A B C E F L M N K

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

\(\sqrt{x^2+x+1}=1\left(x\inℝ\right)\Leftrightarrow x^2+x+1=1\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

\(\sqrt{x^2+1}=-3\)dễ thấy phương trình vô nghiệm vì \(\sqrt{x^2+1}>0\forall x\)

\(\sqrt{x^2+x+1}=1\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

b.\(\sqrt{x^2+1}=-3\) vô nghiệm do \(\hept{\begin{cases}\sqrt{x^2+1}\ge0\\-3< 0\end{cases}}\)

Trả lời:

Ta có: \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{3}{4}\)

\(\Rightarrow3\cos\alpha=4\sin\alpha\)

\(\Rightarrow\cos\alpha=\frac{4}{3}\sin\alpha\)

Mà \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\sin^2\alpha+\left(\frac{4}{3}\sin\alpha\right)^2=1\)

\(\Rightarrow\sin^2\alpha+\frac{16}{12}\sin^2\alpha=1\)

\(\Rightarrow\frac{7}{3}\sin^2\alpha=1\)

\(\Rightarrow\sin^2\alpha=\frac{3}{7}\)

\(\Rightarrow\sin\alpha=\sqrt{\frac{3}{7}}\) \(=\frac{\sqrt{21}}{7}\) ( vì \(\alpha>0\))

\(\Rightarrow\cos\alpha=\frac{4}{3}\sin\alpha=\frac{4}{3}.\frac{\sqrt{21}}{7}=\frac{4\sqrt{21}}{21}\)

Vậy \(\sin\alpha=\frac{\sqrt{21}}{7};\cos\alpha=\frac{4\sqrt{21}}{21}\)

ta có :

\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(1+2\sqrt{5}+5\right)}}\)

\(=\sqrt{\sqrt{\left(1+\sqrt{5}\right)^2}}=\sqrt{1+\sqrt{5}}< \sqrt{1+\sqrt{6}}\)

Vậy \(\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)

\(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{\sqrt{9+2.3.2\sqrt{2}+8}}=\sqrt{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\sqrt{3+2\sqrt{2}\sqrt{ }}\)

\(=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(1+\sqrt{2}\right)^2}=\left(1+\sqrt{2}\right)\)

Vậy \(\sqrt{\sqrt{17+12\sqrt{2}}}=1+\sqrt{2}\)

A B C D E K M I N H

a, có BD _|_ AC và CE _|_ AB (gt) => ^BDC = ^CEB = 90

=> E;D thuộc đường tròn đường kính BC

=> BEDC nội tiếp đường tròn đk BC

=> ^ADI = ^ABC 

xét tam giác ADE và tg ABC có : ^BAC chung

=> tg ADE đồng dạng tg ABC (g-g)

=> DE/BC = AD/AB

mà DE = 2DI và BC = 2BM

=> 2DI/2BM = AD/AB

=> DI/BM AD/AB 

xét tg ADI và tg ABM có : ^ADI = ^ABM (cmt)

=> tg ADI đồng dạng tg ABM (c-g-c)

b, chưa nghĩ ra

Trả lời:

Ta có: \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{1}{3}\)

\(\Rightarrow\cos\alpha=3\sin\alpha\)

Mà \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\sin^2\alpha+\left(3\sin\right)^2\alpha=1\)

\(\Rightarrow\sin^2\alpha+9\sin^2\alpha=1\)

\(\Rightarrow10\sin^2\alpha=1\)

\(\Rightarrow\sin^2\alpha=\frac{1}{10}\)

\(\Rightarrow\sin\alpha=\sqrt{\frac{1}{10}}\) \(=\frac{\sqrt{10}}{10}\)( vì \(\alpha>0\))

\(\Rightarrow\cos\alpha=3\sin\alpha=3.\frac{\sqrt{10}}{10}=\frac{3\sqrt{10}}{10}\)

Vậy \(\sin\alpha=\frac{\sqrt{10}}{10};\cos\alpha=\frac{3\sqrt{10}}{10}\)