P= ab-a+1 ; Q=2ab-(ab-a+2)
Tính
a) P - Q
b)P + Q
K
Khách
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N
29 tháng 12 2018
ĐK: \(ab\ge0\)
\(P=\left(\dfrac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}\right):\left(\dfrac{-2a\sqrt{b}-2\sqrt{ab}}{ab-1}\right)\)
\(P=-1.\)
UV
29 tháng 12 2018
\(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)\(P=\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}+\dfrac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}-\dfrac{ab-1}{ab-1}\right]:\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}-\dfrac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}+\dfrac{ab-1}{ab-1}\right]\)\(P=\dfrac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)+\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)-\left(ab-1\right)}{ab-1}:\dfrac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)-\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)+\left(ab-1\right)}{ab-1}\)\(P=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}:\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ }\)\(P=\dfrac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}:\dfrac{-2\sqrt{a}-2}{ab-1}\)
\(P=\dfrac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{ab-1}.\dfrac{ab-1}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)
HT
13 tháng 6 2019
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
\(a+b\ge2\sqrt{ab}\)\(\Rightarrow16-2\sqrt{ab}\ge2\sqrt{ab}\Leftrightarrow16\ge4\sqrt{ab}\)
\(\Leftrightarrow-\sqrt{ab}\ge-4\)
"="\(\Leftrightarrow a=b=4\)
HT
13 tháng 6 2019
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
\(a+b\ge2\sqrt{ab}\Rightarrow-\left(a+b\right)\le-2\sqrt{ab}\)
\(\Leftrightarrow-4\le-2\sqrt{ab}\Leftrightarrow-\sqrt{ab}\ge-2\)
"="\(\Leftrightarrow a=b=2\)
a) P - Q = ab - a +1 - ab+a-2
= (ab-ab) - (-a +a) - (1-2)
= -1
b) P + Q = ab - a +1 +2ab - (ab - a +2 )
= ab + 1 +2ab^2 - 2ab^2+4ab
=(ab + 4ab) +(2ab^2 - 2ab^2) +1
=5ab +1
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