biểu diễn dưới dạng thương 2 căn bậc hai
a, \(\sqrt{\dfrac{3a}{b}}\left(a< 0,b< 0\right)\)
b, \(\sqrt{\dfrac{a}{xy}\left(a< 0,x< 0,y>0\right)}\)
Bài 1: Khử mẫu của biểu thức dưới căn
a) -xy\(\sqrt{\dfrac{y}{x}}\) ( x >0; y\(\ge\)0)
b) \(\sqrt{\dfrac{5a^3}{49b}}\left(a\ge0;b>0\right)\)
c) \(-7xy\sqrt{\dfrac{3}{xy}}\left(x< 0;y< 0\right)\)
Bài 2: Đưa thừa số ra ngoài căn
a)\(\sqrt{\dfrac{1}{25a^2}}\left(a< 0\right)\)
b) \(\dfrac{1}{3}\sqrt{225a^2}\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
Khử mẫu của biểu thức lấy căn
\(a,\sqrt{\dfrac{-2}{3a^2}}\) (a<0)
\(b,\sqrt{\dfrac{1}{200}}\)
\(c,\sqrt{\dfrac{7}{500}}\)
\(d,\sqrt{\dfrac{3}{98}}\)
\(e,\sqrt{\dfrac{\left(1-\sqrt{2}\right)^2}{8}}\)
\(f,a\sqrt{\dfrac{1}{a}}\left(a>0\right)\)
\(g,\sqrt{\dfrac{4a^3}{64b}}\left(a,b< 0\right)\)
\(h,2ab\sqrt{\dfrac{3}{ab}}\left(ab>0\right)\)
a, Vì trong dấu căn là số âm nên biểu thức này vô nghĩa. b)\(\sqrt{\dfrac{1}{200}}=\dfrac{1}{\sqrt{200}}=\dfrac{1}{10\sqrt{2}}=\dfrac{\sqrt{2}}{10\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{2}}{20}\)
c,\(\sqrt{\dfrac{7}{500}}=\dfrac{\sqrt{7}}{\sqrt{500}}=\dfrac{\sqrt{7}}{10\sqrt{5}}=\dfrac{\sqrt{7}.\sqrt{5}}{10\sqrt{5}.\sqrt{5}}=\dfrac{\sqrt{35}}{50}\)
d, \(\sqrt{\dfrac{3}{98}}=\dfrac{\sqrt{3}}{\sqrt{98}}=\dfrac{\sqrt{3}}{\sqrt{49.2}}=\dfrac{\sqrt{3}}{7\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{7\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{14}\)
Chứng minh :
a) \(\dfrac{3x}{2y}+\dfrac{3}{2}\sqrt{\dfrac{3}{5}}-\sqrt{\dfrac{3}{4}}=\dfrac{3\sqrt{x}}{2}.\left(\dfrac{\sqrt{x}}{y}+\sqrt{\dfrac{3}{5x}}-\sqrt{\dfrac{1}{3}}\right)\)
b)\(ab.\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\) , với a ; b > 0
c) \(\left(\dfrac{3}{a}\sqrt{\dfrac{a^3}{b}}-\dfrac{1}{2}\sqrt{\dfrac{4}{ab}}-2\sqrt{\dfrac{b}{a}}\right):\sqrt{\dfrac{1}{ab}}=3a-2b-1\) với a, b >0
d)\(\left(\sqrt{\dfrac{16a}{b}}+3\sqrt{4ab}-a\sqrt{\dfrac{36b}{a}}+2\sqrt{ab}\right):\left(\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{a}{b}}\right)=2\) Với a, b >0
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b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
Bài 1: CMR \(P=\dfrac{a+b}{\sqrt{a\cdot\left(3a+b\right)}+\sqrt{b\cdot\left(3b+a\right)}}>=\dfrac{1}{2}\)
với a, b > 0
Bài 2: cho x, y, z > 0. CMR
\(P=\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{x+z}}+\sqrt{\dfrac{z}{x+y}}>2\)
Cho a,b,c>0 và a+b+c=căn a +căn b +căn c=2.Tính A=
\(\left(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}\right)\left(\sqrt{1+a}\right)\left(\sqrt{1+b}\right)\left(\sqrt{1+c}\right)\)
Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)
\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)
\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)
Tương tự:
$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$
Khi đó:
\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)
\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)
\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)
\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)
\(B=\left(\dfrac{1}{\sqrt{x}-\sqrt{y}}+\dfrac{1}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) rút gọn B
b) chứng minh B >/ 0
c) So sánh B với căn B
Biểu diễn \(\sqrt{ab}\) ở dạng tích các căn bậc hai với a<0 và b<0
Áp dụng tính \(\sqrt{\left(-25\right).\left(-64\right)}\)
\(\sqrt{ab}=\sqrt{-a}.\sqrt{-b}\) (vì a<0 , b<0)
Áp dụng : \(\sqrt{\left(-25\right).\left(-64\right)}=\sqrt{-\left(-25\right)}.\sqrt{-\left(-64\right)}=\sqrt{25}.\sqrt{64}=5.8=40\)