a, \(\dfrac{2}{\sqrt{5}-1}=\dfrac{2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\left(\sqrt{5}+1\right)}{5-1}=\dfrac{\sqrt{5}+1}{2}\)

b, \(\left\{{}\begin{matrix}x-y=4\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}-y=4\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{11}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\)

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

b,\(\left\{{}\begin{matrix}x-y=4\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x=\dfrac{-3}{2}\end{matrix}\right.\)                       \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-11}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm duy nhất là (x;y)=\(\left(\dfrac{-3}{2};\dfrac{-11}{2}\right)\)

-Chúc bạn học tốt-

+ Ta có:

2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)

                   =2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5

                   =2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).

+ Ta có:

3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)

                    =3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7

                    =3(√10−√7)3=√10−√7=3(10−7)3=10−7.

+ Ta có:

1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)

=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y

+ Ta có:

2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)

=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.

\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)

\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)

\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)

\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)

33+1=3(3−1)(3+1)(3−1)=3(3−1)3−1=3(3−1)2.

23−1=2(3+1)(3−1)(3+1)=2(3+1)3−1=3+

2+32−3=(2+3)(2+3)(2−3)(2

b3+b=b(3−b)(3+b)(3−b)=b(3−b)32−(b)2=b(3−b)9−b.

p2.p−1=p(2.p+1)(2.p−1)(2.p+

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

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

\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)

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

ĐS:

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

Bài 3:

a.

\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)

\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

b.

\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)

\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)

a, \(\frac{a}{\sqrt{a}}=\sqrt{a}\)

b, \(\frac{a}{\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)

c, \(\frac{x}{\sqrt{3x^3}}=\frac{x}{x\sqrt{3x}}=\frac{1}{\sqrt{3x}}=\frac{\sqrt{3x}}{3x}\)

d, \(\frac{4y^2}{\sqrt{2y^5}}=\frac{4y^2}{y^2\sqrt{2y}}=\frac{4}{\sqrt{2y}}=\frac{4\sqrt{2y}}{2y}=\frac{2\sqrt{2y}}{y}\)

a)\(\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}\)                b) \(\dfrac{a}{\sqrt{ab}}=\dfrac{a\sqrt{ab}}{\left(\sqrt{ab}\right)^2}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b}\)                                              c) \(\dfrac{x}{\sqrt{3x^3}}=\dfrac{x\sqrt{3x}}{\sqrt{3x^3.\sqrt{3x}}}=\dfrac{x\sqrt{3x}}{\left(\sqrt{3x^2}\right)^2}=\dfrac{x\sqrt{3x}}{\left(3x^2\right)^2}=\dfrac{x\sqrt{3x}}{3x^2}=\dfrac{\sqrt{3x}}{3x}\)    

a, căn a 

b, căn ab/b

c, căn 3x/3x

d, 2 căn 2y/y

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

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

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

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

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

Trước hết, ta cần tính giá trị của a và b trong G và H:
$$G^2 = \frac{1}{a+b} \Rightarrow a+b = \frac{1}{G^2}$$
$$H^2 = 4a - 4\sqrt{ab} + 4b = 4(\sqrt{a} - \sqrt{b})^2 \Rightarrow \sqrt{a} - \sqrt{b} = \frac{H}{2}$$
Từ đó, suy ra được:
$$\sqrt{a} + \sqrt{b} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4}$$
$$\Rightarrow 2\sqrt{a} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H$$
$$\Rightarrow a = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H\right)^2/4$$
$$\Rightarrow b = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} - H\right)^2/4$$

Tiếp theo, ta tính giá trị của F:
$$F = 6\sqrt{3} + \sqrt{2} = 6\sqrt{3} + \sqrt{2}\frac{\sqrt{6}+\sqrt{2}}{2} = 6\sqrt{3} + 3\sqrt{2} + 3\sqrt{6}$$

Cuối cùng, ta tính giá trị của K:
$$K = 2xy\left(2\sqrt{x} + 3\sqrt{y}\right) = 2\sqrt{xy}(4\sqrt{x} + 6\sqrt{y})$$

Vậy, ta đã tính được giá trị của F, G, H và K.

a, căn a trừ b/ a-b^2

b, 2 căn a + 2 căn b /a-b

c, 6 căn a trừ 6 / 4a-1

a) căn a-b/a-b^2

b) 2(căn a+căn b)/a-b

c) 3(2 căn a-1)/4a-1

d)2xy(2 căn x-3 căn y)/4x-9y

a) √a -b /a-b²

b) 2√a+2b / a-b²

c) 6√a -3 / 4a-1

d) 4xy√x -6xy√y / 4x-9y

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

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

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

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

\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)

a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)

\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)

\(B=\dfrac{\left(1+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{-1}=-2-3\sqrt{5}-5=-7-3\sqrt{5}\)

\(C=\dfrac{5\sqrt{x}-x}{2x}\)

\(D=\dfrac{\left(\sqrt{a}+1\right)\left(2\sqrt{a}+1\right)}{4a-1}\)

\(E=\dfrac{15}{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{75}+\sqrt{45}}{2}\)