C=\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{\text{5-x}}{\text{1-x}^{\text{2}}}\right)\):\(\dfrac{1-2x}{\text{x}^{\text{2}}-1}\)
a) Rút gọn
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Khách
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10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
KV
5 tháng 6 2023
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
4 tháng 6 2023
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
10 tháng 11 2021
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
10 tháng 11 2021
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
18 tháng 12 2021
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
4 tháng 1 2023
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
10 tháng 11 2021
\(C=\dfrac{x-\dfrac{1}{x^2}}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}\)
Đk: \(x\ne0\)
\(\Rightarrow C=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{x^3-1}{x^2+x+1}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
10 tháng 11 2021
\(=\dfrac{\dfrac{x^3-1}{x^2}}{\dfrac{x^2+x+1}{x^2}}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)