Ôn tập chương 1: Căn bậc hai. Căn bậc ba

a) \(A=\left(\dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}}+\dfrac{a}{b-a}\right):\left(\dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}}-\dfrac{a}{a+b+2\sqrt[]{ab}}\right)\left(a;b>0;a\ne\pm b\right)\)

\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{a}\left(\sqrt[]{b}-\sqrt[]{a}\right)+a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{\sqrt[]{a}\left(\sqrt[]{a}+\sqrt[]{b}\right)-a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)

\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{ab}-a+a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{a+\sqrt[]{ab}-a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)

\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)

\(\Leftrightarrow A=\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}.\dfrac{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}{\sqrt[]{ab}}\)

\(\Leftrightarrow A=\dfrac{a+b+2\sqrt[]{ab}}{b-a}\)

b) Với \(a=7-4\sqrt[]{3};b=7+4\sqrt[]{3}\)

\(\Leftrightarrow A=\dfrac{7-4\sqrt[]{3}+7+4\sqrt[]{3}+2\sqrt[]{\left(7-4\sqrt[]{3}\right)\left(7+4\sqrt[]{3}\right)}}{7+4\sqrt[]{3}-\left(7-4\sqrt[]{3}\right)}\)

\(\Leftrightarrow A=\dfrac{14+2\sqrt[]{\left[\left(7^2-\left(4\sqrt[]{3}\right)^2\right)\right]}}{7+4\sqrt[]{3}-7+4\sqrt[]{3}}\)

\(\Leftrightarrow A=\dfrac{14+2\sqrt[]{\left(49-48\right)}}{8\sqrt[]{3}}=\dfrac{16}{8\sqrt[]{3}}=\dfrac{2}{\sqrt[]{3}}=\dfrac{2\sqrt[]{3}}{3}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\)

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

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

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

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

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

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

Khi \(a=2+\sqrt{3};b=2-\sqrt{3}\) thì \(a\cdot b=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\)

=>\(P=-\sqrt{ab}=-1\)

a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)

Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)

=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)

c) Từ câu b suy ra:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)\ge0\)

Vì \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)(câu a)

Nên:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)

a: Khi x=64 thì \(A=\dfrac{2}{8-2}=\dfrac{2}{6}=\dfrac{1}{3}\)

b: \(P=B:A\)

\(=\dfrac{3\sqrt{x}+\sqrt{x}-2-2\left(\sqrt{x}+2\right)}{x-4}:\dfrac{2}{\sqrt{x}-2}\)

\(=\dfrac{4\sqrt{x}-2-2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{2\sqrt{x}-6}{2\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

c: P<0

=>căn x-3<0

=>0<=x<9

mà x nguyên và x<>4

nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)

19: ĐKXĐ: \(\dfrac{\sqrt{23}-2\sqrt{6}}{-x+5}>=0\)

=>-x+5<0

=>-x<-5

=>x>5

20: ĐKXĐ: 2011-m>=0

=>m<=2011

21: ĐKXĐ: x-7>0

=>x>7

22; ĐKXĐ: 4z^2+4z+1>=0

=>(2z+1)^2>=0(luôn đúng)

23: ĐKXĐ: 49x^2-24x+4>=0

=>x^2-24/49x+4/49>=0

=>x^2-2*x*12/49+144/2401+52/2401>=0

=>(x-12/49)^2+52/2401>=0(luôn đúng)

24: ĐKXĐ: 12x+5>=0

=>x>=-5/12

\(\sqrt{81a^2}\)

\(=\sqrt{9^2a^2}\)

\(=\left|9a\right|\) 

Mà: \(a< 0\Rightarrow\left|9a\right|=-9a\)

\(=-9a\)

⇒ Chọn B 

d nha

tick cho mình đi

\(a,\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)

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

\(=\sqrt{2}\)

#Urushi☕

\(P=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}}{x-4}\)

\(=\dfrac{2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{\sqrt{x}}\)

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

\(P=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}}{x-4}\) (ĐK: \(x>0;x\ne4\))

\(P=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}}{x-4}\)

\(P=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{\sqrt{x}}\)

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

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

\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}}\)

\(P=\dfrac{2}{1}\)

\(P=2\)