B= \(\frac{a^2+\sqrt{a}}{2-\sqrt{2}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(a>0;a\ne1\right)\)
a. Rút gọn B
b. Tìm a để B=2
c.Tìm GTNN của B
Những câu hỏi liên quan
\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{a^2-2a+1}}{\sqrt{1-a^2}-\sqrt{a^2-2a+1}}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)ĐK:0< a< 1\)Dk 0
Rút gọn biểu thức:
a) A frac{sqrt{5-2sqrt{6}}+sqrt{8-2sqrt{15}}}{sqrt{7+2sqrt{10}}}
b) B left(1+frac{a+sqrt{a}}{sqrt{a}+1}right).left(1-frac{a-sqrt{a}}{sqrt{a}-1}right) a0 va a # 1
c) C frac{asqrt{a}-8+2a-4sqrt{a}}{a-4}
d) D frac{1}{2a-1}.sqrt{5a^4.left(-4a+4a^2right)}
e) E frac{2}{x^2-y^2}.sqrt{frac{3x^2+6xy+3y^2}{4}}
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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 = \(\frac{2}{x^2-y^2}.\sqrt{\frac{3x^2+6xy+3y^2}{4}}\)
Bleft(frac{asqrt{a}+1}{sqrt{a}+1}right):left(a-1right)+frac{2a+sqrt{a}+1}{sqrt{a}+1}-frac{sqrt{a}}{a-1}vớia1 Cleft(frac{X-1}{sqrt{X}-1}+frac{sqrt{X^3}-1}{1-X}right)-left(frac{left(X-1right)^2+sqrt{X}}{sqrt{X}+1}right)vớiX0,Xne1Dfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-frac{2x+sqrt{x}}{sqrt{x}}+frac{2left(x-1right)}{sqrt{x}-1}vớix0,xne1
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\(B=\left(\frac{a\sqrt{a}+1}{\sqrt{a}+1}\right):\left(a-1\right)+\frac{2a+\sqrt{a}+1}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}vớia>1\)
\(C=\left(\frac{X-1}{\sqrt{X}-1}+\frac{\sqrt{X^3}-1}{1-X}\right)-\left(\frac{\left(X-1\right)^2+\sqrt{X}}{\sqrt{X}+1}\right)vớiX>0,X\ne1\)
\(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}vớix>0,x\ne1\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
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Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah
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Xem thêm câu trả lời
Bai 1)Aleft(frac{sqrt{a}}{2}-frac{1}{2sqrt{a}}right)^2left(frac{sqrt{a}-1}{sqrt{a}+1}-frac{sqrt{a}+1}{sqrt{a}-1}right),a)rút gọn A 2) B1+left(frac{2a+sqrt{a}-1}{1-a}-frac{2asqrt{a}-sqrt{a}-1}{1-asqrt{a}}right).left(frac{a-sqrt{a}}{2sqrt{a}-1}right),a) rút gọn B b) tìm a để Bfrac{sqrt{6}}{1+sqrt{6}}c) chứng minh rằng B2/3 Giúp mk với nếu đúng mk tick cho nha ,cảm ơn
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Bai 1)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
2) B=1+\(\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}-1}{1-a\sqrt{a}}\right).\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right),\)
a) rút gọn B
b) tìm a để B=\(\frac{\sqrt{6}}{1+\sqrt{6}}\)
c) chứng minh rằng B>2/3
Giúp mk với nếu đúng mk tick cho nha ,cảm ơn
chứng minh câu đẳng thức1)frac{sqrt{a}+sqrt{b}}{2sqrt{a}-2sqrt{b}}-frac{sqrt{a}-sqrt{b}}{2sqrt{a}+2sqrt{b}}-frac{2b}{b-a}frac{2sqrt{b}}{sqrt{a}-sqrt{b}}2)left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}+sqrt{b}}-sqrt{ab}right)left(frac{sqrt{a}+sqrt{b}}{a-b}right)^213)frac{sqrt{a}}{sqrt{a}-sqrt{b}}-frac{sqrt{b}}{sqrt{a}+sqrt{b}}-frac{2b}{a-b}1(a lớn hơn bằng 0,b lớn hơn bằng 0)4)left(1+frac{a+sqrt{a}}{sqrt{a}+1}right)left(1-frac{a-sqrt{a}}{sqrt{a}-1}right)1-a(a lớn hơn bằng 0,a khác 1)help me:
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chứng minh câu đẳng thức
1)\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
2)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
3)\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}=1\)(a lớn hơn bằng 0,b lớn hơn bằng 0)
