Dạ dạ

a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)

e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)

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

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

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

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

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

38x2=76

26x4=104

42x5=210

77x3=231

54x6=324

48x3=144

a)\(\dfrac{x}{3}=\dfrac{4x}{12}\)

\(\dfrac{x-1}{4}=\dfrac{3x-3}{12}\)

\(\dfrac{x^2}{6}=\dfrac{2x^2}{12}\)

b)\(\dfrac{1}{6x^2y}=\dfrac{2xy}{12x^3y^2}\)

\(\dfrac{1}{4x^3y^2}=\dfrac{3}{12x^3y^2}\)

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

\(\dfrac{1}{x+1}giữnguyên\)

d)\(\dfrac{1}{3\left(x-y\right)}=\dfrac{x-y}{3\left(x-y\right)^2}\)

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

e)\(\dfrac{x}{x+3}=\dfrac{x^2-3x}{x^2-9}\)

\(\dfrac{1}{3-x}=-\dfrac{x+3}{x^2-9}\)

\(\dfrac{1}{x^2-9}giữnguyên\)

f)\(\dfrac{1}{x^2+xy}=\dfrac{y\left(x-y\right)}{xy\left(x^2-y^2\right)}\)

\(\dfrac{1}{xy-y^2}=\dfrac{x\left(x+y\right)}{xy\left(x^2-y^2\right)}\)

\(\dfrac{2}{y^2-x^2}=-\dfrac{2xy}{xy\left(x^2-y^2\right)}\)

\(A+B=90^0\Rightarrow sinB=cosA=\dfrac{1}{3}\)

\(\Rightarrow sinA=\sqrt{1-cos^2A}=\sqrt{1-\dfrac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)

\(\Rightarrow TanA=\dfrac{sinA}{cosA}=2\sqrt{2}\)

Vậy chọn C

Giả sử điểm cố định là\(A\left(x_0,y_0\right)\)

\(⇒y_0 =mx_0+1 ∀ m\)

\(⇔ − mx_0 + y_0 − 1 = 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_0=0\\y_0-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=0\\y_0=1\end{matrix}\right.\)

Vậy d luôn đi qua điểm cố định\(A\left(0,1\right)\)với mọi m