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Ta có : x2 + 7x + 12 = 0
=> x2 + 4x + 3x + 12 = 0
=> x(x + 4) + 3(x + 4) = 0
=> (x + 3)(x + 4) = 0
=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=-4\end{cases}}\)
Vậy \(x\in\left\{-3;-4\right\}\)
x2 + 7x + 12 = 0
<=> x2 + 3x + 4x + 12 = 0
<=> x( x + 3 ) + 4( x + 3 ) = 0
<=> ( x + 3 )( x + 4 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
Vậy S = { -3 ; -4 }
a(b-c) - c(c-b )
=> a ( b -c ) - c [ - ( b - c ) ]
=> a ( b -c ) + c ( b - c )
= ( a + c ) ( b - c )
Ta có
x 2 – 4 x y + 4 y 2 – 4 = x 2 – 2 . x . 2 y + 2 y 2 – 4 = x – 2 y 2 – 2 2 = x – 2 y – 2 x – 2 y + 2
Vậy m = 2.
Đáp án cần chọn là: B
Bài 1 : (x + 5)3 - x3 - 125
= (x + 5 - x)[(x + 5)2 + x(x + 5) + x2] - 125
= 5(x2 + 10x + 25 + x2 + 5x + x2)
= 5(3x2 + 15x + 25) - 125
= 5(3x2 + 15x + 25 - 25)
= 5(3x2 + 15x)
a) (x-y)2-(x2-2xy)
=y2-2xy+x2-x2+2xy
=y2-(-2xy+2xy)+(x2-x2)
=y2
b)(x-y)2+x2+2xy-(x+y)2
=y2-2xy+x2+x2+2xy-y2-2xy-x2
=(y2-y2)-(2xy+2xy-2xy)+(x2+x2-x2)
=x2-2xy
\(R\left(x\right)=x^2-x=x\left(x-1\right)\)
\(\Leftrightarrow S=\dfrac{1}{3\left(3-1\right)}+\dfrac{1}{4\left(4-1\right)}+\dfrac{1}{5\left(5-1\right)}+...+\dfrac{1}{2023\left(2023-1\right)}+\dfrac{1}{2.2023}\)
\(\Leftrightarrow S=\dfrac{1}{3.2}+\dfrac{1}{4.3}+\dfrac{1}{5.4}+...+\dfrac{1}{2023.2022}+\dfrac{1}{4046}\)
\(\Leftrightarrow S=\dfrac{3-2}{3.2}+\dfrac{4-3}{4.3}+\dfrac{5-4}{5.4}+...+\dfrac{2023-2022}{2023.2022}+\dfrac{1}{4046}\)
\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2022}-\dfrac{1}{2023}+\dfrac{1}{4046}=\dfrac{1}{2}-\dfrac{1}{2023}+\dfrac{1}{4046}=\dfrac{2023-2+1}{4046}=\dfrac{2022}{4046}=\dfrac{1011}{2023}\)