Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+1\right)\)

=(1-căn 5)(1+căn 5)

=1-5

=-4

\(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+1\right)\)

\(=\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)

\(=1^2-\left(\sqrt{5}\right)^2\)

\(=1-5\)

\(=-4\)

`(1+(5-\sqrt{5})/(1-\sqrt{5})).((5+\sqrt{5})/(1+\sqrt{5})+1)`

`=[1-(\sqrt{5}(\sqrt{5}-1))/(\sqrt{5}-1)].[(\sqrt{5}(\sqrt{5}+1))/(1+\sqrt{5})+1]`

`=(1-\sqrt{5})(1+\sqrt{5})`

`=1-5`

`=-4`

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

\(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)\)

=-3

\(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}:\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

\(=\dfrac{-\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}:\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

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

\(=-\sqrt{2}-5+\sqrt{10}\)

(√6-√2)/(1-√3)-5/√5):1/(√5-√2)

\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)

\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)

Mình sửa lại chút nha bạn

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+8\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+8\sqrt{2}}{6}\)

38.

$\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}< \frac{1}{\sqrt{2005}+\sqrt{2004}}=\frac{2005-2004}{\sqrt{2005}+\sqrt{2004}}=\sqrt{2005}-\sqrt{2004}$

39.

$\sqrt{1998}+\sqrt{2000}-2\sqrt{1999}=(\sqrt{2000}-\sqrt{1999})-(\sqrt{1999}-\sqrt{1998})$

$=\frac{1}{\sqrt{2000}+\sqrt{1999}}-\frac{1}{\sqrt{1999}+\sqrt{1998}}$

$< 0$

$\Rightarrow \sqrt{1998}+\sqrt{2000}<2\sqrt{1999}$

Bài 40 bạn làm tương tự câu 38.

(2-căn 2)^2=6-4*căn 2=1/4+23/4-4căn 2

(1/2)^2=1/4

mà 23/4-4căn 2>0

nên 2-căn 2>1/2

\(\dfrac{5}{2}=\sqrt{\dfrac{25}{4}}>\sqrt{\dfrac{20}{4}}=\sqrt{5}\)

=>\(\sqrt{5}< \dfrac{5}{2}\)

=>căn 5-5/2<0

`5/2 = 2,5 = sqrt(6,25)`

Do `5 < 6,25` nên `sqrt 5 - 5/2 < 0`.

Lời giải:

Đặt $A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}$. Hiển nhiên $A<0$

$A^2=12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{(12-3\sqrt{7})(12+3\sqrt{7})}$

$=24-2\sqrt{12^2-(3\sqrt{7})^2}=24-2\sqrt{81}=6$

$\Rightarrow A=-\sqrt{6}$ (do $A<0$)

Lần sau  bạn lưu ý:

- Ghi đây đủ yêu cầu đề bài

- Gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Người đọc đọc đề dễ thì khả năng hỗ trợ cũng cao hơn.

\(2\sqrt{5}-3\sqrt{125}+\sqrt{52}-2\sqrt{8}\)

\(=2\sqrt{5}-15\sqrt{5}+2\sqrt{13}-4\sqrt{2}\)

\(=-13\sqrt{5}+2\sqrt{13}-4\sqrt{2}\)

a: căn 14<4

=>7+căn 14<4+7=11

b: -căn 5<-2

=>-căn 5+9<-2+9=7

d: \(\sqrt{145}< 13\)
=>-11+căn 145<-11+13=2

e: \(7-4\sqrt{5}+2=9-4\sqrt{5}>0\)

=>7-4căn 5>-2

f: -4căn 5>-9

=>-9-4căn 5>-9-9=-18