Cm
\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)
\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=1\)với a>= 0, b >=0, a≠b
Những câu hỏi liên quan
Chứng minh :
a) dfrac{3x}{2y}+dfrac{3}{2}sqrt{dfrac{3}{5}}-sqrt{dfrac{3}{4}}dfrac{3sqrt{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}}-2sqrt{dfrac{b}{a}}right):sqrt{dfrac{1}{ab}}3a-2b-1 với a, b 0
d)left(sqrt{dfrac{16a}{b}}+3sqrt{4ab}-asqrt{dfrac{36b}{a}}+2sqrt{ab}right):left(sqrt{ab}+dfrac{a}{b}sqrt{dfrac{b}{a}}+sqrt{dfrac{a}{b}}rig...
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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
Mọi người giúp tớ với ạ !!!!!! Mình thật sự cần gấp vào ngày mai !!!!
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
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1. Rút gọn các biểu thức sau:
a, dfrac{1}{4}sqrt{180}+sqrt{20}-sqrt{45}+5 ; b,3sqrt{dfrac{1}{3}}+dfrac{1}{4}sqrt{48}-2sqrt{3}
c,sqrt{2a}-sqrt{18a^3}+4sqrt{dfrac{a}{2}} ; d,sqrt{dfrac{a}{1+2b+b^2}}.sqrt{dfrac{4a+8ab+4ab^2}{225}}
2. Chứng minh các hằng đẳng thức sau:
a, sqrt{dfrac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{dfrac{2+sqrt{3}}{2-sqrt{3}}}4
b,dfrac{sqrt{a}}{sqrt{a}-sqrt{b}}-dfrac{sqrt{b}}{sqrt{a}+sqrt{b}}-dfrac{2b}{a-b}1 với a≥0, b≤0, a≠ b...
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1. Rút gọn các biểu thức sau:
a, \(\dfrac{1}{4}\sqrt{180}+\sqrt{20}-\sqrt{45}+5\) ; b,\(3\sqrt{\dfrac{1}{3}}+\dfrac{1}{4}\sqrt{48}-2\sqrt{3}\)
c,\(\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}\) ; d,\(\sqrt{\dfrac{a}{1+2b+b^2}}.\sqrt{\dfrac{4a+8ab+4ab^2}{225}}\)
2. Chứng minh các hằng đẳng thức sau:
a, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)
b,\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=1\) với a≥0, b≤0, a≠ b
c, \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\) với a>0, a≠1
3. Chứng minh rằng giá trị của biểu thức M không phụ thuộc vào a:
M= \(\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\) với a >0; a≠ 1
Giúp em với e cần gấp lắm ạ
chứng minh các đẳng thức sau:
a) sqrt{dfrac{2-sqrt{3}}{2+sqrt{3}}} + sqrt{dfrac{2+sqrt{3}}{2-sqrt{3}}} 4
b) dfrac{sqrt{a}}{sqrt{a}-sqrt{b}} - dfrac{sqrt{b}}{sqrt{a}+sqrt{b}} - dfrac{2b}{a-b} 1 với ≥ 0, b ≥ 0, a ≠ b;
c) left(1+dfrac{a+sqrt{a}}{sqrt{a}+1}right)left(1-dfrac{a-sqrt{a}}{sqrt{a}-1}right) 1 - a với a 0, a ≠ 1
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chứng minh các đẳng thức sau:
a) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\) + \(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\) = 4
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) - \(\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) - \(\dfrac{2b}{a-b}\) = 1 với ≥ 0, b ≥ 0, a ≠ b;
c) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\)\(\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) = 1 - a với a > 0, a ≠ 1
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
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khúc \(\dfrac{a}{a-b}\) sai nhé
\(=\dfrac{a-b}{a-b}=1\)
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Câu a : \(VT=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{2\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)
\(=\sqrt{\dfrac{3-2\sqrt{3}+1}{3+2\sqrt{3}+1}}+\sqrt{\dfrac{3+2\sqrt{3}+1}{3-2\sqrt{3}+1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{3-2\sqrt{3}+1+3+2\sqrt{3}+1}{3-1}\)
\(=\dfrac{8}{2}=4\) ( đpcm )
Câu c : \(VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\) ( đpcm )
