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1) A=4*\(\frac{10^{2n}-1}{9}\) B=\(2\cdot\frac{10^{n+1}-1}{9}\) C=\(8\cdot\frac{10^n-1}{9}\)
đặt 10^n=X => A+B+C+7=(4*x^2-4+2*10*x-2+8x-8+63)/9=(4x^2+28x+49)/9
=> A+B+C+7=\(\frac{\left(2x+7\right)^2}{3^2}\)
2) = 4mn((m^2-1)-(n^2-1))=4mn(m+1)(m-1)-4mn(n-1)(n+1)
mà m,n nguyên => m-1,m,m+1 và n-1,n,n+1 là 3 số nguyên liên tiếp nên chia hết cho 6
do đó 4mn(m^2-n^2) chia hết 6*4=24
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
a^2 + b^2 + c^2= ab + bc + ca
2 ( a^2 + b^2 + c^2 ) = 2 ( ab + bc + ca)
2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca
a^2 + a^2 + b^2 + b^2 + c^2+ c^2 – 2ab – 2bc – 2ca = 0
a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c² + a² – 2ca = 0
(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca) = 0
(a – b)^2 + (b – c)^2 + (c – a)^2 = 0
Vì (a-b)^2 lớn hơn hoặc bằng 0 với mọi a và b
(b-c)^2 lớn hơn hoặc bằng 0 với mọi c và b
(c-a)^2 lớn hơn hoặc bằng 0 với mọi a và c
=> (a-b)^2 =0 ; (b-c)^2=0 ; (c-a)^2=0
=> a=b ; b=c ; c=a
=>a=b=c
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
\(B=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
I don't now
or no I don't
..................
sorry
1a) \(A+B+C\)
\(=\left(x-y\right)^2+4xy-\left(x+y\right)^2\)
\(=\left(x^2-2xy+y^2\right)+4xy-\left(x^2+2xy+y^2\right)\)
\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(4xy-2xy-2xy\right)=0\left(đpcm\right)\)