cho a/2b=b/2c=c/2d=d/2a (a;b;c;d>0)
tinh A=2013a-2012b/c+d+2013b-2012c/a+d+2013c-2012d/a+b+2013d-2012a/b+c
giup mik nha sai cung duoc miem la co bai
cho a/2b=b/2c=c/2d=d/2a (a;b;c;d>0)
tinh A=2013a-2012b/c+d+2013b-2012c/a+d+2013c-2012d/a+b+2013d-2012a/b+c
giup mik nha sai cung duoc miem la co bai
Ta có
$\dfrac{a}{2b} . \dfrac{b}{2c} . \dfrac{c}{2d} . \dfrac{d}{2a} = \left( \dfrac{a}{2b} \right)^4$
$\Leftrightarrow \dfrac{1}{2.2.2.2} = \left( \dfrac{a}{2b} \right)^4$
$\Leftrightarrow \dfrac{1}{2} = \dfrac{a}{2b}$
$\Leftrightarrow 1 = \dfrac{a}{b}$
$\Leftrightarrow a = b$
CMTT ta suy ra $a = b = c = d$.
Vậy ta tính
$A = 2013a – 2012 . \dfrac{b}{c+d} + 2013b – 2012 . \dfrac{c}{a+d} + 2013c – 2012.\dfrac{d}{a+b} + 2013d – 2012 . \dfrac{a}{b+c}$
$= 2013 a + 2013b + 2013c + 2013d – 2012 \left( \dfrac{b}{c+d} + \dfrac{c}{a+d} + \dfrac{d}{a+b} + \dfrac{a}{b+c}\right)$
Do $a = b = c = d$ nên
$\dfrac{b}{c+d} = \dfrac{a}{b+c} = \dfrac{c}{a+d} = \dfrac{d}{a+b} = \dfrac{1}{2}$
Thay vào ta có
$A = 2013.4 – 2012.\dfrac{1}{2} . 4$
$= 2013.4 – 1006.4$
$= 4(2013-1006)$
$= 4.1007 = 4028$
Vậy $A = 4028$.
Ta có
a2b.b2c.c2d.d2a=(a2b)4a2b.b2c.c2d.d2a=(a2b)4
⇔12.2.2.2=(a2b)4⇔12.2.2.2=(a2b)4
⇔12=a2b⇔12=a2b
⇔1=ab⇔1=ab
⇔a=b⇔a=b
CMTT ta suy ra a=b=c=da=b=c=d.
Vậy ta tính
A=2013a−2012.bc+d+2013b−2012.ca+d+2013c−2012.da+b+2013d−2012.ab+cA=2013a−2012.bc+d+2013b−2012.ca+d+2013c−2012.da+b+2013d−2012.ab+c
=2013a+2013b+2013c+2013d−2012(bc+d+ca+d+da+b+ab+c)=2013a+2013b+2013c+2013d−2012(bc+d+ca+d+da+b+ab+c)
Do a=b=c=da=b=c=d nên
bc+d=ab+c=ca+d=da+b=12bc+d=ab+c=ca+d=da+b=12
Thay vào ta có
A=2013.4−2012.12.4A=2013.4−2012.12.4
=2013.4−1006.4=2013.4−1006.4
=4(2013−1006)=4(2013−1006)
=4.1007=4028=4.1007=4028
Vậy A=4028A=4028.