làm phần transformation thôi
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Lời giải:
Lời giải:
$372802=300000+70000+2000+800+2$
$430279=400000+30000+200+70+9$
$920300=900000+20000+300$
$704753=700000+4000+700+50+3$
\(P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}.\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\)
\(=1+\dfrac{9}{\sqrt{x}-3}\le1+\dfrac{9}{0-3}=1-3=-2\)
\(maxP=-2\Leftrightarrow x=0\)
\(1,x=16\Leftrightarrow A=\dfrac{4-1}{4-3}=\dfrac{3}{1}=3\\ 2,B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\\ 3,P=AB=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\\ P=1+\dfrac{9}{\sqrt{x}-3}\\ Vì.\sqrt{x}-3\ge-3\Leftrightarrow\dfrac{9}{\sqrt{x}-3}\le-3\\ \Leftrightarrow P=1+\dfrac{9}{\sqrt{x}-3}\le1-3=-2\\ P_{max}=-2\Leftrightarrow x=0\)
\(b,\left(x+5\right)^2=100\)
TH1:\(\left(x+5\right)^2=10^2\)
\(\Rightarrow x+5=10\\ \Rightarrow x=10-5\\ \Rightarrow x=5\)
TH2:\(\left(x+5\right)^2=\left(-10\right)^2\)
\(\Rightarrow x+5=-10\\ \Rightarrow x=-10-5\\ \Rightarrow x=-15\)
\(c,\left(2x-4\right)^2=0\\ \Rightarrow2x-4=0\\ \Rightarrow2x=0+4\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(d,\left(x-1\right)^3=-27\\ \Rightarrow\left(x-1\right)^3=\left(-3\right)^3\\ \Rightarrow x-1=-3\\ \Rightarrow x=-3+1\\ \Rightarrow x=-2\)
b) (x+5)2=100
(x+5)2=102
x+5=10
x=10-5
x=5
c) (2x-4)2=0
2x-4=0
2x=4
x=4:2
x=\(\pm\)2
d)(x-1)3=-27
(x-1)3=-33
x-1=-3
x=-3+1
x=-2
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