a)\(\sqrt{\dfrac{a^2}{25+10b+b^2}}\) với a < 0, b >0
b)\(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}\)với a khác b
c)\(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}\)với x >= 0
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H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left(a-\sqrt{ab}+\sqrt{b}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a+\sqrt{ab}-b}{a-b}\)
G
8 tháng 8 2023
a) \(\sqrt{2}A=\sqrt{2x-2\sqrt{x-2}.\sqrt{x+2}}+\sqrt{2x+2\sqrt{x-2}.\sqrt{x+2}}\) (\(x\ge2\) )
\(=\sqrt{\left(x+2\right)-2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}+\sqrt{\left(x+2\right)+2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}\)
\(=\sqrt{\left(\sqrt{x+2}-\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}\)
\(=\left|\sqrt{x+2}-\sqrt{x-2}\right|+\sqrt{x+2}+\sqrt{x-2}\)
\(=\sqrt{x+2}-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\) ( do \(x+2>x-2\ge0\Leftrightarrow\sqrt{x+2}>\sqrt{x-2}\) )
\(=2\sqrt{x+2}\)
\(\Leftrightarrow A=\sqrt{2}.\sqrt{x+2}\)
Vậy...
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}.\dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{a-\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-b}{a-b}\)
Vậy...
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
TL
30 tháng 4 2018
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
TL
30 tháng 4 2018
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6
GD
30 tháng 8 2017
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
7 tháng 10 2023
a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)
\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)
b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)
\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)
c:
\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)
d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
e:
\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)
f:
\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)
\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)
(a) \(\sqrt{\dfrac{a^2}{25+10b+b^2}}=\sqrt{\dfrac{a^2}{\left(5+b\right)^2}}=\dfrac{\sqrt{a^2}}{\sqrt{\left(5+b\right)^2}}\)
\(=\dfrac{\left|a\right|}{\left|5+b\right|}=\dfrac{-a}{b+5}\) (do \(a< 0,b>0\Rightarrow b+5>0\))
(b) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\sqrt{\dfrac{\left(ab\right)^2}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}\)
\(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}\).
(c) \(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}=\sqrt{x}+2.\)