\(M< \sqrt{M}\Leftrightarrow M< 1\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+3}< 1\Leftrightarrow3\sqrt{x}< \sqrt{x}+3\left(do.\sqrt{x}+3>0\right)\Leftrightarrow2\sqrt{x}< 3\Leftrightarrow\sqrt{x}< \dfrac{3}{2}\Leftrightarrow0\le x< \dfrac{9}{4}\)

1.2 :

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

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

2b.

\(\left\{{}\begin{matrix}\left|x\right|+y=m\\2\left|x\right|-y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|=m+1\\y=2\left|x\right|-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left|x\right|=\dfrac{m+1}{3}\\y=\dfrac{2m-1}{3}\end{matrix}\right.\)

Hệ phương trình có 2 nghiệm phân biệt khi và chỉ khi

\(\dfrac{m+1}{3}\ne0\Leftrightarrow m\ne-1\)

2. a) 

Thay m = -1 vào hệ phương trình ta có:

\(\left\{{}\begin{matrix}\left|x\right|+y=-1\\2\left|x\right|-y=1\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|=0\\\left|x\right|+y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\)

Câu III :

Xét phương trình hoành độ giao điểm parabol (P) : y=x^2 và đường thẳng (d) : \(y=\sqrt{3}x-\sqrt{3}+1\) ta có:

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

\(\Delta=\left(-\sqrt{3}\right)^2-4\cdot1\left(\sqrt{3}-1\right)=3-4\sqrt{3}+4=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2>0\)

Suy ra phương trình (1) có 2 nghiệm 

\(x_1=\dfrac{\sqrt{3}+2-\sqrt{3}}{2}=1;x_2=\dfrac{\sqrt{3}-2+\sqrt{3}}{2}=\sqrt{3}-1\)

Nên tọa độ 2 giao điểm là \(A\left(1;1\right);B\left(\sqrt{3}-1;4-2\sqrt{3}\right)\)

b)\(A=\dfrac{1}{3}x^2y+x^3y^3-y-\left(4x^3y^3-3y+\dfrac{1}{3}x^2y\right)=\dfrac{1}{3}x^2y+x^3y^3-y-4x^3y^3+3y-\dfrac{1}{3}x^2y=-3x^3y^3+2y\)Thay x=2/3 ; y=-1/2 vào A ta có:

\(A=-3\cdot\left(\dfrac{2}{3}\cdot\dfrac{-1}{2}\right)^3+2\cdot\dfrac{-1}{2}=-3\cdot\left(-\dfrac{1}{3}\right)^3-1=-\dfrac{3}{-27}-1=\dfrac{1}{9}-1=-\dfrac{8}{9}\)

Đơn thức trên có bậc là :10+21=31 

a) 

\(\left(\dfrac{1}{3}x^3y\right)\left(-xy^4\right)^3\left(-\dfrac{3}{2}x^2y^4\right)^2=\left(\dfrac{1}{3}x^3y\right)\left(-x^3y^{12}\right)\left(\dfrac{9}{4}x^4y^8\right)=\left(\dfrac{1}{3}\cdot\left(-1\right)\cdot\dfrac{9}{4}\right)\left(x^3\cdot x^3\cdot x^4\right)\left(y\cdot y^{12}\cdot y^8\right)=-\dfrac{3}{4}x^{10}y^{21}\)

\(lim\left(x->1\right)\dfrac{-\sqrt{4x-3}+\sqrt[3]{6x-5}}{\left(x-1\right)^2}\)

Đặt \(\sqrt{4x-3}=f\left(x\right);\sqrt[3]{6x-5}=g\left(x\right)\Rightarrow g\left(x\right)^6-f\left(x\right)^6=4\left(x-1\right)^2\left(16x-13\right)\)

\(f\left(1\right)=1;g\left(1\right)=1\)

Ta có 

\(lim\left(x->1\right)\dfrac{-f\left(x\right)+g\left(x\right)}{\left(x-1\right)^2}=lim\left(x->1\right)\dfrac{g\left(x\right)^6-f\left(x\right)^6}{\left(x-1\right)^2}\cdot\dfrac{1}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)

\(=lim\left(x->1\right)\dfrac{4\left(16x-3\right)}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)

\(=\dfrac{4\left(16-3\right)}{1^5+1^4\cdot1+1^3\cdot1^2+1^2\cdot1^3+1\cdot1^4+1^5}=\dfrac{26}{3}\)

Ta có : \(3^{70}+5^{70}=9^{35}+25^{35}\)

Vì 25 là số lẻ suy ra

 \(9^{35}+25^{35}⋮9+25\)

Suy ra A chia hết cho 34