Áp dụng bất đẳng thức AM - GM có:

\(A=\dfrac{2\sqrt{a}}{a+1}\le\dfrac{2\sqrt{a}}{2\sqrt{a}}=1\)

Dấu " = " khi a = 1

Vậy \(MAX_A=1\) khi a = 1

\(a-b=20\Rightarrow a^2-2ab+b^2=400\)

\(\Leftrightarrow a^2+b^2=400+2ab\)

\(\Rightarrow a^2+b^2=426\) ( do ab = 3 )

Vậy \(a^2+b^2=426\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

Ta có: \(M=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=8\)

Vậy M = 8

Ta có: \(99\equiv1\left(mod2\right)\)

\(\Rightarrow99^{20}\equiv1\left(mod2\right)\)

\(11\equiv1\left(mod2\Rightarrow\right)11^9\equiv1\left(mod2\right)\)

\(\Rightarrow99^{20}-11^9\equiv0\left(mod2\right)\)

\(\Rightarrow99^{20}-11^9⋮2\)

Áp dụng bài này: Câu hỏi của Đỗ Lê Tú Linh - Toán lớp 7 - Học toán với OnlineMath

\(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\)

\(=\left(a^2b+ab^2+abc\right)+\left(a^2c+ac^2+abc\right)+\left(b^2c+bc^2+abc\right)\)

\(=ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)\)

\(=\left(ab+bc+ca\right)\left(a+b+c\right)\)

\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(=1-\dfrac{1}{n^2+2n+1}\)

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

Giải các phương trình logarit sau:  log x +  log x 2 =  log 9 x

Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\z+x=2b\end{matrix}\right.\)

\(\Rightarrow2A=\left(y+z\right)x^2+\left(x+z\right)y^2+\left(x+y\right)z^2+2xyz\)

\(=x^2y+x^2z+xy^2+y^2z+xz^2+yz^2+2xyz\)

\(=\left(x^2y+x^2z+xyz+xy^2\right)+\left(xz^2+yz^2+xyz+y^2z\right)\)

\(=x\left(xy+xz+yz+y^2\right)+z\left(xy+yz+xz+y^2\right)\)

\(=\left(x+z\right)\left[x\left(y+z\right)+y\left(y+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(=8abc\Rightarrow A=4abc\)

Vậy...