Question

tìm x biết

\(x+1-\left(x+1\right)^2=0\)

\(x^2+2x+1-y^2-4y-4=0\)

Answer 1

\(x+1-\left(x+1\right)^2=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

\(x^2+2x+1-y^2-4y-4=0\)

\(\Rightarrow\left(x+1\right)^2-\left(y+2\right)^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+1=y+2\\x+1=-y-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=1\\x+y=-3\end{matrix}\right.\)

Answer 2

a)Ta có:

x + 1 - (x + 1)² = 0 => ( x + 1) - ( x + 1)( x + 1) = 0

=> ( x + 1) .[ 1 - (x + 1 )] = 0

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

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

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

=> \(\left[{}\begin{matrix}x+1=0\\-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{-1;0\right\}\)

B) Ta có:

x² + 2x + 1 - y² - 4y - 4 = 0

=>( x² + 2x.1 + 1² )- ( y² + 2.2.y + 2²) = 0

=> (x + 1)² - ( y + 2)² = 0

(Áp dụng tính chất: (a + b)² = a² + 2ab + b²

Giải tính chất:

Ta có:( a + b )² = (a + b) ( a + b) = ( a + b) .a + ( a + b ) .b

= aa + ab + ab + bb= a² + 2ab + b²

Vậy.....)

=>\(\left[{}\begin{matrix}\left(x+1\right)^2=\left(y+2\right)^2=0\\\left(x+1\right)^2=\left(y+2\right)^2\ne0\end{matrix}\right.\)

Xét TH1:(x + 1)² = ( y + 2)² = 0

=>\(\left\{{}\begin{matrix}x+1=0\\y+2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

Xét TH2: \(\left(x+1\right)^2=\left(y+2\right)^2\ne0\)

=>\(\left[{}\begin{matrix}x+1=y+2\\x+1=-y-2\end{matrix}\right.=>\left[{}\begin{matrix}x=y+2-1\\x=-y-2-1\end{matrix}\right.=>\left[{}\begin{matrix}x=y+1\\x=-y-3\end{matrix}\right.\)

Vậy\(\left[{}\begin{matrix}x=-1;y=-2\\x=y+1\\x=-y-3\end{matrix}\right.\)

Answer 3

\(x+1-\left(x+1\right)^2=0\)

\(\Rightarrow1\left(x+1\right)-\left(x+1\right)\left(x+1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}-x=0\Rightarrow x=0\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

\(x^2+2x+1-y^2-4y-4=0\)

\(\Rightarrow\left(x+1\right)^2-\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=y+2\\x+1=-y-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y+1\\x=-y-3\end{matrix}\right.\)