bài 1 : a +b , rút gọn và tính

(-a+b-c)-(a-b-c)= -a+b -c-a+b+c= -2a+2b= -2.1+2.-1=-2+-2 = -4

\(\left(m-n\right)\left(m^2+mn+n^2\right)-\left(m+n\right)\left(m^2-mn+n^2\right)\)

\(=m^3-n^3-m^3-n^3\)

\(=-2n^3\)

Phương trình đã cho tương đương : m³ - n³ - ( m³ + n³ )

Tương đương vưới : m³ - n³ - m³ - n³ = 2n³ = 0

bn bỏ số 0 hộ mk nha

cảm ơn

Ta có: \(m+n+k=0\)

\(\Leftrightarrow m+n=-k\)

\(\Leftrightarrow\left(m+n\right)^2=\left(-k\right)^2\)

\(\Leftrightarrow m^2+2mn+n^2=k^2\)

\(\Leftrightarrow m^2+n^2-k^2=-2mn\)

Tương tự, ta có: \(n^2+k^2-m^2=-2nk\)

\(k^2+m^2-n^2=-2km\)

Thay \(m^2+n^2-k^2=-2mn;n^2+k^2-m^2=-2nk;\)\(k^2+m^2-n^2=-2km\) vào biểu thức M ta có:

M = \(\dfrac{1}{-2mn}+\dfrac{1}{-2nk}+\dfrac{1}{-2km}=\dfrac{-1}{2}\left(\dfrac{1}{mn}+\dfrac{1}{nk}+\dfrac{1}{km}\right)\)

M = \(\dfrac{-1}{2}\left(\dfrac{nk^2m+m^2nk+mn^2k}{m^2n^2k^2}\right)\)

\(M=\dfrac{-1}{2}\left(\dfrac{mnk\left(k+m+n\right)}{m^2n^2k^2}\right)\)

M = \(\dfrac{-1}{2}.\dfrac{0}{mnk}\)\(=0\)

\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)

Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)

\(=\dfrac{27}{2}\)

\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào bt N được

\(N=\left(-13\right)\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Vậy: Giá trị của N tại \(a=\dfrac{1}{2};b=-3\) là \(\dfrac{31}{2}\)

\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)

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

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)