Chứng minh:
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) với a,b dương và \(a\ne b\)
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
Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{b}+\sqrt{a}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Chứng minh các đẳng thức (với a, b không âm và \(a\ne b\))
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
b) \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)(đpcm)
a) Ta có:
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{b-a}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
(với a, b không âm và a ≠b )
Vế trái bằng vế phải nên đẳng thức được chứng minh.
b) Ta có:
\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left[\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)
\(=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right]\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}-\sqrt{ab}\right)\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
(với a, b không âm và a ≠b )
Vế trái bằng vế phải nên đẳng thức được chứng minh.
chứng minh
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(VT=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2b}{a-b}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
chứng minh
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) (đpcm)
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
Ta có VT =\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
=\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) =\(\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{a-b}=1=VP\)
Chứng minh đẳng thức:
a) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
( với a > hoặc bằng 0; b > hoặc bằng 0; a khác b )
a: \(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}-2b}{a-b}\)
\(=\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
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
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
khúc \(\dfrac{a}{a-b}\) sai nhé
\(=\dfrac{a-b}{a-b}=1\)
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 )
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) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{-3}{2}\)