a, Với \(x=3\)\(=>A=\frac{x-1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)

Vậy A = 1 khi x = 3

b, Ta có : \(B=\frac{1}{x}-\frac{x}{2x+1}+\frac{2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

Ta có : \(A=\frac{x-1}{2};B=\frac{x-1}{2x+1}\)

\(=>C=A:B=\frac{x-1}{2}:\frac{x-1}{2x+1}=\frac{2x+1}{2}=x+\frac{1}{2}\)

đề sai bạn ơi 

a: Thay x=5 vào B, ta được:

\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)

b:  \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)

a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

\(a,B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\\ B=x-\sqrt{x}+1-\sqrt{x}=\left(\sqrt{x}-1\right)^2\)

Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

b: \(A=\dfrac{2x+1-x+\sqrt{x}}{x\sqrt{x}-1}\cdot\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)

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

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

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

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

1,

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

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

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

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

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

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

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

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

Ta có: \(A=\dfrac{2x}{1-x^3}+\dfrac{1}{x^2-x}+\dfrac{1}{x^2+x+1}\)

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

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

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

Thay x=10 vào A, ta được:

\(A=\dfrac{1}{10\cdot\left(10^3-1\right)}=\dfrac{1}{10\cdot999}=\dfrac{1}{9990}\)