Chương I - Căn bậc hai. Căn bậc ba

\(a,A=\left(\dfrac{\sqrt{x}-9}{x-3\sqrt{x}}+\dfrac{8}{\sqrt{x}-3}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right)\left(ĐKXĐ:x>0;x\ne9\right)\)

\(=\left[\dfrac{\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}+\dfrac{8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(=\dfrac{9\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{x-9-x}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{9\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-9}\)

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

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

\(b,\) Khi đó: \(A=\dfrac{1}{2}\Leftrightarrow1-\sqrt{x}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(c,\) Ta có: \(\dfrac{1}{A}>\dfrac{2}{3}\)

\(\Leftrightarrow A< \dfrac{3}{2}\)

\(\Leftrightarrow1-\sqrt{x}< \dfrac{3}{2}\)

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

\(\Leftrightarrow x>\dfrac{1}{4}\)

Kết hợp với điều kiện, ta được: \(x>\dfrac{1}{4};x\ne9\)

#\(Toru\)

a) Để P<0 \(\left(Đk:x\ge0;x\ne4\right)\)

=> \(\left\{{}\begin{matrix}\sqrt{x}+2>0\\\sqrt{x}-2< 0\end{matrix}\right.\)\(\Rightarrow0\le x< 4\)

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

c) \(E=\dfrac{x-\sqrt{x}+3-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}=\dfrac{x-2\sqrt{x}+5}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-1\right)^2+4}{\sqrt{x}+2}\)

E>0 \(\forall\ge0\)

d) \(\dfrac{x-5\sqrt{x}+7}{\sqrt{x}+2}\le\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(\Leftrightarrow x-5\sqrt{x}+7\le\sqrt{x}-2\)

\(\Leftrightarrow x-6\sqrt{x}+9\le0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2\le0\)

\(\Leftrightarrow x=9\left(\text{vì }\left(\sqrt{x}-3\right)^2\ge0\forall x\ge0\right)\)

đề 2 ạ

a: \(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: A=1/2

=>\(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)

=>-10căn x+4=căn x+3

=>-11căn x=-1

=>căn x=1/11

=>x=1/121

d: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(=\dfrac{-10\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=-\dfrac{12\sqrt{x}}{3\left(\sqrt{x}+3\right)}< =0\)

=>A<=2/3

a) \(x-3\sqrt[]{x}=\sqrt[]{x}\left(\sqrt[]{x}-3\right)\)

b) \(2x+4\sqrt[]{x}=2\sqrt[]{x}\left(\sqrt[]{x}+2\right)\)

c) \(x-3\sqrt[]{x}+2=x-2\sqrt[]{x}-\sqrt[]{x}+2=\sqrt[]{x}\left(\sqrt[]{x}-2\right)-\left(\sqrt[]{x}-2\right)=\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}-1\right)\)

d) \(x+5\sqrt[]{x}+6=x+2\sqrt[]{x}+3\sqrt[]{x}+6=\sqrt[]{x}\left(\sqrt[]{x}+2\right)+3\left(\sqrt[]{x}+2\right)=\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}+3\right)\)

e) \(x-\sqrt[]{x}-6=x-3\sqrt[]{x}+2\sqrt[]{x}-6=\sqrt[]{x}\left(\sqrt[]{x}-3\right)+2\left(\sqrt[]{x}-3\right)=\left(\sqrt[]{x}-3\right)\left(\sqrt[]{x}+2\right)\)

f) \(x+3\sqrt[]{x}-28=x+7\sqrt[]{x}-4\sqrt[]{x}-28=\sqrt[]{x}\left(\sqrt[]{x}+7\right)-4\left(\sqrt[]{x}+7\right)=\left(\sqrt[]{x}+7\right)\left(\sqrt[]{x}-4\right)\)

g) \(x+\sqrt[]{x}-12=x+4\sqrt[]{x}-3\sqrt[]{x}-12=\sqrt[]{x}\left(\sqrt[]{x}+4\right)-3\left(\sqrt[]{x}+4\right)=\left(\sqrt[]{x}+4\right)\left(\sqrt[]{x}-3\right)\)

h) \(x+3\sqrt[]{x}-40=x+8\sqrt[]{x}-5\sqrt[]{x}-40=\sqrt[]{x}\left(\sqrt[]{x}+8\right)-5\left(\sqrt[]{x}+8\right)=\left(\sqrt[]{x}+8\right)\left(\sqrt[]{x}-5\right)\)

a: \(=\left(0.2-1.2+11\right)\cdot9=12\cdot9=108\)

b: \(=\dfrac{75}{\sqrt{9+16}}-3\cdot4\)

\(=\dfrac{75}{5}-12=15-12=3\)

c: \(=4-\sqrt{15}+\sqrt{15}=4\)

d: \(=2-\sqrt{3}+\sqrt{3}-1=2-1=1\)

e: \(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)

\(=3\sqrt{5}-2-3\sqrt{5}-2\)

=-4

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

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

=6

c) \(\sqrt[]{\left(4-\sqrt[]{15}\right)^2}+\sqrt[]{15}\)

\(=\left|4-\sqrt[]{15}\right|+\sqrt[]{15}\)

