1D

2B

4B

5D

8B

9C

10A

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

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

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

b) \(A< 0\rightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}< 0\)

\(\sqrt{x}\left(\sqrt{x}-1\right)< 0\) ( do \(\left(\sqrt{x}+1\right)^2>0\) )

6) \(=-16\dfrac{2}{7}:\dfrac{3}{5}+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\left(-16\dfrac{2}{7}+28\dfrac{2}{7}\right):\dfrac{3}{5}\)

\(=12.\dfrac{5}{3}=20\)

7) \(=\left(4.\dfrac{3}{4}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)

\(=\left(\dfrac{6}{2}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}.\dfrac{6}{5}-17\)

\(=3-17=-14\)

9) \(=\dfrac{3}{5}:\dfrac{-7}{30}+\dfrac{3}{5}:\dfrac{-7}{5}\)

\(=\dfrac{3}{5}.\dfrac{-30}{7}+\dfrac{3}{5}.\dfrac{-5}{7}\)

\(=\dfrac{3}{5}.\left(\dfrac{-30}{7}-\dfrac{5}{7}\right)\)

\(=\dfrac{3}{5}.\left(-5\right)=-3\)

a) \(P=\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\) \(\left(đk:x>1,x\ne0\right)\)

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

b) \(x=\dfrac{4}{9}\left(tmđk\right)\)

\(\Rightarrow\sqrt{x}=\dfrac{2}{3}\)

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

\(x=3+\sqrt{8}\left(tmđk\right)\)

a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x-6y-1\right)\)

b) \(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c) \(=2\left(x-y\right)^2-18\)

\(=2\left[\left(x-y\right)^2-3^2\right]\)

\(=2\left(x-y+3\right)\left(x-y-3\right)\)

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

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

\(B=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

a) \(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)

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

b) \(\left[{}\begin{matrix}x-2=0\\5-x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

c) \(\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x^2=-1\end{matrix}\right.\)

x=1

d) \(2x=20\)

x=10

e) \(\left|x-3\right|-12=5\)

\(\left|x-3\right|=17\)

\(\left[{}\begin{matrix}x-3=17\\x-3=-17\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=20\\x=-14\end{matrix}\right.\)

\(=\left[\left(x+y\right)-1\right]^2\)