Chương I - Căn bậc hai. Căn bậc ba

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

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

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

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

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

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

\(P=\dfrac{x-1}{\sqrt{x}}\) 

b) \(P< 2\) khi và chỉ khi:

\(\dfrac{x-1}{\sqrt{x}}< 2\)

\(\Leftrightarrow\dfrac{x-1-2\sqrt{x}}{\sqrt{x}}< 0\)

\(\Leftrightarrow x-2\sqrt{x}-1< 0\)

\(\Leftrightarrow x-2\sqrt{x}+1-2< 0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-2< 0\) 

\(\Leftrightarrow\sqrt{x}-1< 4\)

\(\Leftrightarrow x< 25\)

Kết hợp với đk:

\(0< x< 25\)  

c) Ta có: 

\(Q=P\cdot\dfrac{\sqrt{x}\left(x+7\right)}{\left(\sqrt{x}-3\right)\left(x-1\right)}\)

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

\(Q=\dfrac{x+7}{\sqrt{x}-3}\) 

\(Q=\dfrac{x-9+16}{\sqrt{x}-3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}-3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}-3}+6\)

Theo BĐT Cauchy cho các số dương ta có: 

\(\sqrt{x}-3+\dfrac{16}{\sqrt{x}-3}\ge2\sqrt{\left(\sqrt{x}-3\right)\cdot\dfrac{16}{\sqrt{x}-3}}=2\sqrt{16}=2\cdot4=8\)

\(\Rightarrow\sqrt{x}-3+\dfrac{16}{\sqrt{x}-3}+6\ge8+6=14\)

Dấu "=" xảy ra khi: \(x=49\)

Vậy \(Q_{min}=14\) khi \(x=49\)  

Câu 3:

1: x^2-5x+4=0

=>\(\left(x-1\right)\left(x-4\right)=0\)

=>\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

2: \(\left\{{}\begin{matrix}x-2y=1\\2x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=1\\4x+2y=14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=15\\2x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7-2x=1\end{matrix}\right.\)

\(3\sqrt{45}-7\sqrt{125}+\sqrt{500}+16\sqrt{9-4\sqrt{5}}\\ =9\sqrt{5}-35\sqrt{5}+10\sqrt{5}+16\sqrt{\left(\sqrt{5}-2\right)^2}\\ =-16\sqrt{5}+16\left(\sqrt{5}-2\right)\\ =-16\sqrt{5}+16\sqrt{5}-32\\ =-32\)

a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)

\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)

b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)

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

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

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

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

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

c: P=A/B

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

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

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

=>P<2

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(Đk:x\ge0;x\ne1\right)\)

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

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

c) Khi \(x=\dfrac{1}{9}\Rightarrow\sqrt{x}=\dfrac{1}{3}\)

\(P=\dfrac{\dfrac{1}{3}-1}{\dfrac{1}{3}+2}=-\dfrac{2}{7}\)

d) \(P< -\dfrac{1}{3}\)

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}< -\dfrac{1}{3}\)

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{1}{3}< 0\)

\(\dfrac{3\left(\sqrt{x}-1\right)+\sqrt{x}+2}{3\left(\sqrt{x}+2\right)}< 0\)

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

Vì \(\sqrt{x}+2>0\Rightarrow4\sqrt{x}-1< 0\Rightarrow\sqrt{x}< \dfrac{1}{4}\)

\(\Rightarrow0\le x< \dfrac{1}{16}\)

e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Vì \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow P=1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

\(MinP=-\dfrac{1}{2}\Leftrightarrow x=0\)

1: ĐKXĐ: 3-2x>=0

=>2x<=3

=>\(x< =\dfrac{3}{2}\)

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

=>3-2x=25

=>2x=3-25=-22

=>\(x=-\dfrac{22}{2}=-11\left(nhận\right)\)

2: ĐKXĐ: 4x-1>=0

=>4x>=1

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

\(\sqrt{4x-1}+5=7\)

=>\(\sqrt{4x-1}=2\)

=>4x-1=4

=>4x=5

=>\(x=\dfrac{5}{4}\left(nhận\right)\)

3: ĐKXĐ: 3x+5>=0

=>3x>=-5

=>\(x>=-\dfrac{5}{3}\)

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

=>\(2\sqrt{3x+5}=6\)

=>\(\sqrt{3x+5}=3\)

=>3x+5=9

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

4: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2-28x+49}=2\)

=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot7+7^2}=2\)

=>\(\sqrt{\left(2x-7\right)^2}=2\)

=>|2x-7|=2

=>\(\left[{}\begin{matrix}2x-7=2\\2x-7=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(nhận\right)\\x=\dfrac{5}{2}\left(nhận\right)\end{matrix}\right.\)

6: ĐKXĐ: \(x\in R\)

\(5\sqrt{4x^2-12x+9}-3=7\)

=>\(5\cdot\sqrt{\left(2x\right)^2-2\cdot2x\cdot3+3^2}=10\)

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

=>|2x-3|=2

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

5: \(\sqrt{36+24x+4x^2}-5=1\)(ĐKXĐ: \(x\in R\))

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot6+6^2}=6\)

=>\(\sqrt{\left(2x+6\right)^2}=6\)

=>|2x+6|=6

=>\(\left[{}\begin{matrix}2x+6=6\\2x+6=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

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

\(B=\frac{\sqrt{4x^2-12x+9}}{2x-3}=\frac{\sqrt{(2x-3)^2}}{2x-3}=\frac{|2x-3|}{2x-3}=\frac{-(2x-3)}{2x-3}=-1\) (do $2x-3<0$)

\(C=20\sqrt{2}+4\sqrt{2}-24\sqrt{2}-4\sqrt{2}=\sqrt{2}(20+4-24-4)=-4\sqrt{2}\)

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

\(=\frac{4}{\sqrt{2^2-3}}=4\)

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

$=|\sqrt{x-4}+2|=\sqrt{x-4}+2$
---------------

$H\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+4\sqrt{2x-4}+4}$

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

$\Rightarrow H=\sqrt{x-2}+\sqrt{2}$

\(E=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{-(\sqrt{2}-1)}+\frac{\sqrt{5}(\sqrt{3}-1)}{-(\sqrt{3}-1)}\right].(\sqrt{7}-\sqrt{5})\)

\(=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=-(7-5)=-2\)

\(F\sqrt{2}=\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}=\sqrt{11-2\sqrt{11}+1}-\sqrt{11+2\sqrt{11}+1}\)

\(=\sqrt{(\sqrt{11}-1)^2}-\sqrt{(\sqrt{11}+1)^2}=|\sqrt{11}-1|-|\sqrt{11}+1|=-2\)

$\Rightarrow F=-\sqrt{2}$