\(=\left(\sqrt{ab}+\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\right)\cdot\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}\right)^2\)
\(=\left(a+2\sqrt{ab}\right)\cdot\dfrac{1}{a+2\sqrt{ab}+b}=\dfrac{a+2\sqrt{ab}}{a+2\sqrt{ab}+b}\)
câu1 : a) A= \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
Câu 2 :
a) A= \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) B= \(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\dfrac{2}{a+1}\right)^2\)
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) \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Chứng minh các đẳng thức sau :
a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{6}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=-2\)
c) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b\) với a, b dương và \(a\ne b\)
d) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\) với \(a\ge0\) và \(a\ne1\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
CM \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=a-b\) (a > 0; b > 0)
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
\(P=\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{x+z}}+\sqrt{\dfrac{z}{x+y}}>2\)
rút gọn
\(\left(\dfrac{1}{\sqrt{a}-\sqrt{a-b}}+\dfrac{1}{\sqrt{a}+\sqrt{a+b}}\right):\left(1+\dfrac{\sqrt{a+b}}{\sqrt{a-b}}\right)\)
\(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\) với a > 0, b > 0.
a) Rút gọn P
b) Tính giá trị của P khi \(a=2\sqrt{3},b=\sqrt{3}\).
Rút gọn biểu thức:
a) A = \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) B = \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\), với a > 0, a ≠ 4
