\(\hept{\begin{cases}x^2+9y^2=10\\x+3y+8=12xy\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+6xy+9y^2=10+6xy\\x+3y=12xy-8\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+3y\right)^2=10+6xy\\\left(x+3y\right)^2=\left(12xy-8\right)^2\end{cases}}\)

\(\Rightarrow10+6xy=\left(12xy-8\right)^2\)

\(\Leftrightarrow144x^2y^2-198xy+54=0\)

\(\Leftrightarrow\orbr{\begin{cases}xy=1\\xy=\frac{3}{8}\end{cases}}\)

- Với \(xy=1\): 

\(\hept{\begin{cases}xy=1\\x+3y+8=12\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{1}{x}\\x+\frac{3}{x}-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3,y=\frac{1}{3}\\x=1,y=1\end{cases}}\)

- Với \(xy=\frac{3}{8}\):

\(\hept{\begin{cases}xy=\frac{3}{8}\\x+3y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{3}{8x}\\x+\frac{9}{8x}+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-7-\sqrt{31}}{4},y=\frac{-7+\sqrt{31}}{12}\\x=\frac{-7+\sqrt{31}}{4},y=\frac{-7-\sqrt{31}}{12}\end{cases}}\)

Thử lại ta đều thấy thỏa mãn. 

Đặt giá tiền một quyển vở và một cái bút lần lượt là \(x,y\)(đồng), \(x,y>0\).

Theo bài ra, ta có hệ phương trình: 

\(\hept{\begin{cases}8x+15y=88500\\13x+9y=90000\end{cases}}\Leftrightarrow\hept{\begin{cases}24x+45y=265500\\65x+45y=450000\end{cases}\Leftrightarrow\hept{\begin{cases}41x=184500\\y=\frac{90000-13x}{9}\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4500\\y=3500\end{cases}}\left(tm\right)\)

học lớp 9 chưa mà đòi đăng ? :))

a) Ta có : \(A=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

Để A nhận giá trị = 0 thì \(\sqrt{x}=0\)<=> x = 0 ( tmđk )

Vậy với x = 0 thì A = 0

b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)

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

\(=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)

c) P = B : A = \(\frac{\frac{\sqrt{x}}{\sqrt{x}+3}}{\frac{\sqrt{x}}{\sqrt{x}-5}}=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}}{\sqrt{x}+3}\times\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

Xét hiệu P - 1 ta có :

\(\frac{\sqrt{x}-5}{\sqrt{x}+3}-1=\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{-8}{\sqrt{x}+3}\)

Vì \(\hept{\begin{cases}-8< 0\\\sqrt{x}+3>0\end{cases}}\Rightarrow\frac{-8}{\sqrt{x}+3}< 0\)hay P - 1 < 0

=> P < 1 

a) \(A=0\Rightarrow\frac{x+5\sqrt{x}}{x-25}=0\Rightarrow x+5\sqrt{x}=0\Leftrightarrow x=0\)(thỏa mãn).

b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(B=\frac{\sqrt{x}}{\sqrt{x}+3}\)

c) \(P=B\div A=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}=1-\frac{8}{\sqrt{x}+3}< 1\)

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OMG !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a) Dễ chứng minh: \(\Delta AEF~\Delta ABC\left(c.g.c\right)\Rightarrow\widehat{AEF}=\widehat{ABC}\)

Tương tự: \(\widehat{CED}=\widehat{CBA}\)từ đó suy ra \(\widehat{AEF}=\widehat{CED}\)

Mà \(BE\perp AC\Rightarrow\widehat{FEH}=\widehat{DEH}\)hay EH là phân giác của \(\widehat{FED}\)

Tương tự: DH là phân giác của  \(\widehat{EDF}\) 

                 FH là phân giác của\(\widehat{EFD}\)

​Do đó H là giao điểm ba đường phân giác trong\(\Delta DEF\)​hay H là tâm đường tròn nội tiếp tam giác DEF (đpcm)

b) AK là đường kính nên \(\widehat{ACK}=90^0\)(góc nội tiếp chắn nửa đường tròn)

\(\widehat{ABC}=\widehat{AKC}\)(hai góc nội tiếp cùng chắn cung AC) \(\Rightarrow\Delta ADB~\Delta ACK\left(g.g\right)\)

\(\Rightarrow\frac{AB}{AK}=\frac{AD}{AC}\)hay \(2R=\frac{AB.AC}{AD}=\frac{AB.AC.BC}{BC.AD}=\frac{AB.AC.BC}{2S}\)

\(\Rightarrow S=\frac{AB.BC.AC}{4R}\)(đpcm)

????????????????????????????????????????????

hỏi đi men

Ta có: \(\frac{a}{1+4b^2}=\frac{a\left(1+4b^2\right)-4ab^2}{1+4b^2}=a-\frac{4ab^2}{1+4b^2}\ge a-\frac{4ab^2}{2\sqrt{4b^2.1}}=a-\frac{2ab^2}{2b}=a-ab\)(bđt cosi)

CMTT: \(\frac{b}{1+4a^2}\ge b-ab\)

=> P \(\ge a+b-2ab=4ab-2ab=2ab\)

Mặt khác ta có: \(a+b\ge2\sqrt{ab}\)(cosi)

=> \(4ab\ge2\sqrt{ab}\) <=> \(2ab\ge\sqrt{ab}\)<=> \(4a^2b^2-ab\ge0\) <=> \(ab\left(4ab-1\right)\ge0\)

<=> \(\orbr{\begin{cases}ab\le0\left(loại\right)\\ab\ge\frac{1}{4}\end{cases}}\)(vì a,b là số thực dương)

=> P \(\ge2\cdot\frac{1}{4}=\frac{1}{2}\)

Dấu "=" xảy ra <=> a = b = 1/2

Vậy MinP = 1/2 <=> a = b= 1/2

Ta có: \(a+b=4ab\le\left(a+b\right)^2\Leftrightarrow\left(a+b\right)\left[\left(a+b\right)-1\right]\ge0\)

Mà \(a+b>0\Rightarrow a+b\ge1\)

Áp dụng BĐT Cô-si, ta có: \(P=\frac{a}{1+4b^2}+\frac{b}{1+4a^2}=\left(a-\frac{4ab^2}{1+4b^2}\right)+\left(b-\frac{4a^2b}{1+4a^2}\right)\)\(\ge\left(a-\frac{4ab^2}{4b}\right)+\left(b-\frac{4a^2b}{4a}\right)=\left(a+b\right)-2ab=\left(a+b\right)-\frac{a+b}{2}=\frac{a+b}{2}\ge\frac{1}{2}\)

Đẳng thức xảy ra khi a = b = ¹/₂