chứng minh p=\(\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)(x>0)không phụ thuộc vào x
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8 tháng 5 2022
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 8 2023
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
\(A-1=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-1=\dfrac{\sqrt{x}-2-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+1}\)
Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\forall x\in D\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)
\(\Rightarrow A-1< 0\Rightarrow A< 1\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
30 tháng 6 2021
c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
5 tháng 8 2022
\(C=\dfrac{1}{\left(\dfrac{x+2\sqrt{xy}+y-x-y}{\left(\sqrt{x+y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)^2}-\dfrac{x+y}{2\sqrt{xy}}-\dfrac{\left(x+y\right)^2}{4xy}\)
\(=\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}{4xy}-\dfrac{\left(x+y\right)^2}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{\left(x+y\right)\left(x+y+2\sqrt{xy}\right)-\left(x+y\right)^2}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{2\sqrt{xy}\left(x+y\right)}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{x+y-x-y}{2\sqrt{xy}}=0\)
25 tháng 4 2017
Hướng dẫn trả lời:
ĐKXĐ: 0 < x ≠ 1.
Đặt √x = a (a > 0 và a ≠ 1)
Ta có:
(2+√xx+2√x+1−√x−2x−1).x√x+x−√x−1√x=[2+aa2+2a+1−a−2a2−1].a3+a2−a−1a=[(2+a)(a−1)−(a−2)(a+1)(a+1)(a2−1)].(a+1)(a2−1)a=2a(a+1)(a2−1).(a+1)(a2−1)a=2
14 tháng 5 2021
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx-4)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
14 tháng 5 2021
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
\(P=\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt[4]{x}\left(\sqrt[4]{x}-1\right)}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{x}}+1\right)^2}}{1+\sqrt{x}}\)
\(=\left(-\sqrt[4]{x}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{1}{\sqrt{x}}+1}{1+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}=0\)