gọi I là trọng tâm của tam giác ABC ta có AI vuông góc với BI

dễ thấy \(AB^2=BI\cdot BN\)

mà \(BI=\frac{2}{3}BN\)(I là trọng tâm)

\(\Rightarrow a^2=\frac{2}{3}BN^2\)

dễ thấy \(AN^2=IN\cdot BN=\frac{1}{3}BN\cdot BN=\frac{1}{3}BN^2=\frac{a^2}{2}\)

suy ra \(AC=\sqrt{2}a\)

\(BC^2=AB^2+AC^2=a^2+2a^2=3a^2\Rightarrow BC=\sqrt{3}a\)

Bài làm:

a) đkxđ: \(a\ne1;a>0\)

b) Ta có: 

\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right)\div\frac{\sqrt{a}+1}{a}\)

\(A=\left[\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right].\frac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}\)

\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)

\(A=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-1\right)\sqrt{a}\)

\(A=a-\sqrt{a}\)

\(A=\frac{\cos57}{\cos57}+\frac{\cot58}{\cot58}-2\left(1+1\right)\)\()\)

=1+1-4

=-2

với mọi x thuộc D ta có:

\(f\left(-x\right)=\frac{\left|-x+1\right|+\left|-x-1\right|}{\left|-x+1\right|-\left|-x-1\right|}=\frac{\left|-\left(x-1\right)\right|+\left|-x\left(x+1\right)\right|}{\left|-\left(x-1\right)\right|-\left|-\left(x+1\right)\right|}=\frac{\left|x-1\right|+\left|x+1\right|}{\left|x-1\right|-\left|x+1\right|}\)

\(=-\frac{\left|x+1\right|+\left|x-1\right|}{\left|x+1\right|-\left|x-1\right|}=-f\left(x\right)\)

\(3.\sqrt{\frac{1}{3}}-\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{9.\frac{1}{3}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)

\(=\sqrt{3}-\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{3}-\sqrt{3}+\sqrt{2}=\sqrt{2}\)

\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)

\(\Leftrightarrow a+b=a+c+b+c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)

\(\Leftrightarrow2c+2\sqrt{ab+bc+ca+c^2}=0\)

Theo giả thiết \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)

Khi đó \(c=0?\)

Nhầm chỗ nào nhắc mình với nha mình cảm ơn nhiều

mình vẫn không phát hiện bạn nhầm chỗ nào

áp dụng bđt Min-cốp-xki ta có \(\sqrt{x^2+xy+y^2}+\sqrt{x^2+xz+z^2}=\sqrt{\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}}+\sqrt{\left(x^2+xz+\frac{z^2}{4}\right)+\frac{3z^2}{4}}\)\(=\sqrt{\left(x+\frac{y}{2}\right)^2+\left(\frac{\sqrt{3}y}{2}\right)^2}+\sqrt{\left(-x-\frac{z}{2}\right)^2+\left(\frac{\sqrt{3}z}{2}\right)^2}\)\(\ge\sqrt{\left(x+\frac{y}{2}-x-\frac{z}{2}\right)^2+\left(\frac{\sqrt{3}y}{2}+\frac{\sqrt{3}z}{2}\right)^2}=\sqrt{\frac{y^2}{4}-\frac{yz}{2}+\frac{z^2}{4}+\frac{3y^2}{4}+\frac{3yz}{2}+\frac{3z^2}{4}}\)

\(=\sqrt{y^2+yz+z^2}\)

Ai giúp em với ạ

Bài này thầy em bảo dùng BĐT Bunhiacopxki