a, có DM _|_ EF và EN _|_ DF (gt)

=> ^IMF = ^INF = 90

=> M;N thuộc đường tròn đường kính IF (Định lí)

=> F;N;I;M thuộc đường tròn đk IF

b, có DM _|_ EF và EN _|_ DF (gt)

=> ^END = ^DME  = 90

=> N;M thuộc đường tròn đk DE

=> D;N;M;E cùng thuộc đường tròn đk DE

có : \(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\Leftrightarrow x-2\sqrt{xy}+y\ge0\)

\(\Leftrightarrow2x+2y\ge x+2\sqrt{xy}+y\)

\(\Leftrightarrow2\left(x+y\right)\ge\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(\Leftrightarrow\sqrt{2}\cdot\sqrt{x+y}\ge\sqrt{x}+\sqrt{y}\)

\(\Leftrightarrow\sqrt{x+y}\ge\frac{\sqrt{x}+\sqrt{y}}{\sqrt{2}}\)

áp dụng vào ta có :  

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

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

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

\(\Rightarrow VT\ge\frac{1}{\sqrt{2}}\left[\sqrt{a}\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\sqrt{b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)+\sqrt{c}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\right]\)

áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có :

\(VT\ge\frac{1}{\sqrt{2}}\left(\sqrt{a}\cdot\frac{4}{\sqrt{b}+\sqrt{c}}+\sqrt{b}\cdot\frac{4}{\sqrt{a}+\sqrt{c}}+\sqrt{c}\cdot\frac{4}{\sqrt{a}+\sqrt{b}}\right)\)

có \(\hept{\begin{cases}\sqrt{b}+\sqrt{c}\le\sqrt{2\left(b+c\right)}\\\sqrt{a}+\sqrt{c}\le\sqrt{2\left(a+c\right)}\\\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\end{cases}}\)  nên \(\hept{\begin{cases}\frac{4}{\sqrt{b}+\sqrt{c}}\ge\frac{4}{\sqrt{2\left(b+c\right)}}\\\frac{4}{\sqrt{a}+\sqrt{c}}\ge\frac{4}{\sqrt{2\left(a+c\right)}}\\\frac{4}{\sqrt{a}+\sqrt{b}}\ge\frac{4}{\sqrt{2\left(a+b\right)}}\end{cases}}\)

\(\Rightarrow vt\ge2\cdot\left(\sqrt{\frac{a}{c+b}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\right)\)

dấu = xảy ra khi a=b=c

sao dòng 9 cs vậy bạn, sao VT lại >=1/ căn 2

ĐKXĐ : a > 1

\(\frac{\sqrt{a+1}}{\sqrt{a^2-1}-\sqrt{a^2+a}}+\frac{1}{\sqrt{a-1}+\sqrt{a}}+\frac{\sqrt{a^3}-a}{\sqrt{a}-1}\)

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

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

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

\(=-2\sqrt{a-1}+a+\sqrt{a}+1\)

Theo định lí Pytago 4 điểm ta có:

\(KB^2-KL^2=MB^2-ML^2\) vì \(MK\perp BL\) 

\(KC^2-KL^2=NC^2-NL^2\) vì \(NK\perp CL\)

Suy ra \(KB^2-KC^2=MB^2-NC^2+NL^2-ML^2\)

\(=\frac{1}{4}\left(BF^2-CE^2+CF^2-BE^2\right)=\frac{1}{4}\left(BC^2-BC^2\right)=0\)

Vậy \(KB=KC.\)

\(A=\left(1+\frac{2c}{a+b}\right)\left(1+\frac{2b}{c+a}\right)\left(1+\frac{2a}{b+c}\right)\)

\(A=\frac{\left(a+c\right)+\left(c+b\right)}{a+b}\cdot\frac{\left(b+a\right)+\left(b+c\right)}{c+a}\cdot\frac{\left(a+b\right)+\left(a+c\right)}{b+c}\)

vì a;b;c dương, theo cô si ta có : 

\(\left(a+c\right)+\left(c+b\right)\ge2\sqrt{\left(a+c\right)\left(c+b\right)}\Rightarrow1+\frac{2c}{a+b}\ge\frac{2\sqrt{\left(a+c\right)\left(c+b\right)}}{a+b}\)

