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1:
1: Khi x=36 thì \(A=\dfrac{36-4}{3+3}=\dfrac{32}{6}=\dfrac{16}{3}\)
2: \(B=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
3: \(P=A:B=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{x-4}{\sqrt{x}+3}=\dfrac{\sqrt{x}}{x-4}\)
P<0
=>x-4<0
=>0<=x<4
=>\(x\in\left\{0;1;2;3\right\}\)
2:
BC^2*MB
\(=\dfrac{BH^2}{BA}\cdot BC^2=\left(\dfrac{BA^2}{BC}\right)^2\cdot\dfrac{BC^2}{BA}\)
\(=\dfrac{BA^4}{BA}\cdot\dfrac{BC^2}{BC^2}=BA^3\)
=>\(MB=\dfrac{BA^3}{BC^2}\)
Câu 1:
b: Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-2\sqrt{15}\)
\(=6-\sqrt{15}\)
c: Ta có: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}\)
\(=4-\sqrt{7}-2\sqrt{7}\)
\(=4-2\sqrt{7}\)
1:
a: =12/10-7/10=5/10=1/2
b: \(=\dfrac{4}{13}-\dfrac{4}{13}+\dfrac{-5}{11}-\dfrac{6}{11}=-\dfrac{11}{11}=-1\)
2:
a: x+2/7=-11/7
=>x=-11/7-2/7=-13/7
b: (x+3)/4=-7/2
=>x+3=-14
=>x=-17
1b) \(C=\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\left(a\ge0\right)=8\sqrt{a}-12\sqrt{a}+6\sqrt{a}=2\sqrt{a}\)
Bài 2:
a),b) \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}}+1\right)\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{2\sqrt{a}}{1-\sqrt{a}}.\dfrac{1}{\sqrt{a}}=\dfrac{2}{1-\sqrt{a}}\)
c) \(P=\dfrac{2}{1-\sqrt{a}}=\dfrac{2}{1-\sqrt{4}}=\dfrac{2}{1-2}=-2\)
d) \(P=\dfrac{2}{1-\sqrt{a}}=9\)
\(\Rightarrow-9\sqrt{a}+9=2\Rightarrow\sqrt{a}=\dfrac{7}{9}\Rightarrow a=\dfrac{49}{81}\left(tm\right)\)
(p) đi qua A(-1;2)
=> 2 = (m - 2).(-1)2
<=> m - 2 = 2
<=> m = 4
Vậy m = 4 thì (p) đi qua A(-1 ; 2)