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a) x4 + 3x3 - 7x2 - 27x - 18
= x4 + x3 + 2x3 + 2x2 - 9x2 - 9x - 18x - 18
= x3 . (x + 1) + 2x2 . (x + 1) - 9x . (x + 1) - 18(x + 1)
= (x + 1)(x3 + 2x2 - 9x - 18)
= (x + 1)[x2 .(x + 2) - 9.(x + 2)]
= (x + 1)(x + 2)(x2 - 32)
= (x + 1)(x + 2)(x + 3)(x - 3)
b) x4 + 3x3 + 3x2 + 3x + 2
= x4 + x3 + 2x3 + 2x2 + x2 + x + 2x + 2
= x3 (x + 1) + 2x2 . (x + 1) + x(x + 1) + 2(x + 1)
= (x + 1)(x3 + 2x2 + x + 2)
= (x + 1)[x2 .(x + 2) + (x + 2)]
= (x + 1)(x + 2)(x2 + 1)
\(x^4+3x^3-7x^2-27x-18\)
\(=\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(9x^2+9x\right)-\left(18x-18\right)\)
\(=x^3\left(x+1\right)+2x^2\left(x+1\right)-9x\left(x+1\right)-18\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+2x^2-9x-18\right)\)
\(=\left(x+1\right)\left[\left(x^3-3x^2\right)+\left(5x^2-15x\right)+\left(6x-18\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)+5x^2\left(x-3\right)+6\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)^2\)
Âu Mai Gớt :)) Bài này là cả giờ sinh hoạt của t.
Đặt: \(L=1.2.3+2.3.4+100.101.102\)
\(4L=1.2.3.4+2.3.4.\left(5-1\right)+...+100.101.102.\left(103-99\right)\)
\(4L=1.2.3.4+2.3.4.5-1.2.3.4+...+100.101.102.103-99.100.101.102\)
\(4L=100.101.102.103\Leftrightarrow L=\dfrac{100.101.102.103}{4}\)(1)
Mặt khác( Kiểu người 2 mặt ý) :
\(L=\left(2-1\right).2.\left(2+1\right)+\left(3-1\right).3.\left(3+1\right)+...+\left(101-1\right).101.\left(101+1\right)\)
\(L=2\left(2^2-1\right)+3\left(3^2-1\right)+...+101\left(101^2-1\right)\)
\(L=2^3-2+3^3-3+...+101^3-101\)
\(L=\left(1^3+2^3+3^3+...+100^3\right)-\left(1+2+3+...+100\right)+101^3-101\)(2)
Từ (1) và (2) ta có: \(\left(1^3+2^3+3^3+...+100^3\right)-\left(1+2+3+...+100\right)+101^3-101=\dfrac{100.101.102.103}{4}\)
\(\Rightarrow A-\dfrac{100.101}{2}+101^3-101=25.101.102.103\)
\(\Rightarrow A=25.101.102.103+101-101^3+\dfrac{100.101}{2}\)
\(A=25502500\)
\(\)Mà: \(B=1+2+3+...+100=\dfrac{100.101}{2}=5050\)
\(\Rightarrow\dfrac{A}{B}=5050\Leftrightarrow A⋮B\)
ta có điều phải chứng minh.