\(=4-\sqrt[]{15}+\sqrt[]{15}\left(4>\sqrt[]{15}\right)\)

\(=4\)

d) \(\sqrt[]{\left(2-\sqrt[]{3}\right)^2}+\sqrt[]{\left(1-\sqrt[]{3}\right)^2}\)

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

\(=2-\sqrt[]{3}+\sqrt[]{3}-1\left(2>\sqrt[]{3};\sqrt[]{3}>1\right)\)

\(=1\)

e) \(\sqrt[]{49-12\sqrt[]{5}}-\sqrt[]{49+12\sqrt[]{5}}\)

\(=\sqrt[]{45-2.2.3\sqrt[]{5}+4}-\sqrt[]{45+2.2.3\sqrt[]{5}+4}\)

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

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

\(=3\sqrt[]{5}-2-3\sqrt[]{5}-2\left(3\sqrt[]{5}>2\right)\)

\(=-4\)

f) \(\sqrt[]{29+12\sqrt[]{5}}-\sqrt[]{29-12\sqrt[]{5}}\)

\(=\sqrt[]{20+2.3.2\sqrt[]{5}+9}-\sqrt[]{20-2.3.2\sqrt[]{5}+9}\)

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

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

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

\(=0\)

a: Khi x=121 thì \(A=\dfrac{121+3}{11+3}=\dfrac{124}{14}=\dfrac{62}{7}\)

b: \(B=\left(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

c: P=A:B

\(=\dfrac{x+3}{\sqrt{x}+3}:\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}\)

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

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

Dấu = xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)

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

=>x=1(nhận)

\(\dfrac{x\sqrt{x}}{\sqrt{x}+2}-2\sqrt{x}\left(dk:x\ge0\right)\\ =\dfrac{x\sqrt{x}-2\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\\ =\dfrac{x\sqrt{x}-2x-4\sqrt{x}}{\sqrt{x}+2}\)

\(\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\)

\(tan=3\\ cot=\dfrac{1}{3}\)

Ta có : \(1+tan^2=\dfrac{1}{cos^2}\Rightarrow1+3^2=\dfrac{1}{cos^2}\Rightarrow cos=\dfrac{\sqrt{10}}{10}\)

\(sin=\sqrt{1-cos^2}=\sqrt{1-\left(\dfrac{\sqrt{10}}{10}\right)^2}=\dfrac{3\sqrt{10}}{10}\)

\(B=\dfrac{sin+cos}{sin^3+cos^3}=\dfrac{sin+cos}{\left(sin+cos\right)\left(sin^2+cos^2-sincos\right)}=\dfrac{1}{1-sincos}\)

\(=\dfrac{1}{1-\dfrac{3\sqrt{10}}{10}.\dfrac{\sqrt{10}}{10}}=\dfrac{10}{7}\)

Vậy \(B=\dfrac{10}{7}\)

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

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

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

b; Để E>1 thì E-1>0

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

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

=>x>1

c: \(E=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\)

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

=>\(E>=2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}+2=4\)

Dấu = xảy ra khi \(\left(\sqrt{x}-1\right)^2=1\)

=>\(\left[{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=0\left(loại\right)\end{matrix}\right.\)

d: Để E là số nguyên thì \(x⋮\sqrt{x}-1\)

=>\(x-1+1⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{2;0\right\}\)

=>\(x\in\left\{4;0\right\}\)

Kết hợp ĐKXĐ, ta được: x=4

e: E=9/2

=>\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)

=>\(2x=9\sqrt{x}-9\)

=>\(2x-3\sqrt{x}-6\sqrt{x}+9=0\)

=>\(\left(2\sqrt{x}-3\right)\left(\sqrt{x}-3\right)=0\)

=>x=9 hoặc x=9/4

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

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

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

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

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

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

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

b) \(E>1\) khi:

\(\dfrac{x}{\sqrt{x}-1}>1\)

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

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

\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

Mà: 

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

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\)

\(\Rightarrow x>1\)   

c) Ta có:

\(E=\dfrac{x}{\sqrt{x}-1}\) với \(x>1\) 

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

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

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

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

\(\Rightarrow E\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}+2=2\cdot1+2=4\) 

Dấu "=" xảy ra: 

\(\left(\sqrt{x}-1\right)^2=1\) 

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\left(ĐK:x>1\right)\)

Vậy: ... 

d) \(E\in Z\) khi:

\(\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\in Z\) 

\(\Rightarrow1\) ⋮ \(\sqrt{x}-1\)

\(\Rightarrow\sqrt{x}-1\) \(\in\) Ư(1)

Mà: \(Ư\left(1\right)=\left\{1;-1\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;0\right\}\)

\(\Rightarrow x\in\left\{4;0\right\}\)

Vậy: ...

e) \(E=\dfrac{9}{2}\) khi:

\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)

\(\Leftrightarrow2x=9\sqrt{x}-9\)

\(\Leftrightarrow2x-9\sqrt{x}+9=0\)

\(\Leftrightarrow2x-6\sqrt{x}-3\sqrt{x}+9\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-3\right)\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\2\sqrt{x}-3=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3^2\\x=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=\dfrac{9}{4}\left(tm\right)\end{matrix}\right.\)