a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)

\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2x+1-3x-1+x^2}{3x}\)

\(A=\frac{x^2-x}{3x}\)

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

\(A=\frac{x-1}{3}\)

b) Thay x = 4 ta có :

\(A=\frac{4-1}{3}=\frac{3}{3}=1\)

c) Để A thuộc Z thì \(x-1⋮3\)

\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)

\(\Rightarrow x\in\left\{1;4;7;...\right\}\)

Vậy.....

Cho Bt 

a,Tìm điều kiện xác định và rút gọn bt A

b,Tính giá trị bt A tại x=4

c,tìm x thuộc Z để a thuộc Z

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{25}{2}\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

\(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(-x^2+1\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^3\right)}{\left(x+3\right)x}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^2\right)+3x^2-14x+3}{\left(x+3\right)x}\)

\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)

\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)

\(A=\frac{-x^3+x^2+7x-15}{x+3}\)

\(A=-\frac{\left(x+3\right)\left(x^2-4x+5\right)}{x+3}\)

\(A=-\left(x^2-4x+5\right)\)

\(A=-x^2+4x-5\)

Trình độ hơi thấp, có gì sai sót xin bỏ qua cho :)

umk cảm ơn bạn trước nhé

Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-3\end{cases}}}\)

\(A=\frac{1-x^2}{x}\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(1-x\right)\left(1+x\right)}{x}\cdot\frac{x^2-x-3}{x+3}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{\left(1-x\right)\left(1+x\right)\left(x^2-x-3\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{-x^4+x^3+4x^2-x-3+3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)

\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)

\(A=\frac{-x^3+x^2+7x-15}{x+3}\)

Phân tích tử thức : 

\(-x^3+x^2+7x-15\)

\(=-x^3-3x^2+4x^2+12x-5x-15\)

\(=-x^2\left(x+3\right)+4x\left(x+3\right)-5\left(x+3\right)\)

\(=\left(x+3\right)\left(-x^2+4x-5\right)\)

\(\Rightarrow A=\frac{\left(x+3\right)\left(-x^2+4x-5\right)}{x+3}=-x^2+4x-5\)

\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)

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

\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

\(=-2\left(x^2-2x-\frac{3}{2}\right)\)

\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)

\(=-2\left(x-1\right)^2+5\)

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)