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

a: Khi x=49 thì \(A=\dfrac{\sqrt{49}-5}{\sqrt{49}}=\dfrac{7-5}{7}=\dfrac{2}{7}\)

b: P=A*B

\(=\dfrac{\sqrt{x}-5}{\sqrt{x}}\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{3\sqrt{x}}{x-25}\right)\)

\(=\dfrac{\sqrt{x}-5}{\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\dfrac{\sqrt{x}-5}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)-3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

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

c: Vì \(\sqrt{x}+2>=2>0\forall x\) thỏa mãn ĐKXĐ
và \(\sqrt{x}+5>=5>0\forall x\) thỏa mãn ĐKXĐ

nên \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}>0\) với mọi x thỏa mãn ĐKXĐ

Vì \(\sqrt{x}+2< \sqrt{x}+5\)

nên \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}< 1\)

=>\(0< P< 1\)

=>P không thể là số nguyên

9:

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

a: Khi x=25 thì \(A=\dfrac{7}{5+8}=\dfrac{7}{13}\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

c: Để B là số nguyên thì \(\sqrt{x}+8⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3+5⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3\inƯ\left(5\right)\)

=>\(\sqrt{x}+3\in\left\{1;-1;5;-5\right\}\)

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

=>x=4(nhận)

d: \(P=A\cdot B=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\cdot\dfrac{7}{\sqrt{x}+8}=\dfrac{7}{\sqrt{x}+3}\)

Để P là số nguyên thì \(7⋮\sqrt{x}+3\)

mà \(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+3=7\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

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

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

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

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

\(=-2\)

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

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

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

Đề thiếu rồi bạn

a: \(\dfrac{4\sqrt{6}-2\sqrt{10}}{2\sqrt{2}}+\dfrac{4}{\sqrt{3}-\sqrt{5}}+3\sqrt{6-2\sqrt{5}}\)

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

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

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

=-3

b: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)

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

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

a, \(\dfrac{4\sqrt{6}-2\sqrt{10}}{2\sqrt{2}}+\dfrac{4}{\sqrt{3}-\sqrt{5}}+3\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{2\sqrt{2}\left(2\sqrt{3}-\sqrt{5}\right)}{2\sqrt{2}}+\dfrac{4\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}+3\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{3}-\sqrt{5}+\dfrac{4\sqrt{3}+4\sqrt{5}}{3-5}+3\left|\sqrt{5}-1\right|\)
\(=2\sqrt{3}-\sqrt{5}-2\sqrt{3}-2\sqrt{5}+3\sqrt{5}-3\)
\(=-3\)
b, \(với\left(x\ne1;y\ne1;y\ge0\right)\)
\(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}=\dfrac{x-1}{\sqrt{y}-1}\dfrac{\sqrt{\left(\sqrt{y}-1\right)^2}}{\left(x-1\right)^2}=\dfrac{\left|\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}\left(1\right)\)
\(TH1:y>1\)
\(\left(1\right)=\dfrac{\sqrt{y}-1}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
\(TH2:0\le y< 1\)
\(\left(1\right)=\dfrac{1-\sqrt{y}}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{-1}{x-1}\)
 

3: ĐKXĐ: x^2-2>=0

=>\(\left[{}\begin{matrix}x>=\sqrt{2}\\x< =-\sqrt{2}\end{matrix}\right.\)

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

=>x^2+x+1=x^2-2

=>x+1=-2

=>x=-3(nhận)

4: ĐKXĐ: x>=4

\(6\sqrt{x-4}=\sqrt{9\left(x+8\right)}\)

=>\(36\left(x-4\right)=9\left(x+8\right)\)

=>\(4\left(x-4\right)=x+8\)

=>4x-16=x+8

=>3x=24

=>x=8(nhận)

A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)

\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)

\(=-5\sqrt{2}+6\sqrt{2}=\sqrt{2}\)

b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)

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

\(=2-\sqrt{3}+\sqrt{3}+1-4\sqrt{3}\)

\(=3-4\sqrt{3}\)

\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)

a: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

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

b: Khi \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\) thì

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

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

a) \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

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

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

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

b) Ta có:

\(x=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)

Thay vào Q ta có:

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

a: ĐKXĐ: \(2x^2-2x+1>=0\)

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

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

Vậy: ĐKXĐ là \(x\in R\)

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

=>\(\left\{{}\begin{matrix}2x-1>=0\\4x^2-4x+1=2x^2-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\2x^2-2x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\2x\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow x=1\)

b: ĐKXĐ: x>=2

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

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

=>\(\sqrt{x-2}\cdot\dfrac{5}{6}=0\)

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

=>x-2=0

=>x=2