Chọn C

ĐKXĐ: \(x< 2\)

\(m\sqrt{2-x}=\dfrac{x^2-2mx+2}{\sqrt{2-x}}\Rightarrow m\left(2-x\right)=x^2-2mx+2\)

\(\Leftrightarrow x^2+2=m\left(x+2\right)\Rightarrow m=\dfrac{x^2+2}{x+2}\)

Xét hàm \(f\left(x\right)=\dfrac{x^2+2}{x+2}\) với \(0< x< 2\)

\(f'\left(x\right)=\dfrac{2x\left(x+2\right)-\left(x^2+2\right)}{\left(x+2\right)^2}=\dfrac{x^2+4x-2}{\left(x+2\right)^2}=0\Rightarrow x=-2+\sqrt{6}\)

\(f\left(0\right)=1;f\left(2\right)=\dfrac{3}{2};f\left(-2+\sqrt{6}\right)=-4+2\sqrt{6}\)

\(\Rightarrow-4+2\sqrt{6}\le m< \dfrac{3}{2}\)

\(\Rightarrow m=1\)

Có đúng 1 giá trị nguyên m thỏa mãn

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\left(2x+1\right)+2x+1-\sqrt[3]{6x+1}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{4x^2}{\sqrt{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{4}{\sqrt{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-4}{1+1}+\dfrac{12}{1+1+1}=2\)

\(A=log_m\left(8m\right)=log_mm+log_m8\)

\(=1+log_m8\)

\(=1+\dfrac{1}{log_8m}=1+\dfrac{1}{log_{2^3}m}=1+\dfrac{1}{\dfrac{1}{3}\cdot log_2m}\)

\(=1+\dfrac{1}{\dfrac{1}{3}a}=1+1:\dfrac{a}{3}=1+\dfrac{3}{a}=\dfrac{a+3}{a}\)

=>Chọn A

Đặt A = 1/2+1/4+1/8+1/18+1/32+1/64+1/128+1/256

=> 2A = 1+1/2+1/4+1/8+1/18+1/32+1/64+1/128

=> 2A - A = 1 - 1/256

=>       A = 255/256 nhé!

ỳbuybg

đúng nha 

sai nhA 

thế tóm lại là đúng hay sai :v

ABC vuông cân \(\Rightarrow\left\{{}\begin{matrix}AC=AB=3\\BC=AB\sqrt{2}=3\sqrt{2}\end{matrix}\right.\)

\(\overrightarrow{BC}.\overrightarrow{CA}=-\overrightarrow{CB}.\overrightarrow{CA}=-BC.AC.cos\left(\overrightarrow{CB};\overrightarrow{CA}\right)\)

\(=-3\sqrt{2}.3.cos45^0=-9\)

Chụp thẳng lại đc ko pẹn !

bạn làm rồi còn gì