1a,5(x-2)-3x=4

5x-10-3x=4

2x-10=4

2x=4+10

2x=14

x=7

1a,5.(x-2)-3x=-4
5x-10-3x=-4
2x-10=-4
2x= -4+10
2x=6
x=3

a) x² - 4x - 5 = 0

=> x² - 5x + x - 5 = 0

=> x(x - 5) + (x - 5) = 0

=> (x + 1)(x - 5) = 0

=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

b) 4x² + 7x - 11 = 0

=> 4x² + 11x - 4x - 11 = 0

=> x(4x + 11) - (4x + 11) = 0

=> (x - 1)(4x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)

c) -7x² + 6x + 1 = 0

=> -7x² + 7x - x + 1 = 0

=> -7x(x - 1) - (x - 1) = 0

=> (-7x - 1)(x - 1) = 0

=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)

d) -10x² + 7x + 3 = 0

=> -10x² + 10x - 3x + 3 = 0

=> -10x(x - 1) - 3(x - 1) = 0

=> (-10x - 3)(x - 1) = 0

=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(a,x^2-4x-5=0\)

\(\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(b,4x^2+7x-11=0\)

\(\Rightarrow4x^2-4x+11x-11=0\)

\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)

\(c,-7x^2+6x+1=0\)

\(\Rightarrow-7x^2+7x-x+1=0\)

\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

\(d,-10x^2+7x+3=0\)

\(\Rightarrow-10x^2+10x-3x+3=0\)

\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)

a, \(M=7.\left(x-y\right)+4a.\left(x-y\right)-5\)

Theo bài ra ta có: x-y=0

=> \(M=0+0-5\)

\(\Rightarrow M=-5\)

b,

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

\(\Rightarrow N=0+3=3\)

lớp 7 lên 8 à làm quen nhá :)

a) \(M=7x-7y+4ax-4ay-5\)

\(M=7\left(x-y\right)+4a\left(x-y\right)-5\)

\(M=0+0-5=-5\)

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

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

\(N=0+3=3\)

  Ta có   \(x-y=0\) 

a)    M = 7x - 7y +4ax - 4ay - 5

           = 7( x - y )+ 4a( x - y ) - 5

           = 7.0 + 4a . 0 - 5

           = -5

b)  N = x( x²+y² ) - y( x² + y² ) + 3

         =  ( x - y ).( x² + y² ) +3

         = 0. ( x² + y² ) + 3

         = 3

\(PT\Leftrightarrow2\left(x+7\right)-x\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-7;2\right\}\)