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15 tháng 3 2021
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3-x^2-2x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3+x^2+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
30 tháng 8 2021
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
30 tháng 8 2021
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
28 tháng 8 2020
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
28 tháng 8 2020
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
HM
1
11 tháng 3 2022
\(4x-4x^2-8=1-4x^2-3\)
\(\Leftrightarrow4x-8=-2\Leftrightarrow x=\dfrac{3}{2}\)
8x(x - 3) - 8(x - 1)(x + 1) = 20
=> 8x2 - 24x - 8(x2 - 1) - 20 = 0
=> 8x2 - 24x - 8x2 + 8 - 20 = 0
=> -24x = -12
=> x = 1/2
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8-20=0\Leftrightarrow-24x-12=0\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=\frac{-1}{2}\)