bài 1 : rút gọn d, \(3\sqrt{2a}-4\sqrt{8a}-\sqrt{32a}\) \(\left(a\ge0\right)\)
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24 tháng 6 2021
`(1-8asqrta)/(1-2sqrta)`
`=(1-(2sqrta)^3)/(1-2sqrta)`
`=((1-2sqrta)(4a+2sqrta+1))/(1-2sqrta)`
`=4a+2sqrta+1`
19 tháng 8 2023
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
18 tháng 9 2021
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
18 tháng 9 2021
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
5 tháng 9 2020
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
5 tháng 9 2020
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
8 tháng 3 2022
a, Với x >= 0 ; x khác 16
\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)
b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)
\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
26 tháng 6 2023
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
\(3\sqrt{2a}-4\sqrt{8a}-\sqrt{32a}=3\sqrt{2a}-8\sqrt{2a}-4\sqrt{2a}=-9\sqrt{2a}\left(a\ge0\right)\)
\(3\sqrt{2a}-4\sqrt{8a}-\sqrt{32a}=3\sqrt{2a}-8\sqrt{2a}-4\sqrt{2a}=-9\sqrt{2a}\)