\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
Những câu hỏi liên quan
Cho biểu thức A = \(\left(\dfrac{\sqrt{ab}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}+1\right):\left(\dfrac{\sqrt{ab}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{a}-\sqrt{b}}-1\right)\)
Cho \(\sqrt{ab}+1=4.\sqrt{b}\), tìm max của biểu thức A.
Đăt\(\sqrt{a}\)=x, \(\sqrt{b}\)=y (x,y>0)
=>xy+1=4y => 4y≥ \(2\sqrt{xy}\)=>\(2\sqrt{y}\)≥\(\sqrt{x}\)=> 4y≥x=> 4≥ \(\dfrac{x}{y}\)=> \(\dfrac{1}{4}\)≤\(\dfrac{y}{x}\)=>\(\dfrac{-1}{4}\)≥\(\dfrac{-y}{x}\)
Xét:A=(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{y-x}\)+1):(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{x-y}\)-1)
= \(\dfrac{-2y^2\left(x+1\right)}{\left(x-y\right)\left(x+y\right)}\).\(\dfrac{\left(x-y\right)\left(x+y\right)}{2xy\left(x+1\right)}\)
=> A= \(\dfrac{-y}{x}\)≤\(\dfrac{-1}{4}\)
Dấu "=" xảy ra <=> xy=1 và x=4y <=> x=2, y=\(\dfrac{1}{2}\) <=> a =4, b=\(\dfrac{1}{4}\)
Vậy Max A =\(\dfrac{-1}{4}\) <=> a=4, b=\(\dfrac{1}{4}\)
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Chứng minh các đẳng thức sau: a) left(1-a^2right):left[left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(frac{1
+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right]+1frac{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}}{2sqrt{ab}}left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a
+sqrt{ab}}right)frac{sqrt{a}}{a}d) left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}...
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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) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Chứng minh các đẳng thức sau:
a) left(1-a^2right):left(left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right).left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right)+1frac{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}}{2sqrt{ab}}.left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a+sqrt{ab}}right)frac{sqrt{a}}{a}
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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) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
Pleft(frac{sqrt{a}-sqrt{b}}{sqrt{a}+sqrt{b}}-frac{sqrt{a}+sqrt{b}}{sqrt{a}-sqrt{b}}right)sqrt{frac{1}{a}-frac{1}{b}}left(frac{left(sqrt{a}-sqrt{b}right)^2}{a-b}-frac{left(sqrt{a}+sqrt{b}right)^2}{a-b}right).sqrt{frac{b-a}{ab}}frac{a-2sqrt{ab}+b-a-2sqrt{ab}-b}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{2017-2018}.sqrt{frac{2018-2017}{ab}}4sqrt{ab}.sqrt{frac{1}{ab}}sqrt{frac{16ab}{ab}}4
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\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
sao tổng lại lớn hơn hiệu
Cho P = \(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)RútgonP.Cho\sqrt{a}+\sqrt{b}=4.Tim.GTNNcuaP\)
sorry, câu b nhầm \(\sqrt{a}+\sqrt{b}=4\) thành \(a+b=4\)
Sửa:
Có \(\sqrt{a}+\sqrt{b}=4\Rightarrow a+b+2\sqrt{ab}=16\Leftrightarrow a+b=16-2\sqrt{ab}\)
Áp dụng BĐT cô si cho 2 số ko âm
\(a+b\ge2\sqrt{ab}\)\(\Rightarrow16-2\sqrt{ab}\ge2\sqrt{ab}\Leftrightarrow16\ge4\sqrt{ab}\)
\(\Leftrightarrow-\sqrt{ab}\ge-4\)
"="\(\Leftrightarrow a=b=4\)
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a/ ĐKXĐ: a,b\(\ge\) 0, ab\(\ne\) 1
\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)+\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)-ab+1}{ab-1}\right]:\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)-\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)+ab-1}{ab-1}\right]\)
\(P=\left(\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}\right):\left(\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ab-1}\right)\)
\(P=\frac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}.\frac{ab-1}{-2\sqrt{a}-2}=\frac{2\sqrt{ab}\left(\sqrt{a+1}\right)}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)
b/ BĐT cô si cho 2 số ko âm
\(a+b\ge2\sqrt{ab}\Rightarrow-\left(a+b\right)\le-2\sqrt{ab}\)
\(\Leftrightarrow-4\le-2\sqrt{ab}\Leftrightarrow-\sqrt{ab}\ge-2\)
"="\(\Leftrightarrow a=b=2\)
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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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Cho B=(\(\frac{\sqrt{a}+1}{\sqrt{a}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\)1):\(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}-\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
a.rút gọn B
b.tính B biết a=\(2-\sqrt{3}\),b=\(\frac{\sqrt{3}-1}{1-\sqrt{3}}\)
Tui cx đang có câu như thế mà k bt làm đây
Hu hu
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B=\(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+-\right)\)Rút gọn biểu thức
P=\(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}-\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
a) Rút gọn
b) Tính P khi a=16 và b=4
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
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Cho biểu thức I = \(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\).\(\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn I
a) Tính giá trị của I với a = 16, b = 4
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
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