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\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)
\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=1\)
1. \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)
\(\Leftrightarrow\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}+\sqrt{\dfrac{\left(2+\sqrt{3}\right)^2}{1}}=4\)
\(\Leftrightarrow2-\sqrt{3}+2+\sqrt{3}=4\)
\(\Leftrightarrow4=4\left(đpcm\right)\)
2. \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=1\)
\(\Leftrightarrow\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=1\)
\(\Leftrightarrow\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{a-b}1\)
\(\Leftrightarrow\dfrac{a-b}{a-b}=1\left(đpcm\right)\)
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Bài : Tính
a)A= \(\dfrac{2}{\sqrt{2}-1}\) - \(\sqrt{\dfrac{3}{4}}\) : \(\sqrt{\dfrac{3}{2}}\) +\(\sqrt{8}\)
b)B= \(\sqrt{9x^2}\) +\(\sqrt{x}\) (\(\sqrt{x}\) +1) với x > 0
a: \(=\dfrac{2\left(\sqrt{2}+1\right)}{2-1}-\sqrt{\dfrac{3}{4}:\dfrac{3}{2}}+2\sqrt{2}\)
\(=2\sqrt{2}+2+2\sqrt{2}-\sqrt{\dfrac{1}{2}}\)
\(=4\sqrt{2}+2-\dfrac{\sqrt{2}}{2}=\dfrac{7}{2}\sqrt{2}+2\)
b: \(B=\left|3x\right|+x+\sqrt{x}\)
\(=3x+x+\sqrt{x}=4x+\sqrt{x}\)
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Rút gọn biểu thức :
a) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\) ( a > 0 , b > 0 )
b) \(\dfrac{1-8a\sqrt{a}}{1-2\sqrt{a}}\) ( a ≥ 0 , a ≠ \(\dfrac{1}{4}\) )
c) \(\dfrac{1-a}{1+\sqrt{a}}\) ( a ≥ 0 )
d) \(\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\) ( a ≥ 0 , a ≠ 9 )
a. \(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
b. \(=\dfrac{1-\left(2\sqrt{a}\right)^3}{1-2\sqrt{a}}=\dfrac{\left(1-2\sqrt{a}\right)\left(1+2\sqrt{a}+4a\right)}{1-2\sqrt{a}}=1+2\sqrt{a}+4a\)
c. \(=\dfrac{1-\left(\sqrt{a}\right)^2}{1+\sqrt{a}}=\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}{1+\sqrt{a}}=1-\sqrt{a}\)
d. \(=\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}=\sqrt{a}\)
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rút gọn biểu thức
Adfrac{sqrt{a}-1}{asqrt{a}-a+sqrt{a}}:dfrac{1}{a^2+sqrt{a}} với a 0
Bdfrac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}}+dfrac{sqrt{a}-sqrt{b}}{2sqrt{ab}}left(dfrac{sqrt{b}}{a-sqrt{ab}}+dfrac{sqrt{b}}{a+sqrt{ab}}right) với a0 b0 và a khác b
Cdfrac{asqrt{b}+b}{a-b}.sqrt{dfrac{ab+b^2-2sqrt{ab^3}}{aleft(a+2sqrt{b}right)+b}}:dfrac{1}{sqrt{a}+sqrt{b}} với ab0
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rút gọn biểu thức
A=\(\dfrac{\sqrt{a}-1}{a\sqrt{a}-a+\sqrt{a}}:\dfrac{1}{a^2+\sqrt{a}}\) với a >0
B=\(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\) với a>0 b>0 và a khác b
C=\(\dfrac{a\sqrt{b}+b}{a-b}.\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\) với a>b>0
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
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Chứng minh đẳng thức:
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{2b}{\sqrt{a}-\sqrt{b}}\)
b) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{-3}{2}\)
Chứng minh các đẳng thức sau:
c) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) ( với a,b > 0 và a \(\ne\) b )
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Tick plz
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Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{b}+\sqrt{a}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
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