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\(2x.\left(9-x\right)+\left(2x+5\right).\left(x+1\right)\)
\(=2x.9+2x.\left(-x\right)+2x.x+2x.1+5x+5.1\)
\(=18x-2x^2+2x^2+2x+5x+5\)
\(=25x+5\)
\(\left(x-3\right)^2+\left(x+3\right)^2+2.\left(3-x\right).\left(3+x\right)\)
\(=\left(x-3\right)^2-2.\left(x-3\right).\left(x+3\right)+\left(x+3\right)^2\)
\(=[\left(x-3\right)-\left(x+3\right)]^2\)
\(=\left(x-3-x-3\right)^2\)
\(=36\)
Gọi đa thức đó là x2 + cx + d
Ta có : P = (x2 + cx + d)2 = x4 + c2x2 + d2 + 2cx3 + 2dx2 + 2cdx
=> x4 + ax3 + bx2 - 8x + 1 = x4 + 2cx3 + (c2 + 2d)x2 + 2cdx + d2
=> a = 2c ; b = c2 + 2d ; 2cd = -8 ; d2 = 1
Vì d2 = 1 => \(\orbr{\begin{cases}d=1\\d=-1\end{cases}}\)
Khi d = 1 => c = -4 ; b = 18 ; a = -8
Khi d = -1 => c = 4 ; b = 14 ; a = 8
Vậy các cặp (a;b) thỏa mãn là (-8 ; 18) ; (8 ; 14)
Bài 1:
Đặt \(t=2x^2+3x-1\) ta có:
\(t^2-5\left(t+4\right)+24=0\)
\(\Rightarrow t^2-5t-20+24=0\)
\(\Rightarrow t^2-5t+4=0\)
\(\Rightarrow\left(t-4\right)\left(t-1\right)=0\)\(\Rightarrow\left[\begin{matrix}t=4\\t=1\end{matrix}\right.\)
*)Xét \(2x^2+3x-1=4\)
\(\Rightarrow\left(x-1\right)\left(2x+5\right)=0\)\(\Rightarrow\left[\begin{matrix}x=1\\x=-\frac{5}{2}\end{matrix}\right.\)
*)Xét \(2x^2+3x-1=1\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-2\\x=\frac{1}{2}\end{matrix}\right.\)
Bài 2:
\(\left(x^2-4\right)\left(x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Rightarrow\left(x^2-4\right)\left(x+3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left[x+3-\left(x-1\right)\right]=0\)
\(\Rightarrow4\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
x2 + 5y2 + 2x - 6y - 4xy + 2 = 0
⇔ ( x2 - 4xy + 4y2 + 2x - 2y + 1 ) + ( y2 - 2y + 1 ) = 0
⇔ ( x - 2y + 1 )2 + ( y - 1 )2 = 0
Vì \(\hept{\begin{cases}\left(x-2y+1\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{cases}}\)⇒ ( x - 2y + 1 )2 + ( y - 1 )2 ≥ 0 ∀ x, y
Dấu "=" xảy ra khi x = y = 1
Khi đó : S = x2020 + ( y - 2 )2021 = 12020 + ( 1 - 2 )2021 = 1 - 1 = 0
\(\left(2-n\right)\left(n^2-3n+1\right)+n\left(n^2+12\right)+8\)
\(=2n^2-6n+2-n^3+3n^2-n+n^3+12n+8\)
\(=5\left(n^2+n+2\right)⋮5\)