Rút gọn đa thúc sau:M=(\(12\text{x}^8+8\text{x}^2+6\text{x}-7\))-(\(12\text{x}^8+2\text{x}-8\))+\(\left(5-8\text{x}^2\right)\)
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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.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{2\text{x}3+4\text{x}6+6\text{x}9+8\text{x}12}\)
\(\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{2\text{x}3+4\text{x}6+6\text{x}9+8\text{x}12}\)
\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{\text{1x}2\text{x}3+2\text{x}4\text{x}3+3\text{x}6\text{x}3+4\text{x}8\text{x}3}\)
\(=\frac{1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8}{3\left(1\text{x}2+2\text{x}4+3\text{x}6+4\text{x}8\right)}\)
\(=\frac{1}{3}\)
Ta có: \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}=\frac{1.2}{2.3}+\frac{2.4}{4.6}+\frac{3.6}{6.9}+\frac{4.8}{8.12}.\)
\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{3}.4=\frac{4}{3}\)
\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
Bài làm:
\(M=\left(12x^8+8x^2+6x-7\right)-\left(12x^8+2x-8\right)+\left(5-8x^2\right)\)
\(M=4x+6\)
\(M=4x+6\)
Học tốt