a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)

\(=1\)

b)ĐK: \(x>0,x\ne9\)

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

\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)

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

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

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

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

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

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

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

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

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

Ukm ko để ý

ĐỀ THI VÀO 10 ĐÓ CẢM ƠN MN TRƯỚC NHA:))

a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)

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

b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)

\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)

Th1: x-3 < 0

\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)

Th2: x-3 > 0

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

c)

Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

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

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

Th1: \(x-2\ge0\Leftrightarrow x\ge2\)

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

Th2: \(x-2\le0\Leftrightarrow x\le2\)

kết hợp với đk, ta được: 1 \< x \< 2

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

d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)

chẳng biết có sai sót gì 0 nữa, xin lỗi tớ 0 xem lại đâu vì chán quá!

\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)

\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)

\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

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

\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)

\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).

\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)

\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)

\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)

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

Sai đề

a: \(M=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

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

b: Khi a=9/25 thì \(M=\dfrac{\dfrac{3}{5}-4}{\dfrac{3}{5}-2}=\dfrac{-17}{5}:\dfrac{-7}{5}=\dfrac{17}{7}\)

c: Để |M|=1/6 thì M=1/6 hoặc M=-1/6

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{1}{6}\\\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6\sqrt{a}-24=\sqrt{a}-2\\6\sqrt{a}-24=-\sqrt{a}+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5\sqrt{a}=22\\7\sqrt{a}=26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\left(\dfrac{22}{5}\right)^2\\a=\left(\dfrac{26}{7}\right)^2\end{matrix}\right.\)

\(P=\left(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\right)\)

\(P=\dfrac{x\left(\sqrt{x}-3\right)-\left(x+1\right)\left(\sqrt{x}+3\right)+6x+\sqrt{x}}{x-9}:\dfrac{\left(\sqrt{x}-3\right)^2-x+9}{x-9}\)

\(P=\dfrac{x\sqrt{x}-3x-x\sqrt{x}-3x-\sqrt{x}-3+6x+\sqrt{x}}{x-9}:\dfrac{x-6\sqrt{x}+9-x+9}{x-9}\)

\(P=-\dfrac{3}{x-9}:\dfrac{-6\sqrt{x}+18}{x-9}=-\dfrac{3}{x-9}.\dfrac{x-9}{-6\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{1}{2\sqrt{x}-6}\)

b. \(x=12+6\sqrt{3}=9+2.3.\sqrt{3}+3=\left(3+\sqrt{3}\right)^2\Rightarrow\sqrt{x}=3+\sqrt{3}\)

\(P=\dfrac{1}{2.3+2\sqrt{3}-6}=\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{6}\)