P/S: Có thể nhận thấy rằng: \(A=B^2\).Công thức tổng quát:
\(1^3+2^3+...+l^3=\left(1+2+3+...+l\right)^2\)
B= 1+2+3+...+100=\(\dfrac{100\left(100+1\right)}{2}\)
=50 x 101
Ta lại có A =13+23+33+.....+1003
= (13+1003) + (23 + 993) + .....+ (503 +513)
vì\(\left\{{}\begin{matrix}\text{1^3+100^3⋮100+1=101}\\\text{2^3+99^3⋮2+99=101}\\............................\\\text{50^3+51^3⋮50+51=101}\end{matrix}\right.\)
=> A \(⋮\)101(1)
mặt khác
A = (13+993)+(23 + 983) + .....+ (493 +513)+(503 +1003)
vì\(\left\{{}\begin{matrix}1^3+99^3⋮1+99=100⋮50\\...................................\\49^3+51^3⋮49+51=100⋮50\\50^3+100^3⋮100+50=150⋮50\end{matrix}\right.\)
=> A\(⋮\)50(2)
Từ (1) và (2) => A\(⋮\)101 và\(⋮\)50
Mà (101 ,50)=1 => A\(⋮\)101x50=B
KL : A\(⋮\)B
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
Bạn chỉ cần để ý điều này thôi: \(\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=x^2-2+\frac{1}{x^2}\)
Do đó giả thiết viết lại thành:
\(\left(a^2-2+\frac{1}{a^2}\right)+\left(b^2-2+\frac{1}{b^2}\right)+\left(c^2-2+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-\frac{1}{a}=0\\b-\frac{1}{b}=0\\c-\frac{1}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{a}\\b=\frac{1}{b}\\c=\frac{1}{c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b^2=1\\c^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2\right)^{1010}=1^{1010}\\\left(b^2\right)^{1010}=1^{1010}\\\left(c^2\right)^{1010}=1^{1010}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{2020}=1\\b^{2020}=1\\c^{2010}=1\end{matrix}\right.\) \(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}=3\)
Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)
(vì \(xy\ne0\Rightarrow x,y\ne0\))
\(\Rightarrow x-1\ne0;y-1\ne0\)
\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)
\(x^3-1=\left(x-1\right)\left(x^2-x+1\right)=-y\left(x^2-x+1\right)\Rightarrow\frac{y}{x^3-1}=\frac{-1}{x^2+x+1}\)
\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)
\(=-\left(\frac{x^2+x+1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\right)=-\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)+2}{x^2y^2+\left(x+y\right)^2-2xy+xy\left(x+y\right)+xy+\left(x+y\right)+1}\right)\)
\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
Đặt \(\hept{\begin{cases}x=2b+2c-a\\y=2c+2a-b\\z=2a+2b-c\end{cases}}\)
Vì a,b,c là độ dài ba cạnh của 1 tam giác nên \(x,y,z>0\)
Khi đó :
\(\Rightarrow\hept{\begin{cases}a=\frac{2y+2z-x}{9}\\b=\frac{2z+2x-y}{9}\\c=\frac{2x+2y-z}{9}\end{cases}}\)
Ta có bất đẳng thức mới theo ẩn x,y,z :
\(\frac{2y+2z-x}{9x}+\frac{2z+2x-y}{9y}+\frac{2x+2y-z}{9z}\ge1\)
\(\Leftrightarrow\frac{2}{9}\left(\frac{y}{x}+\frac{z}{x}\right)+\frac{2}{9}\left(\frac{z}{y}+\frac{x}{y}\right)+\frac{2}{9}\left(\frac{x}{z}+\frac{y}{z}\right)-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{2}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{2}{9}\left(\frac{y}{z}+\frac{z}{y}\right)+\frac{2}{9}\left(\frac{z}{x}+\frac{x}{z}\right)-\frac{1}{3}\ge1\)
Ta chứng minh bất đẳng thức phụ sau :
\(\frac{a}{b}+\frac{b}{a}\ge2\forall a,b>0\)
Thật vậy : \(\frac{a}{b}+\frac{b}{a}\ge2\)
\(\Leftrightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\)
\(\Leftrightarrow\frac{a^2+b^2}{ab}-2\ge0\)
\(\Leftrightarrow\frac{a^2+b^2-2ab}{ab}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)(luôn đúng \(\forall a,b>0\))
Áp dụng , ta được :
\(\frac{2}{9}.2+\frac{2}{9}.2+\frac{2}{9}.2-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{12}{9}-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{9}{9}\ge1\)(đúng)
Vậy bất đẳng thức được chứng minh
\(A=1^3+2^3+3^3+...+16^3\)
\(=\left(1^3+16^3\right)+\left(2^3+15^3\right)+\left(3^3+14^3\right)+...+\left(8^3+9^3\right)\)
\(=\left(1+16\right)\left(1^2-1.16+16^2\right)+\left(2+15\right)\left(2^2-2.15+15^2\right)+...+\left(8+9\right)\left(8^2-8.9+9^2\right)\)
\(=17.\left(1^2-1.16+16^2\right)+17.\left(2^2-2.15+15^2\right)+...+17.\left(8^2-8.9+9^2\right)\)
\(=17.\left(1^2-1.16+16^2+2^2-2.15+15^2+...+8^2-8.9+9^2\right)⋮17\)
hay : \(A⋮17\) ( đpcm )