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\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\dfrac{1}{2}\right)^2+5x^2\)
\(=\left(x+2\right)\left(x-2+3x-6\right)-\left(9x^2-3x+1\right)+5x^2\) \(=\left(x+2\right)\left(4x-8\right)-9x^2+3x-1+5x^2\) \(=4\left(x+2\right)\left(x-2\right)-4x^2+3x+1\) \(=4\left(x^2-4\right)-4x^2+3x+1\) \(=4x^2-16-4x^2+3x+1\) \(=3x-15\) \(=3\left(x-5\right)\) Với x=2018 thì \(M=3\left(2018-5\right)=3.2015=6045\)
a/ ĐKXĐ ....
A=\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
=\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
=\(\frac{1}{x}-\frac{1}{x-5}\)
=\(-\frac{5}{x^2-5x}\)
b/ \(x^3-x+2=0\Leftrightarrow\left(x+1\right)\left(\left(x-1\right)^2+1\right)=0\)
<=> x=-1, thay vào tính nốt
câu 1:
\(a,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
=> \(25x^2+10x+1-\left(25x^2-9\right)=30\)
=> \(25x^2+10x+1-25x^2+9=30\)
=> \(10x+10=30\)
=> \(10x=20\)
=> \(x=2\)
Vậy..........
\(b,\left(2x+3\right)^2-\left(2x-3\right)^2+4\left(x^2-6x\right)=64\)
=> \(6.4x+4x^2-24x=64\)
=> \(24x+4x^2-24x=64\)
=> \(4x^2=64\)
=> \(x^2=64:4=16\)
=> \(\left|x\right|=\sqrt{16}\)
=> \(x=\pm4\)
Vậy \(x\in\left\{4;-4\right\}\)
a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7
b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)
c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)
d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)
= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)
= -2(x2 - 2x + 1 + x2 - 1 + x2 + 2x + 1) + 6x2 - 6
= -2(3x2 + 1) + 6x2 - 6
= -6x2 - 2 + 6x2 - 6
= -8
e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2
= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2
= (2x + 7 - 2x + 8)2
= 152 = 225
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
a: \(\Leftrightarrow x^2+x+4x+4+m-4⋮x+1\)
=>m-4=0
hay m=4
b: \(\Leftrightarrow2x^2+4x-x-2+m+2⋮x+2\)
=>m+2=0
hay m=-2
c: \(\Leftrightarrow x^4-x^3+5x^2+x^2-x+5+m-5⋮x^2-x+5\)
=>m-5=0
hay m=5
\(\left(9x-1\right)^2-2\left(9x-1\right)\left(5x-1\right)+\left(5x-1\right)^2=\left(9x-1-5x+1\right)^2=\left(14x\right)^2=196x^2\)
M = ( x - 5)( x + 2 ) + ( 3x - 6 )( x + 2 ) - ( 3x - 1/2 )2 + 5x2
= x2-3x-10+3x2-12-(9x2-3x+1/4)+5x2
= x2-3x-10+3x2-12-9x2+3x-1/4+5x2
= 0.x - 89/4
Thay x=2018 => M= -89/4
\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)
\(M=x^2+2x-5x-10+\left(3x^2+6x-6x-12\right)-\left(9x^2-\frac{3}{2}x+\frac{1}{4}\right)+5x^2\)
\(M=x^2-3x-10+3x^2-12-9x^2+\frac{3}{2}x-\frac{1}{4}+5x^2\)
\(M=-\frac{3}{2}x-\frac{41}{4}\)
Thay x = 2018 vào biểu thức \(M=-\frac{3}{2}x-\frac{41}{4}\), ta có:
\(M=-\frac{3}{2}.2018-\frac{41}{4}=-3027-\frac{41}{4}=\frac{-12149}{4}\)
Vậy giá trị của biểu thức \(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)khi x = 2018 là \(-\frac{12149}{4}\)