giúp mk nha mk cần gấp

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

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

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

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

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

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

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

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

b) Để P > 0 => \(\frac{-3}{\sqrt{x}-3}>0\)

mà -3 < 0 nên để P > 0 thì \(\sqrt{x}-3< 0\)<=> \(\sqrt{x}< 3\)<=> \(x< 9\)

Kết hợp với ĐKXĐ => Với \(0\le x< 9\)thì P > 0

Đặt \(\frac{x}{y}+\frac{y}{x}=a\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2=a^2\Leftrightarrow\frac{x^2}{y^2}+\frac{y^2}{x^2}+2=a^2\)

Mà \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\sqrt{\frac{x^2}{y^2}\cdot\frac{y^2}{x^2}}=2\) (BĐT Cauchy)

\(\Rightarrow a^2=\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\ge4\Leftrightarrow\orbr{\begin{cases}a\ge2\\a\le-2\end{cases}}\)

Ta có: \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(\Leftrightarrow a^2-2+4\ge3a\)

\(\Leftrightarrow a^2-3a+2\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)\ge0\)

\(\Rightarrow\orbr{\begin{cases}a\ge2\\a\le1\end{cases}}\) Dấu "=" xảy ra khi: \(\orbr{\begin{cases}a=2\\a=1\end{cases}}\)

Nếu \(a=2\Leftrightarrow\frac{x}{y}=\frac{y}{x}=\frac{2}{2}=1\Rightarrow x=y=1\)

Nếu \(a=1\Leftrightarrow\frac{x}{y}=\frac{y}{x}=\frac{1}{2}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}y\\y=\frac{1}{2}x\end{cases}}\Rightarrow x=y=0\left(ktm\right)\)

Vậy \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\Leftrightarrow x=y=1\)

đk: \(\hept{\begin{cases}x\ne0\\x^2\ge\sqrt{\frac{5}{6}}\end{cases}}\)

Ta có: \(x^2\ge\sqrt{\frac{5}{6}}\Rightarrow\hept{\begin{cases}\frac{5}{x^2}>0\\6x^2-1>0\end{cases}}\)

Áp dụng BĐT Cauchy ta có: \(\sqrt{30-\frac{5}{x^2}}=\sqrt{\frac{5}{x^2}\left(6x^2-1\right)}\le\frac{\frac{5}{x^2}+6x^2-1}{2}\) (1)

Mà \(\sqrt{6x^2-\frac{5}{x^2}}=\sqrt{\left(6x^2-\frac{5}{x^2}\right)\cdot1}\le\frac{6x^2-\frac{5}{x^2}+1}{2}\) (2)

Cộng vế (1) và (2) lại ta được:

\(\sqrt{30-\frac{5}{x^2}}+\sqrt{6x^2-\frac{5}{x^2}}\le\frac{12x^2}{2}=6x^2\) 

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{5}{x^2}=6x^2-1\\6x^2-\frac{5}{x^2}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{-1;1\right\}\)

Theo đề bài ta có: \(\hept{\begin{cases}x>1\\y>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1>0\\y>0\end{cases}}\)

=> \(\frac{1}{\left(x-1\right)^3}>0\) ; \(\left(\frac{x-1}{y}\right)^3>0\) ; \(\frac{1}{y^3}>0\)

Áp dụng BĐT Cauchy ta có:

\(\frac{1}{\left(x-1\right)^3}+1+1\ge3\cdot\frac{1}{x-1}=\frac{3}{x-1}\)

\(\left(\frac{x-1}{y}\right)^3+1+1\ge3\cdot\frac{x-1}{y}=\frac{3\left(x-1\right)}{y}\)

\(\frac{1}{y^3}+1+1\ge3\cdot\frac{1}{y}=\frac{3}{y}\)

Cộng vế 3 BĐT trên lại ta được:

\(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}+6\ge3\left(\frac{1}{x-1}+\frac{x-1}{y}+\frac{1}{y}\right)=3\left(\frac{1}{x-1}+\frac{x}{y}\right)\)

\(\Leftrightarrow\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left[\left(\frac{1}{x-1}-2\right)+\frac{x}{y}\right]=3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=2\\y=1\end{cases}}\)

\(y=\sqrt{\sqrt{x}-1+\sqrt{9-x}}\)

\(y=x^2-10x+27\)

\(|xgiaođiểm\left(\frac{-\sqrt{33}}{2}+\frac{1}{2},0\right)\)

\(|ygiaođiểm\left(0,\sqrt[4]{2}\right)\)

\(|ygiaođiểm\left(0,27\right)\)

\(|\)giá trị bé nhất (5,2)

\(|\)Dạng tiêu chuẩn y=(x-5²)+2

Chèn ảnh kiểu gì mà mình quên rồi

Gọi hình thang đó là ABCD (AB//CD)

AB=15

AD=BC=25 góc

DAB=góc ABC=120 độ.Kẻ AH,BK vuông góc với CD (HK \(\varepsilon\)CD) \(\Rightarrow\)HK=AB=15(cm)

Xét tam giác AHD có:AD=25;Góc D=60 độ \(\Rightarrow\)DH=AD.\(\cos\)=\(\frac{AD}{2}\)=12.5(cm)

Tương tự ta có CK=12.5(cm) \(\Rightarrow CD\)=CK+DH+HK=12.5+12.5+15=40(cm) 

\(\Rightarrow\)Chu vi ABCD=AB+BC+CD+DA=105(cm)

105 cm nhé bạn

Tìm GTLN

Áp dụng BĐT Bunyakovsky ta có:

\(P^2=\left(\sqrt{x}+2\sqrt{y}\right)^2\le\left(1^2+2^2\right)\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]\)

\(=5\left(a+b\right)=5\cdot2020=10100\)

\(\Rightarrow P\le10\sqrt{101}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{1}=\frac{y}{4}\\x+y=2020\end{cases}}\Rightarrow\hept{\begin{cases}x=404\\y=1616\end{cases}}\)

Vậy \(Max_P=10\sqrt{100}\Leftrightarrow\hept{\begin{cases}x=404\\y=1616\end{cases}}\)