\left(\sqrt{13}+2y\right)\)

b: \(=\left(x-\sqrt{3}\right)^2-2y^2\)

\(=\left(x-\sqrt{3}-y\sqrt{2}\right)\left(x-\sqrt{3}+y\sqrt{2}\right)\)

c: \(=x\sqrt{x}-y\sqrt{y}+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

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

a: \(A=\sqrt{15}-\sqrt{14}=\dfrac{1}{\sqrt{15}+\sqrt{14}}\)

\(B=\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

mà căn 15>căn 13

nen A<B

b: \(A=\sqrt{7}-\sqrt{5}=\dfrac{2}{\sqrt{7}+\sqrt{5}}\)

\(B=\sqrt{5}-\sqrt{3}=\dfrac{2}{\sqrt{5}+\sqrt{3}}\)

mà căn 7>căn 3

nen A<B

c: \(A=\dfrac{4}{\sqrt{105}+\sqrt{101}}\)

\(B=\dfrac{4}{\sqrt{101}+\sqrt{97}}\)

mà căn 105>căn 97

nên A<B

Bài `1:`

`a)[3+\sqrt{3}]/[1+\sqrt{3}]=[\sqrt{3}(1+\sqrt{3})]/[1+\sqrt{3}]=\sqrt{3}`

`b)[\sqrt{14}-\sqrt{7}]/[2-\sqrt{2}]=[\sqrt{7}(\sqrt{2}-1)]/[\sqrt{2}(\sqrt{2}-1)]=\sqrt{7}/\sqrt{2}=\sqrt{14}/2`

`c)4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{1/5}`

`=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}.\sqrt{5/5}`

`=8\sqrt{5}-3\sqrt{5}=5\sqrt{5}`

`d)(2\sqrt{8}+3\sqrt{5}-7\sqrt{2})(\sqrt{72}-5\sqrt{20}-2\sqrt{2})`

`=(4\sqrt{2}+3\sqrt{5}-7\sqrt{2})(6\sqrt{2}-10\sqrt{5}-2\sqrt{2})`

`=(3\sqrt{5}-3\sqrt{2})(4\sqrt{2}-10\sqrt{5})`

`=12\sqrt{10}-150-24+30\sqrt{10}`

`=42\sqrt{10}-174`

e: \(=2\sqrt{5}+4+2\sqrt{5}-4=4\sqrt{5}\)

a: =5*2+3*5-8=17

b: =căn 2+căn 2-3căn 2=-căn 2

c: =2-căn 3+5-căn 3

=7-2căn 3

d: =(căn 7+căn 5)(căn 7-căn 5)=7-5=2

1,

1\(1,\dfrac{1}{3}\sqrt{51}=\dfrac{\sqrt{51}}{3}\\ \dfrac{1}{5}\sqrt{150}=\sqrt{6}\)

\(\dfrac{\sqrt{51}}{3};\sqrt{6}=\dfrac{3\sqrt{6}}{3}=\dfrac{\sqrt{54}}{3}\)

\(51< 54=>\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)

2, 

\(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{1}{4}.6}=\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\\ 6\sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{36.1}{2}}=\sqrt{18}=3\sqrt{2}\)

\(\dfrac{\sqrt{6}}{2};3\sqrt{2}=\dfrac{2.3\sqrt{2}}{2}=\dfrac{6\sqrt{2}}{2}=\dfrac{\sqrt{72}}{2}\\ \sqrt{6}< \sqrt{72}=>\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)

1) Có: \(\dfrac{1}{3}\sqrt{51}>\dfrac{1}{3}\sqrt{50}=\dfrac{1}{3}.\sqrt{5}\sqrt{10}=\dfrac{\sqrt{5}}{3}\sqrt{10}\)

\(\dfrac{1}{5}\sqrt{150}=\dfrac{1}{5}.\sqrt{15}\sqrt{10}=\dfrac{\sqrt{15}}{5}\sqrt{10}=\dfrac{\sqrt{3}\sqrt{5}}{\sqrt{5}\sqrt{5}}\sqrt{10}=\sqrt{\dfrac{3}{5}}\sqrt{10}< \dfrac{\sqrt{5}}{3}\sqrt{10}=\dfrac{1}{3}\sqrt{51}\)

2) Có: \(\dfrac{1}{2}\sqrt{6}=\dfrac{\sqrt{6}}{\sqrt{2}\sqrt{2}}=\dfrac{\sqrt{2}\sqrt{3}}{\sqrt{2}\sqrt{2}}=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{3}{2}}\)

\(6\sqrt{\dfrac{1}{2}}=\dfrac{6.1}{\sqrt{2}}=\dfrac{\sqrt{2}\sqrt{2}.3}{\sqrt{2}}=\sqrt{2}.3>\sqrt{\dfrac{3}{2}}=\dfrac{1}{2}\sqrt{6}\)

a: \(=\dfrac{\sqrt{7}}{10}\)

b: \(=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

c: \(=\sqrt{\dfrac{3}{36}}=\dfrac{\sqrt{3}}{6}\)

d: \(=\sqrt{\dfrac{15}{100}}=\dfrac{\sqrt{15}}{10}\)

a) \(\sqrt{\dfrac{7}{100}}=\dfrac{\sqrt{7}}{10}\)

b) \(\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{\sqrt{5}\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

c) \(\sqrt{\dfrac{1}{12}}=\sqrt{\dfrac{3}{36}}=\dfrac{\sqrt{3}}{6}\)

d) \(\sqrt{\dfrac{3}{20}}=\sqrt{\dfrac{15}{100}}=\dfrac{\sqrt{15}}{10}\)

e) \(\sqrt{\dfrac{1}{20}}=\sqrt{\dfrac{5}{100}}=\dfrac{\sqrt{5}}{10}\)

f) \(\sqrt{\dfrac{1}{60}}=\sqrt{\dfrac{15}{900}}=\dfrac{\sqrt{15}}{30}\)

a: \(=2\cdot\left(\sqrt{5}-2\right)-6-2\sqrt{5}\)

\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)

b: \(=\sqrt{5}-\sqrt{3}\)

c: \(=\sqrt{2}+1+\dfrac{2+\sqrt{22}}{18}\)

\(=\dfrac{18\sqrt{2}+20+\sqrt{22}}{18}\)

\(=\sqrt{169\cdot49}=13\cdot7=91\)

căn tất cả 16.9*490

\(\sqrt{16.9}\) `xx` \(490\)

`=` \(\sqrt{144}\) `xx` \(490\)

`= 12 xx 490`

`= 5880` 

\(=\dfrac{6\left(5-2\sqrt{3}\right)}{25-12}=\dfrac{30-12\sqrt{3}}{13}\)