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1) a) \(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(99+98\right)+....\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+.....+2+1\)
\(=\dfrac{100.101}{2}=5050\)
2) a) \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3a^2b+3ab^2-3a^2b+3ab^2=a^3+b^3=VT\)
b) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b+3ab^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)Khi \(a^3+b^3+c^3=3abc\) \(\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
i.i \(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{3}{abc}=3\)iii. \(a^3+b^3+c^3=3abc\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: a=b=c
\(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
TH2: a+b+c=0
\(B=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
a: ĐKXĐ: \(x\notin\left\{2;-2;3\right\}\)
\(A=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{x-2}{x+2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-1}{x-3}=\dfrac{4x}{x-3}\)
b: |x|=1/3 nên x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(A=\dfrac{\dfrac{4}{3}}{\dfrac{1}{3}-3}=\dfrac{4}{3}:\dfrac{-8}{3}=\dfrac{-1}{2}\)
Khi x=-1/3 thì \(A=\dfrac{-\dfrac{4}{3}}{-\dfrac{1}{3}-3}=\dfrac{-4}{3}:\dfrac{-10}{3}=\dfrac{2}{5}\)
c: Để A<1 thi A-1<0
=>\(\dfrac{4x-x+3}{x-3}< 0\)
=>(3x+3)/(x-3)<0
=>-1<x<3
Bài 1:
Ta có: $A=3n^3-5n^2+3n-5=n^2(3n-5)+(3n-5)=(3n-5)(n^2+1)$
Để $A$ là số nguyên tố thì 1 trong 2 thừa số $3n-5$ hoặc $n^2+1$ bằng $1$
Nếu $3n-5=1\Rightarrow n=2$. Thay vào $A=5\in\mathbb{P}$ (thỏa mãn)
Nếu $n^2+1=1\Rightarrow n=0\Rightarrow A=-5$ không phải số nguyên tố (loại)
Vậy $n=2$
Bài 2:
1.
$A=n^3+2n^2-3=(n^3-1)+2(n^2-1)=(n-1)(n^2+n+1)+2(n-1)(n+1)$
$=(n-1)(n^2+n+1+2n+2)=(n-1)(n^2+3n+3)$
Để $A$ là số nguyên tố thì 1 trong 2 số $n-1,n^2+3n+3$ phải bằng $1$
Dễ thấy với $n\in\mathbb{N}$ thì $n^2+3n+3>1$. Do đó $n-1=1\Rightarrow n=2$
Khi đó $A=13$ là snt (thỏa mãn)
2.
$A=n^3+2n^2-3=2013$
$\Leftrightarrow n^3+2n^2-2016=0$
$\Leftrightarrow n^2(n-12)+14n(n-12)+168(n-12)=0$
$\Leftrightarrow (n-12)(n^2+14n+168)=0$
Dễ thấy với $n\in\mathbb{N}$ thì $n^2+14n+168>0$
Do đó $n-12=0\Rightarrow n=12$
1, \(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=a^3+b^3+3a^3b+3ab^3+6a^2b^2\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2\right)\)
\(=a^2-ab+b^2+3ab\left(a+b\right)^2\)
\(=a^2-ab+b^2+3ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(=1\)
Vậy A = 1
Bài 2: ( đặt đề bài là A )
Đặt \(b+c-a=x,a+c-b=y,a+b-c=z\)
\(\Rightarrow a+b+c=x+y+z\)
\(\Leftrightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3.2c.2a.2b=24abc\)
Vậy...
Bài 3:
+) Xét p = 3 có: \(p^2+2=11\in P\) ( t/m )
+) Xét \(p\ne3\) thì:
+ \(p=3k+1\Rightarrow p^2+2=\left(3k+1\right)^2+2=9k^2+6k+3⋮3\notin P\)
+ \(p=3k+2\Rightarrow p^2+2=\left(3k+2\right)^2+2=9k^2+12k+6⋮3\notin P\)
Vậy p = 3
Bài 4:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2c}{abc}+\dfrac{2a}{abc}+\dfrac{2b}{abc}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
\(\Rightarrowđpcm\)
Bài 2:
\(A=\left(2ac-a^2-c^2+b^2\right)\left(2ac+a^2+c^2-b^2\right)\)
\(=\left[b^2-\left(a-c\right)^2\right]\left[\left(a+c\right)^2-b^2\right]\)
\(=\left(b-a+c\right)\left(b+a-c\right)\left(a+c-b\right)\left(a+c+b\right)\)>0
b: \(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
\(=\left(a^2+5a\right)^2+10\left(a^2+5a\right)+25\)
\(=\left(a^2+5a+5\right)^2\)