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14 tháng 6 2018

\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\dfrac{1}{2}\right)^2+5x^2\)

\(=\left(x+2\right)\left(x-2+3x-6\right)-\left(9x^2-3x+1\right)+5x^2\) \(=\left(x+2\right)\left(4x-8\right)-9x^2+3x-1+5x^2\) \(=4\left(x+2\right)\left(x-2\right)-4x^2+3x+1\) \(=4\left(x^2-4\right)-4x^2+3x+1\) \(=4x^2-16-4x^2+3x+1\) \(=3x-15\) \(=3\left(x-5\right)\) Với x=2018 thì \(M=3\left(2018-5\right)=3.2015=6045\)
14 tháng 6 2018

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

14 tháng 6 2018

\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)

\(M=x^2+2x-5x-10+\left(3x^2+6x-6x-12\right)-\left(9x^2-\frac{3}{2}x+\frac{1}{4}\right)+5x^2\)

\(M=x^2-3x-10+3x^2-12-9x^2+\frac{3}{2}x-\frac{1}{4}+5x^2\)

\(M=-\frac{3}{2}x-\frac{41}{4}\)

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 \(-\frac{12149}{4}\)

16 tháng 12 2022

\(M=\dfrac{x}{\left(x-2\right)\cdot\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\)

\(=\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-6}{x^2-4}\)

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

a) Ta có: \(\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)

\(=\left(3x-2+3x+2\right)^2\)

\(=36x^2\)(1)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:

\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)

b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=100^2=10000\)

a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{3\left(x-2\right)}{x+2}\)

\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)

b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)

\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)

=0