Ta có:

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

Ta lại có:

\(x^7+y^7\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^4y^x-x^3y^4\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^3y^3\left(x+y\right)\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)+x^3y^3z\) ( Thay x + y = -z )

Ta sẽ đi tính \(x^3+y^4;x^4+y^4\)
Lại có​:

1/ \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=-z^3+3xyz\)

2/ \(x^2+y^2=\left(x+y\right)^2-2xy=z^2-2xy\)

\(\Rightarrow x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left(z^2-2xy\right)^2-2x^2y^2=z^4-4xyz^2+2x^2y^2\)

Như vậy \(x^7+y^7=\left(-z^3+3xyz\right)\left(z^4-4xyz^2+2x^2y^2\right)+x^3y^3z\)

\(\Rightarrow x^7+y^7=-z^7+7xyz^5-14x^2y^2z^3+7x^3y^3z\)

\(\Rightarrow x^7+y^7+z^7=7xyz^5-14x^2y^2z^3+7x^3y^3z\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left(z^4-2xyz^2+x^2y^2\right)\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left[z^2\left(z^2-2xy\right)+x^2y^2\right]\)

Mà \(z^2-2xy=x^2+y^2\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left[z^2\left(x^2+y^2\right)+x^2y^2\right]\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left(x^2z^2+y^2z^2+x^2y^2\right)\)

​

a) \(12\left(2x-5\right)^2-3\left(1+4x\right)\left(4x-1\right)\)

\(=12\left[\left(2x\right)^2-2.2x.5+5^2\right]-3\left(4x+1\right)\left(4x-1\right)\)

\(=12\left(4x^2-20x+25\right)-3\left[\left(4x\right)^2-1\right]\)

\(=48x^2-240x+300-3\left(16x^2-1\right)\)

\(=48x^2-240x+300-48x^2+3\)

\(=-240x+303\)

a) \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2+25\)

\(A=5x^3-15x+7x^2-5x^3-7x^2+25\)

\(A=-15x+25\)

Thay x = -5 vào A

\(A=-15.\left(-5\right)+25\)

\(A=75+25=100\)

b) \(B=x\left(x-y\right)+y\left(x-y\right)\)

\(B=\left(x-y\right)\left(x+y\right)\)

\(B=x^2-y^2\)

Thay x =1,5 và y = 10 vào B

\(B=\left(1,5\right)^2-10^2\)

\(B=2,25-100=-97,75\)

\(M=\dfrac{1}{3456}.4\dfrac{1}{987}-3\dfrac{986}{987}.\dfrac{5}{3456}-6\dfrac{1}{3456}.\dfrac{1}{987}\)

Đặt \(\dfrac{1}{3456}=a\) và \(\dfrac{1}{987}=b\)

Chỉ cần thay những số bằng chữ đã đặt rồi rút gọn xong thế vào lại là ok

Ta có:

\(A=\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)

\(A=\left(x^2-4x-2x+8\right)-\left(x^2-x-3x+4\right)\)

\(A=\left(x^2-6x+8\right)-\left(x^2-4x+4\right)\)

\(A=x^2-6x+8-x^2+4x-4\)

\(A=-2x+4\)

Thay \(x=1\dfrac{3}{4}=\dfrac{7}{4}\) vào A ta được:

\(A=-2.\dfrac{7}{4}+4\)

\(A=-\dfrac{7}{2}+4\)

\(A=\dfrac{1}{2}\)

\(9x^2-9xy-4y^2\)

\(=9x^2-12xy+3xy-4y^2\)

\(=3x\left(3x-4y\right)+y\left(3x-4y\right)\)

\(=\left(3x-4y\right)\left(3x+y\right)\)

Ta có:

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

\(=mn\left[\left(m^2-1\right)-\left(n^2-1\right)\right]\)

\(=mn\left(m^2-1\right)-mn\left(n^2-1\right)\)

\(=mn\left(m-1\right)\left(m+1\right)-mn\left(n-1\right)\left(n+1\right)\)

\(=n\left(m-1\right)m\left(m+1\right)-m\left(n-1\right)n\left(n+1\right)\)

Ta thấy:

\(n\left(m-1\right)m\left(m+1\right)⋮6\) ( Vì ( m - 1 )m( m + 1 ) là tích của 3 số tự nhiên liên tiếp

Và \(m\left(n-1\right)n\left(n+1\right)⋮6\) ( Vì ( n - 1 )n( n + 1) là tích của 3 số tự nhiên liên tiếp )

\(\Rightarrow n\left(m-1\right)m\left(m+1\right)-m\left(n-1\right)n\left(n+1\right)⋮6\)

Vậy \(mn\left(m^2-n^2\right)⋮6\)

\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)

Sửa đề

\(x^3-7x^2+14x-8\)

\(=x^3-4x^2-3x^2+12x+2x-8\)

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

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

\(=\left(x-4\right)\left(x^2-2x-x+2\right)\)

\(=\left(x-4\right)\left[x\left(x-2\right)-\left(x-2\right)\right]\)

\(=\left(x-4\right)\left(x-2\right)\left(x-1\right)\)

\(A=x^2-2xy+2y^2-4y+5\)

\(A=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)

\(A=\left(x-y\right)^2+\left(y-2\right)^2+1\)

Vì \(\left(x-y\right)^2\ge0\) với mọi x,y

\(\left(y-2\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-y\right)^2+\left(y-2\right)^2\ge0\) với mọi x,y

\(\Rightarrow\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)

\(\Rightarrow Amin=1\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\Rightarrow x=2\\y=2\end{matrix}\right.\)