Cho A=(x^3-3x^2-7x-15)/(x^5-x^4-10x^3-38x^2-51x-45)
a, Rút gọn
b, tìm GTLN của A
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a) \(a^4-4a^3+8a^2-16a+16\)
\(=a^4-4a^3+4a^2+4a^2-16a+16\)
\(=\left(a^4-4a^3+4a^2\right)+\left(4a^2-16a+16\right)\)
\(=a^2\left(a^2-4a+4\right)+4\left(a^2-4a+4\right)\)
\(=a^2\left(a^2-2.a.2+2^2\right)+4\left(a^2-2.a.2+2^2\right)\)
\(=a^2\left(a-2\right)^2+4\left(a-2\right)^2\)
\(=\left(a-2\right)^2\left(a^2+4\right)\)
a ) ( 2x + 1 )2 - 4 ( x + 2 )2 = 9
4x2 + 4x + 1 - 4 ( x2 +4x + 4 ) = 9
4x2 + 4x + 1 - 4x2 -16x -16 = 9
-12x - 15 = 9
-12x = 24
x = -2
b) 3 ( x - 1 )2 - 3x ( x - 5 ) = 1
3 ( x2 - 2x + 1 ) - 3x2 + 15x = 1
3x2 - 6x + 3 - 3x2 + 15x = 1
9x + 3 = 1
9x = -2
x = \(\frac{-2}{9}\)
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)
\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)
\(\Leftrightarrow2x^2+5=0\)(vô lý)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)
\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)
Vậy pt vô nghiệm
a) -5 . (2 - x) + 4(x - 3) = 10x - 15
-10 + 5x + 4x -12 = 10x - 15
5x + 4x - 10x = -15 + 10 + 12
-x = 7
x = -7
b) 5 . (3 - 2x) + 5 . (x - 4) = 6 - 4x
15 - 10x + 5x - 20 = 6 - 4x
-10x + 5x + 4x = 6 - 15 + 20
-x = 11
x = -11
c) - 7 . (3x - 5) + 2 . (7x - 14) = 28
-21x + 35 + 14x - 28 = 28
-21x + 14x = 28 - 35 + 28
-7x = 21
x = 21 : (-7)
x = -3
d) 4 . (x - 5) - 3 . (x + 7) = 5 . (-4)
4x - 20 - 3x - 21 = -20
4x - 3x = -20 + 20 + 21
x = 21
e) 5 . (4 - x) - 7. (-x + 2) = 4 - 9 + 3
20 - 5x + 7x - 14 = -2
-5x + 7x = -2 - 20 + 14
2x = -8
x = -8 : 2
x = -4
Đúng 100%
câu c
- 7 ( 3x - 5 ) + 2 ( 7x - 14 ) = 28
- 21x + 35 + 14x - 28 = 28
21x - 14x = 35 - 28 - 28
7x = - 21
x = ( - 21) : 7
x = - 3
\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1