Chương I - Căn bậc hai. Căn bậc ba

\(P=\dfrac{a+1}{\sqrt{a}}+\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}+\dfrac{a^2-a\sqrt{a}+\sqrt{a}-1}{\sqrt{a}-a\sqrt{a}}\)

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

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

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

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

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

\(P=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}}\)

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

Với đk: \(a>0,x\ne1\)

(a) Ta có: \(\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2\)

\(\Rightarrow A=\sqrt{\left(x^2+\dfrac{1}{x^2}\right)^2+2\left(x^2+\dfrac{1}{x^2}+2\right)-3}:\left(x^2-x+1\right)\)

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

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

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

\(=\dfrac{x^2+\dfrac{1}{x^2}+1}{x^2-x+1}=\dfrac{x^4+x^2+1}{x^4-x^3+x^2}\)

Vậy: \(A=\dfrac{x^4+x^2+1}{x^4-x^3+x^2}\)

(b) \(A=\dfrac{\left(x^4-x^3+x^2\right)+x^3+1}{x^4-x^3+x^2}\)

\(=1+\dfrac{x^3+1}{x^4-x^3+x^2}\)

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

\(\Rightarrow A=1+\dfrac{x+1}{x^2}=1+\dfrac{1}{x}+\dfrac{1}{x^2}\)

\(\Rightarrow A=\left[\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2+1\)

\(\Rightarrow A=\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy: \(MinA=\dfrac{3}{4}\Leftrightarrow x=-2\) (thỏa mãn).

Bài 1:

a)

\(A=\sqrt{2}\left(\sqrt{4.2}-\sqrt{16.2}+3\sqrt{9.2}\right)\\ =\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\\ =\sqrt{2}.7\sqrt{2}\\ =7\)

b)

\(B=\dfrac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{6-1}+\dfrac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{2-3}+\sqrt{\sqrt{2}^2-2.\sqrt{2}.\sqrt{1}+\sqrt{1}^2}\\ =\dfrac{5\left(\sqrt{6}-1\right)^2}{5}-\left(\sqrt{2}-\sqrt{3}\right)^2+\sqrt{\left(\sqrt{2}-1\right)^2}\\ =5\left(6-2\sqrt{6}+1\right)-\left(2-2\sqrt{6}+3\right)+\sqrt{2}-1\\ =30-10\sqrt{6}+5-2+2\sqrt{6}-3+\sqrt{2}-1\\ =29-8\sqrt{6}\)

2:

a: \(\sqrt{x^2-2x+9}=x+2\)

=>x>=-2 và x^2-2x+9=x^2+4x+4

=>x>=-2 và -2x+9=4x+4

=>x>=-2 và -6x=-5

=>x=5/6(nhận)

b: 

ĐKXĐ: x^2-4>=0 và x+2>=0

=>x>=-2 và (x>=2 hoặc x<=-2)

=>x=-2 hoặc x>=2

\(\sqrt{x^2-4}+\sqrt{x+2}=0\)

=>x^2-4=0 và x+2=0

=>x=-2

c: 

ĐKXĐ: x>=1

\(\sqrt{x+3-4\sqrt{x-1}}=3\)

=>\(\sqrt{x-1-2\cdot\sqrt{x-1}\cdot2+4}=3\)

=>\(\left|\sqrt{x-1}-2\right|=3\)

=>\(\left[{}\begin{matrix}\sqrt{x-1}-2=3\\\sqrt{x-1}-2=-3\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=5\)

=>x-1=25

=>x=26

d: \(\sqrt{4-8x}-4\sqrt{1-2x}+\sqrt{\dfrac{9-18x}{4}}+1=0\)

=>\(2\sqrt{1-2x}-4\sqrt{1-2x}+\dfrac{3}{2}\sqrt{1-2x}+1=0\)

=>\(1-\dfrac{1}{2}\sqrt{1-2x}=0\)

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

=>1-2x=4

=>2x=-3

=>x=-3/2

\(\sqrt{4}=2< 3\)

=>\(\sqrt{4+\sqrt{4}}< \sqrt{4+3}< 3\)

\(\Leftrightarrow\sqrt{4+\sqrt{4+\sqrt{4}}}< \sqrt{4+3}=3\)

...

