A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)

A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)

A = 4

Vậy A không phụ thuộc vào x

Chúc bn học tốt!

Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)

\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)

\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)

\(=10\cdot\dfrac{2}{5}=4\)

\(A=\left[\frac{6x^2}{x^3-1}-\frac{2x-2}{x^2+x+1}-\frac{1}{x-1}\right]:\frac{x^2+9}{\left(x-1\right)\left(9-4x\right)}\)

\(=\left[\frac{6x^2}{x^3-1}-\frac{\left(2x-2\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{6x^2-\left(2x^2-4x+2\right)-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{5x^2-2x^2+4x-2-x-1}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

\(=\frac{3x^2+3x-3}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

Biểu thức A bạn viết đúng chưa?

Để \(\frac{2x-4}{x+2}\)nguyên thì

\(2x-4⋮x+2\)

\(\Rightarrow2\left(x+2\right)-8⋮x+2\)

Mà \(2\left(x+2\right)⋮x+2\)

\(\Rightarrow8⋮x+2\)

\(\Rightarrow x+2\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\)

\(\Rightarrow x\in\left\{-1;0;2;6;-3;-4;-6;-10\right\}\)

Học tốt

A=\(\frac{2x-4}{x+2}=\frac{2\left(x+2\right)-8}{x+2}=2-\frac{8}{x+2}\)

Để A nguyên thì \(\frac{8}{x+2}\)nguyên =>\(x+2\inƯ\left(8\right)=\left\{\pm1,\pm2,\pm4,\pm8\right\}\)

Ta có bảng:

Vậy x={-10,-6,-4,-3,1,0,2,6}thì A nguyên

Để A có giá trị nguyên khi \(\frac{2x-4}{x+2}\) nguyên

ta có \(\frac{2x-4}{x+2}=\frac{2x+4-8}{x+2}=\frac{2\left(x+2\right)-8}{x+2}=2-\frac{8}{x+2}\)

vì 2 nguyên nên A nguyên khi \(\frac{8}{x+2}\) nguyên 

<=> x+2 thuộc Ư(8)

ta có bảng

vậy x={-3;-1;-4;0;-6;2;-10;6}

\(A=3x-x^2\)

\(=-\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}\right)\)

\(=-\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)

\(=\frac{9}{4}-\left(x-\frac{3}{2}\right)^2\ge\frac{9}{4}\)

Min A = \(\frac{9}{4}\)khi \(x-\frac{3}{2}=0=>x=\frac{3}{2}\)

\(B=25+2x-x^2\)

\(=-\left(x^2-2x+1-26\right)\)

\(=-\left(\left(x-1\right)^2-26\right)\)

\(=26-\left(x-1\right)^2\ge26\)

Min A = 26 khi \(x-1=0=>x=1\)

\(C=x^2-5x+19\)

\(=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2+\frac{51}{4}\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{51}{4}\ge\frac{51}{4}\)

Min C = \(\frac{51}{4}\)khi \(x+\frac{5}{2}=0=>x=\frac{-5}{2}\)

@@@ nha các bạn . Thanks

cảm ơn bạn nhiều lắm

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

\(=\dfrac{4-x}{x}:\dfrac{x^2+x+1-x+3}{x^2+x+1}\)

\(=\dfrac{4-x}{x}\cdot\dfrac{x^2+x+1}{x^2+4}=\dfrac{\left(4-x\right)\left(x^2+x+1\right)}{x\left(x^2+4\right)}\)

b: x^4-7x^2-4x+20=0

=>(x-2)^2(x^2+4x+5)=0

=>x=2

Khi x=2 thì \(A=\dfrac{\left(4-2\right)\left(4+2+1\right)}{2\left(4+4\right)}=\dfrac{7}{8}\)