Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Rightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}\Leftrightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(a+b\right)^2\ge4ab\left(1\right)\\\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(2\right)\end{cases}}\)

Theo đề bài:

\(a+b+3ab=1\)

\(\Leftrightarrow4\left(a+b\right)+12ab=4\)

\(\Leftrightarrow4\left(a+b\right)+3\left(a+b\right)^2\ge4\left(theo\left(1\right)\right)\)

\(\Leftrightarrow3\left(a+b\right)^2+4\left(a+b\right)-4\ge0\)

\(\Leftrightarrow\left(a+b+2\right)\left[3\left(a+b\right)-2\right]\ge0\)

\(\Leftrightarrow3\left(a+b\right)-2\ge0\left(a,b>0\Rightarrow a+b+2>0\right)\)

\(\Leftrightarrow a+b\ge\frac{2}{3}\)

`\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge\frac{4}{9}\left(theo\left(2\right)\right)\)

Áp dụng các kết quả trên, ta có:

\(\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2\le2\left(1-a^2+1-b^2\right)\)\(=4-2\left(a^2+b^2\right)\le4-\frac{4}{9}=\frac{32}{9}\)

\(\Rightarrow\sqrt{1-a^2}+\sqrt{1-b^2}\le\frac{4\sqrt{2}}{3}\)

Ta có: \(\frac{3ab}{a+b}=\frac{1-\left(a+b\right)}{a+b}=\frac{1}{a+b}-1\le\frac{1}{\frac{2}{3}}-1=\frac{1}{2}\)

\(\Rightarrow A\le\frac{4\sqrt{2}}{3}+\frac{1}{2}\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b+3ab=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\3a^2+2a-1=0\end{cases}\Leftrightarrow}a=b=\frac{1}{3}\left(a,b>0\right)}\)

Vậy max A là \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)

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

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

\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)

\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\) ĐK : \(x\ge0;x\ne1\)

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

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

\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}+3}\)   

\(=\frac{3\sqrt{x}+1-2\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)   

\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)   

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

\(=\frac{1}{\sqrt{x}-1}\)

\(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)

\(=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

\(\sqrt{11+2\sqrt{18}}=\sqrt{11+2\sqrt{9.2}}\)

\(=\sqrt{\left(\sqrt{9}\right)^2+2\sqrt{9.2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}=\left|\sqrt{9}+\sqrt{2}\right|=\sqrt{9}+\sqrt{2}\)

Ta có: \(P=-\left(b\sqrt{a}-2a\sqrt{b}+a\sqrt{a}\right)+a\sqrt{a}=-\left(\sqrt{b+\sqrt{a}}-\sqrt{a+\sqrt{a}}\right)^2+a\sqrt{a}\)

           \(\le a\sqrt{a}\le1\)

Dấu "=" xảy ra khi a=b=1 

Mình làm thế này không biết có đúng ko mn

\(\left(P\right):y=x^2\)

\(d:y=\left(2-2m\right)x+m\)

+) Xét phương trình: \(x^2+\left(2m-2\right)x-m=0\left(1\right)\)có \(\Delta'=m^2-m+1>0\forall m\)

Vậy d luôn cắt (P) tại A,B phân biệt.

+) Giả sử \(x_1,x_2\)là hai nghiệm của (1), ta có: \(A\left(x_1;y_1\right),B\left(x_2;y_2\right)\)hay \(A\left(x_1;x_1^2\right),B\left(x_2;x_2^2\right)\)

Vì \(M\left(\frac{1}{2};1\right)\) là trung điểm AB nên \(\hept{\begin{cases}x_1+x_2=1\\\frac{x_1^2+x_2^2}{2}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x_1+x_2=1\\1-2x_1x_2=2\end{cases}}\)(I)

Theo hệ thức Viet: \(\hept{\begin{cases}x_1+x_2=2-2m\\x_1x_2=-m\end{cases}}\)(II)

Từ (I),(II) suy ra \(\hept{\begin{cases}2-2m=1\\1+2m=2\end{cases}}\Leftrightarrow m=\frac{1}{2}\)

Như vậy \(KH=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(2-2m\right)^2-4\left(-m\right)}=\sqrt{3}.\)

cô-si ngược auto ra @-@

Từ giả thiết ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \(\frac{1}{x}=A;\frac{1}{y}=N;\frac{1}{z}=H\)khi đó : \(A+N+H=1\)

Ta có : \(H.=\frac{H}{9A^2+1}+\frac{A}{9N^2+1}+\frac{N}{9H^2+1}\)

Theo bđt cô si ta có đánh giá sau :

\(\frac{H}{9A^2+1}=\frac{H\left(9A^2+1\right)-9HA^2}{9A^2+1}=H-\frac{HA^2}{9A^2+1}\ge H-\frac{3}{2}AH\)

Tương tự và cộng theo vế ta được :

\(H=A+N+H-\frac{3}{2}\left(AN+NH+HA\right)=1-\frac{3}{2}\left(AN+NH+HA\right)\)

Áp dụng bđt phụ \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)có:

\(1-\frac{3}{2}\left(AN+NH+HA\right)\ge1-\frac{\frac{3}{2}\left(A+N+H\right)^2}{3}=1-\frac{\frac{3}{2}}{3}=\frac{1}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(A=N=H=\frac{1}{3}\)\(< =>x=y=z=\frac{1}{3}\)

=))

2000 à bn

năm 2000 có phải năm nhuận ko?

trả lời đi!