=>\(\sqrt{4+\sqrt{4+\sqrt{4+...+\sqrt{4}}}}< \sqrt{4+3}< 3\)

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

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

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

\(Q=\left(1-\dfrac{1}{\sqrt{x}}\right)^2:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: x > 0)

\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)^2:\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)

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

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

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

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

\(Q=\dfrac{x-1}{x}\)

\(Q=\left(1-\dfrac{1}{\sqrt{x}}\right)^2:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)(ĐKXĐ: x > 0)

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

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

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

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

\(=\dfrac{x-1}{x}\)

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

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

\(=4-2\sqrt{2}\)

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

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

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

a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)

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

\(P=2\)

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

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

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

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

\(N=-4\)

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

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

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

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

\(Q=1\)

a: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)

b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)

a) \(\dfrac{6}{5\sqrt{8}}\)

\(=\dfrac{6}{5\cdot2\sqrt{2}}\)

\(=\dfrac{6}{10\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{10}\)

b) \(\dfrac{7}{5+2\sqrt{3}}\)

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

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

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

\(=\dfrac{35-14\sqrt{3}}{13}\)

c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=3\sqrt{7}+3\sqrt{5}\)

a: \(\sqrt{\dfrac{3}{20}}=\sqrt{\dfrac{15}{100}}=\dfrac{\sqrt{15}}{10}\)

b: \(\sqrt{\dfrac{5}{18}}=\sqrt{\dfrac{10}{36}}=\dfrac{\sqrt{10}}{6}\)

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

d: \(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\dfrac{x}{y}\cdot\dfrac{\sqrt{y}}{\sqrt{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

\(M=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+1\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+1\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(=6+\sqrt{6}-11\sqrt{6}-11=-5-10\sqrt{6}\)

\(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)\)

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

\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+1\right)\)

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

\(M=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+1\right)\)

\(M=\left(5\sqrt{6}-4\sqrt{6}+1-12\right)\left(\sqrt{6}+1\right)\)

\(M=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)

\(M=6+\sqrt{6}-11\sqrt{6}-11\)

\(M=-10\sqrt{6}-5\)

a)

\(=\sqrt{3^2-2.3.\sqrt{7}+\sqrt{7}^2}\\ =\sqrt{\left(3-\sqrt{7}\right)^2}\\ =\left|3-\sqrt{7}\right|\\ =3-\sqrt{7}\)

b)

\(=\sqrt{\sqrt{17}^2-2.\sqrt{17}.2+2^2}\\ =\sqrt{\left(\sqrt{17}-2\right)^2}\\ =\left|\sqrt{17}-2\right|\\ =\sqrt{17}-2\)

c)

\(=\sqrt{\sqrt{5}^2-2.\sqrt{5}.2+2^2}\\ =\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}-2\right|\\ =\sqrt{5}-2\)

d)

\(=\sqrt{\sqrt{9}^2-2.\sqrt{9}.\sqrt{6}+\sqrt{6}^2}\\ =\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\\ =\left|\sqrt{9}-\sqrt{6}\right|\\ =\sqrt{9}-\sqrt{6}\)

e: \(=\sqrt{13+2\cdot\sqrt{40}}\)

\(=\sqrt{13+2\cdot\sqrt{8}\cdot\sqrt{5}}\)

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

f: \(=\sqrt{45+2\cdot3\sqrt{5}\cdot1+1}\)

\(=\sqrt{\left(3\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)

g: \(=\sqrt{25-2\cdot5\cdot2\sqrt{6}+24}\)

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

h: \(=\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}\)

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