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1) Ta có : Đặt M = 3x + 1 + 3x + 2 + ... + 3x + 100
= 3x(3 + 32 + ... + 3100)
= 3x[(3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + ... + (397 398 + 399 + 3100)]
= 3x[(3 + 32 + 33 + 34) + 34.(3 + 32 + 33 + 34) + ... + 396.(3 + 32 + 33 + 34)]
= 3x(120 + 34.120 + .... + 396.120)
= 3x.120.(1 + 34 + .... + 396)
=> \(M⋮120\)(ĐPCM)
2) Ta có \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a + b + c = 0
=> a + b = - c
b + c = -a
c + a = -b
Khi đó P = \(\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy nếu a + b + c = 0 thì P = -3
nếu a + b + c \(\ne\)0 thì P = 6
Ta có :
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...\)\(+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)
\(=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)
\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)
\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)
Vì \(120⋮120\)
\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\inℕ\right)\left(đpcm\right)\)
e, \(x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)
đặt 80=x+1 ta đc
\(x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x+15=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+15=x+15=79+15=94\)
Ta có :
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) vì \(x^2+y^2=1\)
\(\Rightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\frac{x^4.b+y^4.a}{ab}=\frac{\left(x^2+y^2\right)^2}{ab}\)
\(\Leftrightarrow\left(x^4.b+y^4.a\right)\left(a+b\right)=ab\left(x^2+y^2\right)^2\)
\(\Rightarrow x^4ab+x^4b^2+a^2y^4+aby^4\)
\(=ab\left(x^2+y^2\right)\left(x^2+y^2\right)\)
\(\Rightarrow ab\left(x^4+x^2y^2+x^2y^2+y^4\right)\)
\(\Rightarrow abx^4+abx^2y^2+abx^2y^2+abx^2y^2+aby^4\)
\(\Rightarrow b^2x^4+a^2y^4\)
\(=2abx^2y^2\)
\(\Rightarrow\left(bx^2\right)^2+\left(ay^2\right)^2-ax^2.by^2-ax^2-by^2=0\)
\(\Rightarrow\left[\left(bx^2\right)^2-ax^2.by^2\right]+\left[\left(ay^2\right)^2-ax^2.by^2\right]=0\)
\(bx^2\left(bx^2-ay^2\right)+ay^2\left(ay^2-bx^2\right)=0\)
\(bx^2\left(bx^2-ay^2\right)-ay^2\left(bx^2-ay^2\right)\)
\(\left(bx^2-ay^2\right)^2=0\)
\(bx^2-ay^2=0\)
\(bx^2=ay^2\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}\)
Mà \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Rightarrow x^2.\frac{x^2}{a}+y.\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\)
\(\Rightarrow\frac{x^2}{a}\left(x^2+y^2\right)=\frac{x^2+y^2}{a+b}\)
\(\Rightarrow\frac{x^2}{a}=\frac{1}{a+b}\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{1}{a+b}\)
Ta có :
\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{a^{1002}}=\left(\frac{x^2}{a}\right)^{1002}+\left(\frac{y^2}{b}\right)^{1002}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}< đpcm>\)
Hok tốt
P/s : _Làm bừa nên chắc k đúng đâu - - _M bt a hok ngu thek nào r mak (:
_E cóa thý a hok ngu âu >: ?
_Với cả giải vợi lak đầy đủ roy hả ?
_Thank nhìu nhìu <<<:
a)\(A=x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)
Tại \(x=2006\) thì giá trị biểu thức \(A\) là:
\(A=2006^6-2007\cdot2006^5+...-2007\cdot2006+2007\)
\(=2006^6-\left(2006+1\right)\cdot2006^5+...-\left(2006+1\right)\cdot2006+2007\)
\(=2006^6-2006^6+2006^5-...-2006^2-2006+2007\)
\(=-2006+2007=1\)
b)Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Khi đó
\(VT=\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(1\right)\)
\(VP=\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(2\right)\)
Từ \((1) và (2)\) ta có điều phải chứng minh
c)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2004\right|+\left|x-1\right|=\left|2004-x\right|+\left|x-1\right|\)
\(\ge\left|2004-x+x-1\right|=2003\)
Đẳng thức xảy ra khi \(1\le x\le2004\)
Vậy với \(1\le x\le2004\) thì \(A_{Min}=2003\)
Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Áp dụng vào bài toán \(\left|x-2004\right|+ \left|x-1\right|\ge\left|x-2004+1-x\right|=2003\)
Dấu "=" xảy ra khi \(\left(x-2004\right)\left(1-x\right)\ge0\)
.....
Bài 4
\(127^{23}< 128^{23}=\left(2^7\right)^{23}=2^{7.23}=2^{161}\)
\(513^{18}>512^{18}=\left(2^9\right)^{18}=2^{9.18}=2^{161}\)
Vì \(127^{23}< 2^{161}< 513^{18}\)nên \(127^{23}< 513^{18}\)
Khả năng của mình chỉ làm được 2 bài thôi. Các bạn thông cảm!
Bài 3
\(3^{n+2}-2^{n+2}+3^n-2^n=3^n.3^2-2^n.2^2+3^n-2^n.\)
\(=\left(3^n.9+3^n\right)-\left(2^n.4+2^n\right)=3^n.\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1}.2.5=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right).\)chia hết cho 10
Bài 2 :
Ta có : \(S=4+4^2+4^3+...+4^{2004}\)
=> \(4S=4^2+4^3+...+4^{2005}\)
=> \(4S-S=\left(4^2+4^3+...+4^{2005}\right)-\left(4+4^2+...+4^{2004}\right)\)
=> \(3S=-4+4^{2005}\)
=> \(3S+4=-4+4^{2005}+4=4^{2005}\)
Mà \(4^{2005}:4^{2004}=4\)
=> \(4^{2005}⋮4^{2004}\)
=> \(3S+4⋮4^{2004}\) ( đpcm )