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\(P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}.\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\)
\(=1+\dfrac{9}{\sqrt{x}-3}\le1+\dfrac{9}{0-3}=1-3=-2\)
\(maxP=-2\Leftrightarrow x=0\)
\(1,x=16\Leftrightarrow A=\dfrac{4-1}{4-3}=\dfrac{3}{1}=3\\ 2,B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\\ 3,P=AB=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\\ P=1+\dfrac{9}{\sqrt{x}-3}\\ Vì.\sqrt{x}-3\ge-3\Leftrightarrow\dfrac{9}{\sqrt{x}-3}\le-3\\ \Leftrightarrow P=1+\dfrac{9}{\sqrt{x}-3}\le1-3=-2\\ P_{max}=-2\Leftrightarrow x=0\)
a/
b/
Tọa độ giao điểm của 2 đồ thị là:
\(\dfrac{1}{2}x^2=2x-2\\ \Leftrightarrow\dfrac{1}{2}x^2-2x+2=0\\ \Leftrightarrow x=2\)
b) Phương trình hoành độ giao điểm của (P) và (d):
1/2 x² = 2x - 2
⇔x² = 4x - 4
⇔x² - 4x + 4 = 0
⇔(x - 2)² = 0
⇔x - 2 = 0
⇔x = 2
⇔y = 2.2 - 2 = 2
Vậy tọa độ giao điểm của (P) và (d) là (2;2)
bài 7
A=\(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)}+\dfrac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+x+1\right)}\)
A=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
bài 8
P=\(\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2}\right].\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)=\(\dfrac{x-1}{2\sqrt{x}\left(\sqrt{x}-1\right)}\)
P=\(\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
bài 9
P=\(\left[\dfrac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\right].\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{2\sqrt{xy}-x-y}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
bài 10
P=\(\left[\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\left(\sqrt{x}+2\right)^2}\right]:\left[\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\)
P=\(\dfrac{\sqrt{x}+2-2}{\left(\sqrt{x}+2\right)^2}:\dfrac{2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P=\(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{-\sqrt{x}}\)=\(\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\)
Bài 3:
1: Ta có: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{x-4}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\left(đk:x\ge-1\right)\)
\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)
\(\Leftrightarrow2\sqrt{x+1}=4\)
\(\Leftrightarrow\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\left(tm\right)\)
Bài 2:
e) \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=\sqrt{x-2}-12\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}-12.\sqrt{\dfrac{1}{9}}.\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow3\sqrt{x-2}=12\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\Leftrightarrow x=18\left(tm\right)\)
ko biết
\(\tan a=\frac{22,1}{S}\)
\(\cot a=\frac{s}{22,1}\)
b , Khi \(a=1^015'=\frac{22,1}{s}\Rightarrow S=\frac{22,1}{\tan1^015'}=1012,83\left(m\right)\)