a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

a) = . = . = vì x > 0.

Do đó = .

b) = . = ..

Vì y < 0 nên │y│= -y. Do đó = . = .

c) 5xy. = 5xy. = 5xy..

Vì x < 0, y > 0 nên = -x và = .

Do đó: 5xy = 5xy. = -.

d) 0,2 = = 0,2

a: ĐKXĐ: x<>1/2

Sửa đề: \(A=\frac{3}{2\left(2x-1\right)}\cdot\sqrt{8x^4\left(4x^2-4x+1\right)}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot\sqrt8\cdot\sqrt{x^4}\cdot\sqrt{\left(2x-1\right)^2}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot2\sqrt2\cdot x^2\cdot\left|2x-1\right|=\frac{6\sqrt2\cdot x^2}{2\left(2x-1\right)}\cdot\left|2x-1\right|\)

=\(\pm3\sqrt2\cdot x^2\)

b: ĐKXĐ: b<>0

\(B=\frac{a-b}{b^2}\cdot\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{b^2}\cdot\sqrt{a^2}\cdot\frac{\sqrt{b^4}}{\sqrt{\left(a-b\right)^2}}\)

\(=\frac{a-b}{b^2}\cdot\left|a\right|\cdot\frac{b^2}{\left|a-b\right|}=\left|a\right|\cdot\frac{a-b}{\left|a-b\right|}=\pm\left|a\right|\)

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.

Bài 1:

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

\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}-1}\right)\cdot\frac{a-1}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}-1}\cdot\left(\sqrt{a}-1\right)=\sqrt{a}-1\)

b: A<0

=>\(\sqrt{a}-1<0\)

=>\(\sqrt{a}<1\)

=>0<a<1

Bài 2:

\(A=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

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

\(=\frac{3+\sqrt{x}}{\sqrt{x}-2}:\left(\sqrt{x}+3\right)=\frac{1}{\sqrt{x}-2}\)

\(B=3\sqrt8-\sqrt{50}-\sqrt{\left(\sqrt2-1\right)^2}\)

\(=3\cdot2\sqrt2-5\sqrt2-\left|\sqrt2-1\right|\)

\(=\sqrt2-\left(\sqrt2-1\right)=1\)

\(C=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)

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

\(D=\left(\frac{1-a\cdot\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

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

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}\)

=1

a)\(\dfrac{\sqrt{243a}}{\sqrt{3a}}=\dfrac{\sqrt{24}.\sqrt{3a}}{\sqrt{3a}}=2\sqrt{6}\)

b)\(\dfrac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}=3\sqrt{9b^2}=\left[{}\begin{matrix}9b\\-9b\end{matrix}\right.\)

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

\(=\left(\dfrac{2-2a}{a\left(a+1\right)}\right):\dfrac{1-a}{\left(a+1\right)^1}\)

\(=\dfrac{\left(2-2a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}\)

\(=\dfrac{2\left(1-a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}=\dfrac{2\left(a+1\right)}{a}\)

a.\(=\sqrt{2}.\left(\sqrt{25}-\sqrt{9}\right)=\sqrt{2}.\left(5-3\right)=2\sqrt{2}\)

2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)