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

\(a,P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\left(\dfrac{\sqrt{x}+2}{2}\right)^2\left(x\ge0;x\ne4\right)\\ P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\\ P=\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(b,\)Ta có \(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

Thay vào \(P\), ta được:

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

\(c,\)Để \(P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow x< 4\)

Vậy để \(P< 1\) thì \(x< 4\)

Tick nha

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\left(\dfrac{\sqrt{x}+2}{2}\right)^2\)

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

b: Thay \(x=6-2\sqrt{5}\) vào P, ta được:

\(P=\dfrac{\sqrt{5}+1+2}{\sqrt{5}+1-2}=\dfrac{3+\sqrt{5}}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{2}\)

a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)

\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)

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

b: 2x^2+7x+3=0

=>(2x+3)(x+2)=0

=>x=-3/2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)

d: |B|<1

=>B>-1 và B<1

=>B+1>0 và B-1<0

=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)

Đề không hiển thị biểu thức B. Bạn xem lại.

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

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{1}{2x-1}=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(2x-1\right)}=\dfrac{-4x}{2x-1}\)

b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\dfrac{-4\cdot3}{2\cdot3-1}=\dfrac{-12}{5}\)

Khi x=-3 thì \(B=\dfrac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\dfrac{12}{-7}=\dfrac{-12}{7}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

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

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

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