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Bài 1:
a,\(5^{2005}+5^{2003}=5^{2003}(25+1)=26.5^{2003}\vdots13(đpcm)\)
b,\(a^2+b^2+1\ge ab+a+b\)
<=>\(2a^2+2b^2+2\ge2ab+2a+2b\)
<=>\((a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)\ge0\)
<=>\((a-b)^2+(a-1)^2+(b-1)^2\ge0(tm)\)
=> đpcm
a) 52005 + 52003 = 52003 ( 52 + 1 ) = 52003 . 26 = 52003 . 2 .13
=> 52005 + 52003 chia hết cho 13
b) a2 + b2 +1 \(\ge\) ab + a + b
\(\Leftrightarrow\) 2a2 + 2b2 + 2 ≥ 2ab + 2a + 2b
\(\Leftrightarrow\)(a2 − 2ab + b2) + (a2 − 2a + 1) + (b2 − 2b + 1) ≥ 0
\(\Leftrightarrow\) (a − b)2 + (a − 1)2 + (b − 1)2 ≥ 0
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
10x2 - 5x + 1 \(\ge\)x2 + x
Ta có : 10x2 - 5x + 1 - x2 - x\(\ge0\)
9x2 - 6x + 1\(\ge0\)
( 3x - 1 )2 \(\ge0\) ( luôn đúng )
\(\Rightarrow\)10x2 - 5x + 1 \(\ge\) x2 + x
b) a2 + b2 + c2 \(\ge\)ab + ac + bc
Nhân cả 2 vế với 2 ta được :
2a2 + 2b2 + 2c2 \(\ge\) 2ab + 2ac + 2bc
Ta có : 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc \(\ge0\)
( a2 - 2ab + b2 ) + ( a2 - 2ac + c2) + ( b2 - 2bc + c2 ) \(\ge0\)
( a - b ) 2 + ( a - c )2 + ( b - c )2 \(\ge0\) ( luôn đúng )
\(\Rightarrow\) a2 + b2 + c2 \(\ge\) ab + ac + bc
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) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
Ta có : a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
=> (a - b)2 + (b - c)2 + (c - a)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
b) Ta có : 2(x2 + t2) + (y + t)(y - t) = 2x(y + t)
=> 2x2 + 2t2 + y2 - t2 = 2xy + 2t
=> 2x2 + t2 + y2 = 2xt + 2xy
=> 2x2 + t2 + y2 - 2xt - 2xy = 0
=> (x2 - 2xy + y2) + (x2 + t2 - 2xt) = 0
=> (x - y)2 + (x - t)2 = 0
=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)
c) Ta có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = 0
=> a = b = c = 0
Khi đó A = (0 - 1)2003 + 02004 + (0 + 1)2005
= - 1 + 0 + 1 = 0
Vậy A = 0
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\) \(\Rightarrow xyz=1\)
\(P=\frac{1}{\frac{1}{x^3}\left(\frac{1}{y}+\frac{1}{z}\right)}+\frac{1}{\frac{1}{y^3}\left(\frac{1}{z}+\frac{1}{x}\right)}+\frac{1}{\frac{1}{z^3}\left(\frac{1}{x}+\frac{1}{y}\right)}\)
\(P=\frac{x^3yz}{y+z}+\frac{y^3xz}{x+z}+\frac{z^3xy}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)\ge\frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(x;y;z\right)=\left(1;1;1\right)\)
b) Áp dụng bđt bunhiacopxki ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(1.a+1.b+1.c\right)^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow a^2+b^2+c^2+d^2+2\left(ab+bc+dc+ad\right)=4\)(*)
Có 2(ab+bc+dc+ad)<=2(a^2+b^2+c^2+d^2 )(**)
Cộng 2 vế của (**) cho a^2+b^2+c^2+d^2 có
3(a^2+b^2+c^2+d^2)>=4
a,b <0 hiển nhiên a^2 +b^2 >= a+b {VT>0 VP <0}
xét a,b >0
a^2 +b^2 >=2ab>=2
a^2 +b^2 -2a-2b +a^2 +b^2 >= a^2 +b^2 -2a-2b +2 =a^2 +b^2 -2a-2b +1+1 =(a-1)^2 +(b-1)^2 >=0 hiển nhiên => dpcm
đẳng thwucs kh a=b=1
a) Ta có: \(5^{2005}+5^{2003}=5^{2003}\left(5^2+1\right)=5^{2003}\cdot26\)
Vì \(26⋮13\)
nên \(5^{2003}\cdot26⋮13\)
hay \(5^{2005}+5^{2003}⋮13\)