(\(\dfrac{\text{3}}{\text{2}}\).\(\sqrt[]{\dfrac{\text{4}}{\text{25}}+}\)3.\(\sqrt[]{\text{0,04}}\)):\(\sqrt[]{\dfrac{\text{9}}{\text{64}}}\)
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1
5 tháng 11 2021
\(=4\cdot5-2\cdot\dfrac{2}{3}=20-\dfrac{4}{3}=\dfrac{56}{3}\)
20 tháng 10 2023
a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)
\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)
\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)
\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)
\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)
b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)
\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)
5 tháng 11 2021
\(=-5\cdot4+0.5-3\cdot\dfrac{4}{5}=-19.5-\dfrac{12}{5}=-\dfrac{219}{10}\)
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
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
6 tháng 4 2022
\(=\dfrac{8-4\sqrt{3}-3\left(\sqrt{3}-1\right)-2}{\sqrt{3}-1-2}=\dfrac{6-4\sqrt{3}-3\sqrt{3}+3}{\sqrt{3}-3}\)
\(=\dfrac{-7\sqrt{3}+3}{\sqrt{3}-3}=3\sqrt{3}+2\)
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
\(=\left(\dfrac{3}{2}\cdot\dfrac{2}{5}+2\cdot\dfrac{1}{5}\right):\dfrac{3}{8}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}=\dfrac{8}{3}\)