4)\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)(a lớn hơn bằng 0,a khác 1)
help me:<<<
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
C/Minh đẳng thức:
a) left(frac{sqrt{a}+2}{a+2sqrt{a}+1}-frac{sqrt{a}-2}{a-1}right).frac{sqrt{a}+1}{sqrt{a}}frac{2}{a-1} (với a0, b0, a≠b)
b)frac{2}{sqrt{ab}}:left(frac{1}{sqrt{a}}-frac{1}{sqrt{b}}right)^2-frac{a+b}{left(sqrt{a}-sqrt{b}right)^2}-1 (với a0, b0,a≠b)
c) frac{2sqrt{a}+3sqrt{b}}{sqrt{ab}+2sqrt{a}-3sqrt{b}-6}-frac{6-sqrt{ab}}{sqrt{ab}+2sqrt{a}+3sqrt{b}+6}frac{a+9}{a-9} (với a≥0, b≥0,a≠9)
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C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)
Bài 1 :
a, sqrt{45}-2sqrt{frac{4}{3}}+frac{sqrt{18}}{sqrt{6}}-sqrt{5frac{1}{3}}
b, (sqrt{7}-sqrt{3} )2 +sqrt{84}
Bài 2 : Chứng minh đẳng thức
left(frac{sqrt{21}-sqrt{7}}{sqrt{3}-1}frac{sqrt{15}+sqrt{3}}{sqrt{5}+1}right):frac{1}{sqrt{7}+sqrt{3}}4
Bài 3: Cho biểu thức : Aleft(1-frac{2sqrt{2a}}{a+2}right):left(frac{1}{left(sqrt{a}+2right)}-frac{2sqrt{2a}}{left(a+2right)left(sqrt{a}+2right)}right)
a. Rút gọn A
b. Tính A khi a 2009-2sqrt{2008}
Bài 4 : Cho A left(1-frac{4}{sqrt{x}+1}+frac{1}...
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Bài 1 :
a, \(\sqrt{45}-2\sqrt{\frac{4}{3}}+\frac{\sqrt{18}}{\sqrt{6}}-\sqrt{5\frac{1}{3}}\)
b, (\(\sqrt{7}-\sqrt{3}\) )2 +\(\sqrt{84}\)
Bài 2 : Chứng minh đẳng thức
\(\left(\frac{\sqrt{21}-\sqrt{7}}{\sqrt{3}-1}\frac{\sqrt{15}+\sqrt{3}}{\sqrt{5}+1}\right):\frac{1}{\sqrt{7}+\sqrt{3}}=4\)
Bài 3: Cho biểu thức : A=\(\left(1-\frac{2\sqrt{2a}}{a+2}\right):\left(\frac{1}{\left(\sqrt{a}+2\right)}-\frac{2\sqrt{2a}}{\left(a+2\right)\left(\sqrt{a}+2\right)}\right)\)
a. Rút gọn A
b. Tính A khi a =2009-2\(\sqrt{2008}\)
Bài 4 : Cho A =\(\left(1-\frac{4}{\sqrt{x}+1}+\frac{1}{x-1}\right):\frac{x-2\sqrt{x}}{x-1}\) điều kiện x>0 , x≠1,x≠4
a.Rút gọn
b. Tìm x để A =\(\frac{1}{2}\)
Rút gọn biểu thức:
a) \(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab+\sqrt{a}}}{\sqrt{ab}-1}+1\right)\)
b) \(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right)\)
Rút gọn biểu thức:
a, \(\frac{2}{a}\sqrt{\frac{16a^2}{9}}\) với a < 0
b, \(\frac{3}{a-1}\sqrt{\frac{4a^2-8a+4}{25}}\) với a > 1
c, \(\frac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}\) với a ≠ b
d, \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) với a ≠ 1, a ≥ 0
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
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a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
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Cho a , b , c > 0. CM :\(\frac{1}{\sqrt{a}}+\frac{3}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\ge\frac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}\)
Áp dụng BĐT Cauchy - Schwarz ta có :
\(VT=\frac{1}{\sqrt{a}}+\frac{3}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(\ge\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{2\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(\ge\frac{\left(1+2+1+2+2\right)^2}{2\sqrt{3c+2a}+3\sqrt{b}+\sqrt{a}}\)
\(\ge\frac{64}{\sqrt{\left(1+2^2+3\right)\left(a+2a+3c+3b\right)}}\)
\(=\frac{64}{\sqrt{24\left(a+c+b\right)}}=\frac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}=VF\)
Chúc bạn học tốt !!!
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Mình nghĩ là:
a = 1
b = 2
c = 4