\(\left(b+a\right)+\left(b+c\right)\ge2\sqrt{\left(b+a\right)\left(b+c\right)}\Rightarrow1+\frac{2b}{c+a}\ge\frac{2\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}\)

\(\left(a+b\right)+\left(a+c\right)\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\Rightarrow1+\frac{2a}{b+c}\ge\frac{2\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}\)

\(\Rightarrow A\ge8\cdot\frac{\sqrt{\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Rightarrow A\ge8\)

Ta có : \(1+\frac{2c}{a+b}=\frac{\left(a+c\right)\left(b+c\right)}{a+b}\ge\frac{2\sqrt{\left(a+c\right)\left(b+c\right)}}{a+b}\)

tương tự \(1+\frac{2b}{c+a}\ge\frac{2\sqrt{\left(b+c\right)\left(b+a\right)}}{c+a}\); \(1+\frac{2a}{b+c}\ge\frac{2\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}\)

Suy ra : \(\left(1+\frac{2c}{a+b}\right)\left(1+\frac{2b}{c+a}\right)\left(1+\frac{2a}{b+c}\right)\ge\frac{8\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\) 

Dấu "=" xảy ra <=> a = b = c 

\(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3x\)

đặt \(\hept{\begin{cases}\sqrt{2x^2+3x+5}=a\\\sqrt{2x-3x+5}=b\end{cases}}\left(a;b\ge0\right)\)

pt trở thành \(a+b=\frac{a^2-b^2}{2}\)

\(\Leftrightarrow a^2-b^2-2a-2b=0\)

\(\Leftrightarrow\left(a-1\right)^2-\left(b+1\right)^2=0\)

\(\Leftrightarrow\left(a-1-b-1\right)\left(a-1+b+1\right)=0\)

\(\Leftrightarrow\left(a-b-2\right)\left(a+b\right)=0\)

th1 : a + b = 0 hay  \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=0\)

vì \(\sqrt{2x^2+3x+5}\ge0\) và \(\sqrt{2x^2-3x+5}\ge0\)

\(\Rightarrow\hept{\begin{cases}2x^2+3x+5=0\\2x^2-3x+5=0\end{cases}}\)

\(\Rightarrow4x^2+10=0\)

\(\Rightarrow4x^2=-10\left(loai\right)\)

th2 : a - b - 2 = 0 hay \(\sqrt{2x^2+3x+5}-\sqrt{2x^2-3x+5}-2=0\)

\(\Leftrightarrow\sqrt{2x^2+3x+5}=2+\sqrt{2x^2-3x+5}\)

\(\Leftrightarrow2x^2+3x+5=4+4\sqrt{2x^2-3x+5}+2x^2-3x+5\)

\(\Leftrightarrow4\sqrt{2x^2-3x+5}=6x-4\)

\(\Leftrightarrow2\sqrt{2x^2-3x+5}=3x-2\) (x >=2/3)

\(\Leftrightarrow4\left(2x^2-3x+5\right)=9x^2-12x+4\)

\(\Leftrightarrow8x^2-12x+20=9x^2-12x+4\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\left(tm\right)\\x=-4\left(loai\right)\end{cases}}\)

vay x = 4

Căm ơn bạn nhiều nhé =)

sin \(\alpha\) bằng \(\frac{3}{5}\)

Trả lời:

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}\)

\(\Rightarrow\cos\alpha=\sqrt{\frac{16}{25}}=\frac{4}{5}\)

Ta có: \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)

\(\Rightarrow\sin\alpha=\frac{\cos\alpha}{\cot\alpha}=\frac{3}{5}\) (1)

Thay \(\cos\alpha=\frac{4}{5}\) vào (1) ta có:

\(\frac{\frac{4}{5}}{\cot\alpha}=\frac{3}{5}\Rightarrow\cot\alpha=\frac{4}{5}:\frac{3}{5}=\frac{4}{3}\)

Vậy \(\cos\alpha=\frac{4}{5};\cot\alpha=\frac{4}{3}\)