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

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

\(\dfrac{a-3\sqrt{a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-3\sqrt{a}+3}{\sqrt{a^3}+\sqrt{3^3}}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-\sqrt{3a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)nếu cần trục căn thứ ở mẫu thì thế này :

\(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)

chúc bạn học tốt..........

@Nhã Doanh

a) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

a: \(=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

b: \(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}=\pm\left|ab\right|\)

c: \(=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)

a: \(=\sqrt{\dfrac{x^2+2x}{y^4}}=\dfrac{\sqrt{x^2+2x}}{y^2}\)

b: \(=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

c: \(=\left(a-b\right)\cdot\sqrt{\dfrac{a+b}{\left(a-b\right)}}\)

\(=\sqrt{\dfrac{\left(a+b\right)\cdot\left(a-b\right)^2}{a-b}}=\sqrt{a^2-b^2}\)

\(=\sqrt{\left(\dfrac{2\sqrt{5}+2}{3-\sqrt{5}}\right)^2\cdot\dfrac{24-8\sqrt{5}}{3+3\sqrt{5}}}\)

\(=\sqrt{\left(36+16\sqrt{5}\right)\cdot\dfrac{-16+8\sqrt{5}}{3}}\)

\(=\sqrt{\dfrac{64+32\sqrt{5}}{3}}\)

Lời giải:

ĐK: \(x\geq 0; x\neq 9\)

PT tương đương:

\(\frac{(\sqrt{x}+2)(\sqrt{x}+3)-5\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{22}{x-9}\)

\(\Leftrightarrow \frac{-4x+20\sqrt{x}+6}{x-9}=\frac{22}{x-9}\)

\(\Rightarrow -4x+20\sqrt{x}+6=22\)

\(\Leftrightarrow -x+5\sqrt{x}-4=0\)

\(\Leftrightarrow x-5\sqrt{x}+4=0\)

\(\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-4)=0\)

\(\Rightarrow \left[\begin{matrix} \sqrt{x}=1\rightarrow x=1\\ \sqrt{x}=4\rightarrow x=16\end{matrix}\right.\) (đều thỏa mãn)

\(\sqrt{4+2\sqrt{2}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right).\left(2-\sqrt{2+\sqrt{2}}\right)}\)

= \(\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)

=\(\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)

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

= \(\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=\sqrt{4}>\sqrt{3}\)

pt j ạ

Đặt \(\sqrt{15x}=a\)

Pt sẽ là \(\dfrac{5}{3a}-a+11=\dfrac{1}{3a}\)

=>\(\dfrac{4}{3a}=a-11\)

\(\Leftrightarrow3a^2-33a-4=0\)

=>\(a=11.12\)

=>căn 15x=11,12

=>15x=123,6544

hay \(x\simeq8,24\)

a) \(\sqrt{\dfrac{x}{y^3}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy}{y^4}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy+2x}{y^4}}=\dfrac{\sqrt{xy+2x}}{\sqrt{y^4}}=\dfrac{\sqrt{xy+2x}}{\left|y^2\right|}=\dfrac{\sqrt{xy+2x}}{y^2}\)(vì y²\(\ge0\))

b) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}.\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

c) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}\)

Nếu a-b>0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{a-b}=\left|ab\right|\)

Nếu a-b<0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{-\left(a-b\right)}=-\left|ab\right|\)

d) \(\dfrac{a-3\sqrt{a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}\right)^3+\left(\sqrt{3}\right)^3}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-3\sqrt{a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)

Nếu trục căn thức ở mẫu thì \(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{\left(\sqrt{a}+\sqrt{3}\right)\left(\sqrt{a}-\sqrt{3}\right)}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)

đây là b) nhé

cho sửa lại đề bài với ạ đề bài bài 1 là rút gọn biểu thức

2. a)

\(P=\left(\dfrac{3}{x+1}+\dfrac{1}{\sqrt{x+1}}\right):\dfrac{1}{\sqrt{x+1}}\)

\(=\left(\dfrac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x+1}}\right).\sqrt{x+1}\)

\(=\dfrac{3+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\sqrt{x+1}\)

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

b)\(P\left(16\right)=\dfrac{2+\sqrt{16}}{16-1}=\dfrac{6}{15}\)