mình cần gấp mong các bạn giải giùm

c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)

b) (4√x + 4)/(x + 2√x + 5) ≥ 1

⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

Do x ≥ 0 ⇒ x + 2√x + 5 > 0

⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0

⇔ 4√x + 4 - x - 2√x - 5 ≤ 0

⇔ -x + 2√x - 1 ≤ 0

⇔ -(x - 2√x + 1) ≤ 0

⇔ -(√x - 1)² ≤ 0 (luôn đúng)

Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0

a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

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

b: 4(căn x+1)>=4

x+2căn x+5>=5

=>P<=4/5<1

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{2\text{x}3+4\text{x}6+6\text{x}9+8\text{x}12}\)

\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{\text{1x}2\text{x}3+2\text{x}4\text{x}3+3\text{x}6\text{x}3+4\text{x}8\text{x}3}\)

\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{3\left(1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8\right)}\)

\(=\frac{1}{3}\)

Ta có: \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}=\frac{1.2}{2.3}+\frac{2.4}{4.6}+\frac{3.6}{6.9}+\frac{4.8}{8.12}.\)

                                                       \(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{3}.4=\frac{4}{3}\)

1x2 + 2x4 + 3x6 + 4x8

= (1x2) (1 + 2 + 3 + 4) 2x3 + 4x6 + 6x9 + 8x12

= (2x3) (1 + 2 + 3 + 4)

Do đó, (1x2 + 2x4 + 3x6 + 4x8 ) / (2x3 + 4x6 + 6x9 + 8x12)

= (1x2) / (2x3) = 1/3

\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)

\(=\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)