\(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
giải vs ạ
Rút gọn biểu thức:
a) A = \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
b) B = \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) a>0 va a # 1
c) C = \(\frac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
d) D = \(\frac{1}{2a-1}.\sqrt{5a^4.\left(-4a+4a^2\right)}\)
e) E = \(\frac{2}{x^2-y^2}.\sqrt{\frac{3x^2+6xy+3y^2}{4}}\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
Bài 1: CMR:
a, (4+\(\sqrt{3}\)). (4-\(\sqrt{3}\))=13
b, \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=2\)
c, \(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=4\)
d, \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=a-b\)(a>0, b>0, a≠b)
Bài 2: CMR:
a, \(\sqrt{a}+\frac{\sqrt{1}}{\sqrt{a}}\ge2\left(a>0\right)\)
b, a+b+\(\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\left(a,b>0\right)\)
c, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xyz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\left(x,y,z>0\right)\)
d, \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=-8\sqrt{3}\)
e, \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)=a-b(a>0, b>0, a≠b)
Bài 3: Tìm Min hoặc Max(nếu có):
a, \(\sqrt{x^2+9}\)
b, \(\frac{2}{\sqrt{x^2+1}}\)
c, 1-\(\sqrt{5+2x-x^2}\)
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
1,Trục căn thức ở mẫu, rút gọn: ( với \(x\ge0;x\ne1\))
a,\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b,\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
2,Chứng minh các đẳng thức sau:
a,\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=1\)
b,\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
c,\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}+\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=-1\)
d,\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Cho a, b>0 và x=\(\frac{2ab}{b_{ }^2+1}\). Xét: P=\(\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}+\frac{1}{3b}\).
1) Chứng minh: P xác định, rút gọn P.
2) Khi a, b thay đổi. Tìm GTNN của P.
A=\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
a) Rút gọn A
b) Tính A khi x = 4-2\(\sqrt{3}\)
Giúp với ạ !!!
Cho biểu thức:
A=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
a. Rút gọn A