A B D C I

Đặt BC = a , AC = b , AB = c . Ta có :

\(BD=\frac{a+c-d}{2}\)

\(DC=\frac{a+b-c}{2}\)

Do đó , ta giả sử \(\left(b\ge c\right)\)

\(BD.DC=\frac{a+c-b}{2}.\frac{a+b-c}{2}\)

                 \(=\frac{a-\left(b-c\right)}{2}.\frac{a+\left(b-c\right)}{2}\)

                 \(=\frac{a^2-\left(b-c\right)^2}{4}\)

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

                 \(=\frac{a^2-\left(b^2+c^2\right)+2bc}{4}\)

Do \(a^2=b^2+c^2\)nên   \(BD.DC=\frac{2bc}{3}=\frac{bc}{2}=S_{ABC}\)

minh chiu

Trả lời:

\(\sqrt{x}+\sqrt{3x-2}=x^2+1\)\(\left(ĐK:x\ge\frac{2}{3}\right)\)

\(\Leftrightarrow x^2+1-\sqrt{x}-\sqrt{3x-2}=0\)

\(\Leftrightarrow2x^2+2-2\sqrt{x}-2\sqrt{3x-2}=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(x-2\sqrt{x}+1\right)+\left(3x-2-2\sqrt{3x-2}+1\right)=0\)

\(\Leftrightarrow2.\left(x-2x+1\right)+\left(\sqrt{x}-1\right)^2+\left(\sqrt{3x-2}-1\right)=0\)

\(\Leftrightarrow2.\left(x-1\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{3x-2}-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x}-1=0\\\sqrt{3x-2}-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=1\\\sqrt{3x-2}=1\end{cases}}\)

                                                     \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=1\left(TM\right)\\3x-2=1\end{cases}}\)

                                                     \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=1\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)

Vậy nghiệm của phương trình là \(x=1\)

\(\sqrt{x}+\sqrt{3x-2}=x^2+1\)

\(4x+2\sqrt{x\left(3x-2\right)}-2=x^4+2x^2+1\)

\(2\sqrt{x\left(3x-2\right)}=x^4+2x^2-4x+3\)

\(4x\left(3x-2\right)=\left(x^4+2x^2-4x+3\right)^2\)

\(12x^2-8x=\left(x^4+2x^2-4x+3\right)^2\)

\(12x^2-8x-\left(x^4+2x^2-4x+3\right)^2=0\)

Đến đây chỗ \(\left(x^4+2x^2-4x+3\right)^2\)lớn quá mình chưa tìm đc cách giải 

cac cap tam giac co dien h bang nhau la AOB va BOC. Vi co cap song song voi nhau va cat toi diem O

bạn Phạm Thị Thúy Phượng gửi nhầm bài rồi 

Ta có: \(\frac{AC}{AB}=\frac{5}{12}\)

\(\Leftrightarrow\frac{AC^2}{AB^2}=\frac{25}{144}\)

\(\Leftrightarrow\frac{AC^2+AB^2}{AB^2}=\frac{25+144}{144}\)

\(\Leftrightarrow\frac{BC^2}{AB^2}=\frac{169}{144}\)

\(\Leftrightarrow\frac{BC}{AB}=\frac{13}{12}\)\(\Leftrightarrow\frac{26}{AB}=\frac{13}{12}\Leftrightarrow AB=\frac{26\times12}{13}=24\left(cm\right)\)

\(\Rightarrow\frac{AC}{24}=\frac{5}{12}\Leftrightarrow AC=\frac{24\times5}{12}=10\left(cm\right)\)

Xét \(\Delta ABC\), \(\widehat{A}=90^0,AH\perp BC\)

\(AB^2=BH\times BC\)( Hệ thức lượng trong tam giác vuông )

 \(\Leftrightarrow24^2=BH\times26\)

\(\Leftrightarrow576=BH\times26\)

\(\Leftrightarrow BH=\frac{288}{13}\left(cm\right)\)

\(AC^2=CH\times CB\)( hệ thức lượng trong tam giác vuông )

\(\Leftrightarrow10^2=CH\times26\)

\(\Leftrightarrow100=CH\times26\)

\(\Leftrightarrow CH=\frac{50}{13}\left(cm\right)\)

Đáp số \(AB=24cm\), \(AC=10cm\)

            \(BH=\frac{288}{13}cm\), \(CH=\frac{50}{13}cm\)

a)
Cho x, y, z thỏa mãn x^2 + y^2 + z^2 ≤ 3. Tìm GTNN của P = 1/(1 + xy) + 1/(1 + yz) + 1/(1 + xz)
Engel
P≥(1+1+1)^2/[3+(xy+yz+zx)]
BDT co ban
x^2 + y^2 + z^2 ≥xy+yz+zx
=>xy+yz+zx≤3
<=>P≥3^2/(3+3)=3/2
gtnn P=3/2 "= khi x=y=z=1