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\(a,5\sqrt{4a^6}-3a^3=5\left|2a^3\right|-3a^2=-10a^3-3a^3=-13a^3\)(vì a<0)
b)\(\sqrt{9a^4}+3a^2=\left|3a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c)\(\frac{\sqrt{x^2-10x+25}}{x-5}=\frac{\left|x-5\right|}{x-5}\)
Với x-5>0 => x>5 => \(\frac{\sqrt{x^2-10x+25}}{x-5}=1\)
Với x-5<0=>x<5 =>\(\frac{\sqrt{x^2-10x+25}}{x-5}=-1\)
2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)
c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)
d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)
DAT P = Q:R \(Q=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(3\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(3\sqrt{a}-1\right)}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{3\sqrt{a}+1}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(R=1-\dfrac{2\sqrt{a}-a+1}{3\sqrt{a}+1}=\dfrac{a+\sqrt{a}}{3\sqrt{a}+1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{3\sqrt{a}+1}\)
\(\Rightarrow P=Q:R=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\times\dfrac{3\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(P=\dfrac{3}{3\sqrt{a}-1}\)
\(P>\dfrac{3}{\left|1-3\sqrt{5}\right|}\Leftrightarrow\dfrac{3}{3\sqrt{a}-1}>\dfrac{3}{3\sqrt{5-1}}\)
\(3\sqrt{a}-1< 3\sqrt{5}-1\)
\(\Rightarrow0\le\sqrt{a}\le\sqrt{5}\)
\(a=\) 0 ;1 ;2 ;3 ;4
a lớn nhất \(\Rightarrow a\) = 4
ĐK...
đặt \(\sqrt{x^2-x-6}=a\left(a\ge0\right)\)
Ta có pt <=> \(a^2+a-12=0\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\Leftrightarrow a-3=0\left(vi:a+3>0\right)\)
đến đây tự làm nhá
8n
Do a là nghiệm của pt nên
\(a^2-a-1=0\Leftrightarrow a^2=a+1\Leftrightarrow a^6=\left(a+1\right)^3=a^3+3a^2+3a+1\)
Và \(a^2-a-1=0\Leftrightarrow a^2-a=1\)
\(P=\dfrac{a^6-3a^3\left(a^2-a\right)-a^3+2018}{a^6-\left(a^3+3a^2+3a+1\right)+2020}=\dfrac{\left(a+1\right)^3-4a^3+2018}{\left(a+1\right)^3-\left(a+1\right)^3+2020}\)
\(P=\dfrac{-3a^3+3a^2+3a+2019}{2020}=\dfrac{-3a\left(a^2-a-1\right)+2019}{2020}=\dfrac{2019}{2020}\)