Để hai đường thẳng vuông góc :

\(\Leftrightarrow m\left(4m-5\right)=-1\Leftrightarrow4m^2-5m+1=0\Rightarrow\orbr{\begin{cases}m=1\\m=\frac{1}{4}\end{cases}}\)

b ) Gọi điểm cố định mà \(d_2\) đi qua là M \(\left(x_0;y_0\right)\)

\(\Rightarrow y_0=\left(4m-5\right)x_0+3m\forall m\)

\(\Leftrightarrow m\left(4x_0+3\right)-\left(5x_0+y_0\right)=0\)

\(\Rightarrow\hept{\begin{cases}4x_0+3=0\\5x_0+y_0=0\end{cases}\Rightarrow\hept{\begin{cases}x_0=-\frac{3}{4}\\y_0=\frac{15}{4}\end{cases}\Rightarrow}M\left(-\frac{3}{4};\frac{15}{4}\right)}\)

\(P=\frac{2a}{2\sqrt{\left(b+1\right)\left(b^2-b+1\right)}+2}+\frac{2b}{2\sqrt{\left(c+1\right)\left(c^2-c+1\right)}+2}\)\(+\frac{2c}{2\sqrt{\left(a+1\right)\left(a^2-a+1\right)}+2}\)

\(P\ge\frac{2a}{b^2+4}+\frac{2b}{c^2+4}+\frac{2c}{a^2+4}\)

\(2P\ge\frac{4a}{b^2+4}+\frac{4b}{c^2+4}+\frac{4c}{a^2+4}=a-\frac{ab^2}{b^2+4}+b-\frac{bc^2}{c^2+4}+a-\frac{ca^2}{a^2+4}\)

\(2p\ge a+b+c-\left(\frac{ab^2}{4b}+\frac{bc^2}{4c}+\frac{ca^2}{4a}\right)\)

\(2P\ge6-\frac{1}{4}\left(ab+bc+ca\right)\ge6-\frac{1}{12}\left(a+b+c\right)^2=3\)

\(\Rightarrow P\ge\frac{3}{2}\)

Dấu " = " xảy ra khi \(a=b=c=2\)

\(\Leftrightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2=\frac{1}{abc}\)

Đặt : \(\left(\frac{a}{bc};\frac{b}{ac};\frac{c}{ab}\right)=\left(x,y,z\right)\)

\(x+y+z+2=xyz\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)\)

\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+1=1\)

\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2\)

\(\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)

   

Áp dụng bdt ptolemy cho tứ giác OAMB có:

oa.mb+ob.ma>=om.ab

=>2oa.mb + 2ob.ma>=2om.ab

=>3ob.mb+2ob.ma>=2om.ab

=>2ma+3ma>=2om.(ab/ob)

Vì ab/ob không đổi nên min=2om.(ab/ob)

Dấu bằng khi tứ giác oamb nôi tiếp đường tròn.

vì 0 \(\le\)|a|,|b| < 1 nên a² < 1 ; b² < 1

\(\Rightarrow\left(a^2-1\right)\left(b^2-1\right)>0\Rightarrow a^2b^2-a^2-b^2+1>0\)

\(\Leftrightarrow a^2b^2+1>a^2+b^2\Leftrightarrow a^2b^2+2ab+1>a^2+2ab+b^2\)

\(\Leftrightarrow\left(ab+1\right)^2>\left(a+b\right)^2\Leftrightarrow\left|ab+1\right|>\left|a+b\right|\)

Bạn ơi mình nói ngắn gọn thôi 

Quy đồng hai vế với (a+1)(b+1(c+1) phá ngoặc đơn là tìm được đáp án