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Gọi số thứ nhất là a; số thứ hai là b. (a, b \(\inℤ^+\))

Theo đề bài: a : b = \(\frac{a}{b}=\frac{2}{3}\)và b - a = 12

Vì \(\frac{a}{b}=\frac{2}{3}\)nên a = 2m; b = 3m  (m \(\inℕ^∗\)) và b > a.

Ta còn có: b - a = 12

=> 3m - 2m = 12

=> m(3 - 2) = 12

=> m           = 12

=> a = 12.2 = 24 và b = 12.3 = 36

Vậy a = 24 và b = 36

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne\pm2y\end{cases}}\)

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{x+2y}{\left(x+2y\right)\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}=\frac{2x-4y+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(\frac{5+\sqrt{3}-5+\sqrt{3}}{25-3}=\frac{2\sqrt{3}}{22}=\frac{\sqrt{3}}{11}\\ \)

học tốt 

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

\(=\frac{5+\sqrt{3}}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}-\frac{5-\sqrt{3}}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}\)

\(=\frac{5+\sqrt{3}-\left(5-\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}\)

\(=\frac{5+\sqrt{3}-\left(5-\sqrt{3}\right)}{22}\)

\(=\frac{5+\sqrt{3}-5+\sqrt{3}}{22}\)

\(=\frac{2\sqrt{3}}{22}=\frac{\sqrt{3}}{11}\)

a.)Đkxđ bạn tự tìm nha!!!

A=\(\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)

\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}:\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}\)

\(\Leftrightarrow\)\(\frac{x+1}{x-1}\left(tm\text{đ}k\right)\)

b.)Thay \(x=\frac{1}{2}\)vào A \(\Rightarrow\)\(A=-3\)

x² - 5 = 0

=> x² = 5

=> x = \(\sqrt{5}\)

Nguyễn Ngọc Anh             

\(x^2-5=0\Rightarrow x^2=5\Rightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

vậy \(x\in\left\{\pm\sqrt{5}\right\}\)

M = x^2 + y^2 - xy - x + y + 1

12M = 12x^2 + 12y^2 - 12xy - 12x + 12y + 12

12M = 3(4x^2 + y^2 + 1 - 4xy - 4x + 2y) + 9y^2 + 6y + 9

12M = 3(2x - y - 1)^2 + (3y + 1)^2 + 8

12M > 8

tự xét dấu = 

M = x² + y² - xy - x + y +1

2M = 2x² + 2y² - 2xy - 2x + 2y + 2

2M = ( x² - 2xy + y² ) + ( x² -2x +1 ) + ( y² + 2y + 1)

2m = ( x - y )² + ( x-1 )² + ( y + 1 )²

Ta có \(\left(x-y\right)^2\ge\forall x;y\)

          \(\left(x-1\right)^2\ge0\forall x\)

          \(\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow2M\ge0\forall x;y\)

Dấu "=" xảy ra khi x - y = 0; x - 1 = 0; y + 1 = 0

                      <=> x = y ; x = 1; y = -1 ( vô lí )

Vậy không tồn tại giá trị nhỏ nhất nào của biểu thức M