chịu

a: \(B=\frac{1}{x+4}-\frac{3x}{4-x}-\frac{25x-4}{x^2-16}\)

\(=\frac{x-4+3x\left(x+4\right)-25x+4}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3x^2+12x-24x}{\left(x-4\right)\left(x+4\right)}=\frac{3x^2-12x}{\left(x-4\right)\left(x+4\right)}=\frac{3x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{3x}{x+4}\)

b: |3-2x|=5

=>|2x-3|=5

=>\(\left[\begin{array}{l}2x-3=5\\ 2x-3=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=8\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)

Thay x=-1 vào B, ta được:

\(B=\frac{3\cdot\left(-1\right)}{-1+4}=\frac{-3}{3}=-1\)

c: A<=2/3B

=>\(\frac{2x-5}{x+4}\le\frac23\cdot\frac{3x}{x+4}=\frac{2x}{x+4}\)

=>\(-\frac{5}{x+4}\le0\)

=>x+4>0

=>x>-4

Kết hợp ĐKXĐ, ta được: x>-4 và x<>4

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

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

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

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

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

a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\sqrt{x}-1\)

a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

                        Đk: \(x>0\) và \(x\ne1\)

\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

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

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

b) Thay \(x=3+2\sqrt{2}\) vào A ta được:

  \(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)

      \(=\sqrt{2}+1-1=\sqrt{2}\)

(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

$\textbf{a)}$

Điều kiện: $x\ne0,\ x\ne-1,\ x\ne1.$

Ta có $B=\left(\dfrac{x+1}{2(x-1)}+\dfrac{3x-1}{(x-1)(x+1)}-\dfrac{x+3}{2(x+1)}\right):\dfrac3{x+1}.$

Quy đồng các phân thức trong ngoặc:

$\dfrac{x+1}{2(x-1)}=\dfrac{(x+1)^2}{2(x-1)(x+1)},$

$\dfrac{x+3}{2(x+1)}=\dfrac{(x+3)(x-1)}{2(x-1)(x+1)}.$

Do đó \[\begin{aligned}&\dfrac{(x+1)^2+2(3x-1)-(x+3)(x-1)}{2(x-1)(x+1)}\\&=\dfrac{x^2+2x+1+6x-2-(x^2+2x-3)}{2(x-1)(x+1)}\\&=\dfrac{6x+2}{2(x-1)(x+1)}=\dfrac{3x+1}{(x-1)(x+1)}.\end{aligned}\]

Suy ra $B=\dfrac{3x+1}{(x-1)(x+1)}\cdot\dfrac{x+1}{3}=\dfrac{3x+1}{3(x-1)}.$

a) -4x²+2x

b) -4x²+2x=0

x(-4x+2)=0

=> x=0 hoặc -4x+2=0

                     -4x = -2

                        x=1/2(đpcm)

c) Thay x=-1/4 vào -4x²+2x ta có : -4 (-1/4)² +2(-1/4) = ... (tự tính )

a) A = (x - 3)(x + 1) - (2x - 1)^2 - (x + 2)(x - 2)

A = x^2 - 2x - 3 - 4x^2 + 4x - 1 - x^2 + 4

A = -4x^2 + 2x

b) 4x^2 - 2x = 0

<=> 2x(2x - 1) = 0

<=> 2x = 0 hoặc 2x - 1 = 0

<=> x = 0 hoặc x = 1/2

c) với x = -1/4, ta có:

4(-1/4)^2 - 2(-1/4) = 3/4