ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

(bài giải mang tính chất hướng dẩn)

a) ta có : để \(\left(d\right)\) song song với \(đt:y=\sqrt{3}x\)

\(\Leftrightarrow\dfrac{-2m}{m-1}=\sqrt{3}\Leftrightarrow m=\dfrac{\sqrt{3}}{\sqrt{3}+2}\)

ta có góc giữ \(d\) và \(Ox\) cũng chính là góc của đt \(y=\sqrt{3}x\) và \(Ox\)

vì \(\left(đt:y=\sqrt{3}x\backslash\backslash\left(d\right)\right)\)

mà ta có : \(\left(y=\sqrt{3}x\right)\) đi qua 2 điểm \(A\left(0;0\right)vàB\left(1;\sqrt{3}\right)\)

khi đó vẽ hình ra ta dể dàng \(\Rightarrow tan\alpha=\sqrt{3}\) (với \(\alpha\) là góc cần tìm)

\(\Rightarrow\alpha\)

câu b : tương tự câu bênh kia nha

Mysterious Person Nguyễn Thanh Hằng DƯƠNG PHAN KHÁNH DƯƠNG đề bài khó wá

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

Mysterious Person DƯƠNG PHAN KHÁNH DƯƠNG Nguyễn Thanh Hằng help. Thanks

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5

ĐKXĐ: m<>1, m<>0

a: Để hai đường song song thì \(-\dfrac{2m}{m-1}=\sqrt{3}\)

=>\(-2m=\sqrt{3}m-\sqrt{3}\)

\(\Leftrightarrow m\left(-2-\sqrt{3}\right)=-\sqrt{3}\)

hay \(m=-3+2\sqrt{3}\)

tana=căn 3

nên a=60 độ

b: 

\(y=-\dfrac{2m}{m-1}x+\dfrac{2}{m-1}\)

=>\(\dfrac{2m}{m-1}x+y-\dfrac{2}{m-1}=0\)

\(h=d\left(O;d\right)=\dfrac{\left|-\dfrac{2m}{m-1}\cdot0+y\cdot0-\dfrac{2}{m-1}\right|}{\sqrt{\left(\dfrac{2m}{m-1}\right)^2+1^2}}\)

\(=\dfrac{2}{\left|m-1\right|}:\sqrt{\dfrac{4m^2+m^2-2m+1}{\left(m-1\right)^2}}\)

\(=\dfrac{2}{\sqrt{5m^2-2m+1}}\)

Để h lớn nhất thì \(\sqrt{5m^2-2m+1}\) nhỏ nhất

\(5m^2-2m+1=5\left(m^2-\dfrac{2}{5}m+\dfrac{1}{5}\right)\)

\(=5\left(m^2-2\cdot m\cdot\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{4}{25}\right)\)

\(=5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}>=\dfrac{4}{5}\)

=>\(\sqrt{5\left(m-\dfrac{1}{5}\right)^2+\dfrac{4}{5}}>=\dfrac{2}{\sqrt{5}}\)

Dấu = xảy ra khi m=1/5