a: \(a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)\)

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

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

(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)

#Học tốt!!!

\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

\(a\cdot b\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{a\cdot b}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

+ 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+

\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)

a: \(=-\sqrt{a}\)

b: \(=\sqrt{p}\)

a) ĐS: .

b) ĐS: Nếu thì

Nếu ab

c) ĐS:

d)

Nhận xét. Nhận thấy rằng để có nghĩa thì Do đó . Vì thế có thể phân tích tử thành nhân tử.

a) ĐS: .

b) ĐS: Nếu thì

Nếu ab

c) ĐS:

d)

Nhận xét. Nhận thấy rằng để có nghĩa thì Do đó . Vì thế có thể phân tích tử thành nhân tử.

a. \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) = \(3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=-3+3\sqrt{6}\)

b.\(ab\sqrt{1+\dfrac{1}{a^2b^2}}=\sqrt{a^2b^2\left(1+\dfrac{1}{a^2b^2}\right)}=\sqrt{a^2b^2+1}\)

c.\(\sqrt{\dfrac{a}{b^3}+\dfrac{a}{b^4}}=\sqrt{\dfrac{ab+a}{b^4}}=\dfrac{\sqrt{ab+a}}{\sqrt{b^4}}=\dfrac{\sqrt{ab+a}}{b^2}\)

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

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

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

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

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

ĐS:

quá đúng

1234567890-01234567890-=qưertyuiop[]\';;lkjhfgdsazxcvbnm,./\'l;[]7894561230.+-

a) A=\cos x-\cos x+\sin ^{2} x \cdot \dfrac{1}{\cos ^{2} x}-\tan ^{2} x=0A=cosx−cosx+sin2x⋅cos2x1​−tan2x=0

b) B=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{1-\cos x+1+\cos x}{(1-\cos x)(1+\cos x)}}-\sqrt{2}B=sinx1​⋅(1−cosx)(1+cosx)1−cosx+1+cosx​​−2​

\begin{aligned}&=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{1-\cos ^{2} x}}-\sqrt{2}=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{\sin ^{2} x}}-\sqrt{2} \\&=\sqrt{2}\left(\dfrac{1}{\sin ^{2} x}-1\right)=\sqrt{2} \cot ^{2} x\end{aligned}​=sinx1​⋅1−cos2x2​​−2​=sinx1​⋅sin2x2​​−2​=2​(sin2x1​−1)=2​cot2x​

a, = \(\sqrt{a^2b^2.\left(1+\frac{1}{a^2b^2}\right)}\) = \(\sqrt{a^2b^2+1}\)

c, = \(\sqrt{\frac{a+ab}{b^4}}\) = \(\frac{\sqrt{a+ab}}{b^2}\)

k mk nha

a, \(ab\sqrt{1+\frac{1}{a^2b^2}}\)

 \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab\sqrt{\frac{1+a^2b^2}{a^2b^2}}=\frac{ab}{\left|ab\right|}\sqrt{1+a^2b^2}\)

\(=\hept{\begin{cases}\sqrt{1+a^2b^2}ĐK:ab>0\\-\sqrt{1+a^2b^2}ĐKab< 0\end{cases}}\)

b, \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}\)

\(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a+ab}{b^4}}=\frac{1}{b^2}\sqrt{a+ab}\)