Bat dang thuc lop 8
Bắt đầu bởi uk.em_rat_ngoc, 08-11-2010 - 20:50
#1
Đã gửi 08-11-2010 - 20:50
1, Cho a <b c va` x<y z.CM
a, :frac{a+b}{2} . :frac{x+y}{2} :frac{ax+by}{2}
b, :frac{a+b+c}{3} + :frac{x+y+z}{3} :frac{ax+by+cz}{3}
2, (a^{n})^2+( b^{n})^2 gia tri tuyet doi (a^n.b^n)
Moi nguoi` lam` chi tie^'t giu'p e nha'. Thanks nhiu`
a, :frac{a+b}{2} . :frac{x+y}{2} :frac{ax+by}{2}
b, :frac{a+b+c}{3} + :frac{x+y+z}{3} :frac{ax+by+cz}{3}
2, (a^{n})^2+( b^{n})^2 gia tri tuyet doi (a^n.b^n)
Moi nguoi` lam` chi tie^'t giu'p e nha'. Thanks nhiu`
#2
Đã gửi 08-11-2010 - 21:03
Câu 2 :1, Cho $a <b \leq c $va` $x<y \leq z$.CM
a, $\dfrac{a+b}{2} . \dfrac{x+y}{2} \leq \dfrac{ax+by}{2}$
b, $ \dfrac{a+b+c}{3} . \dfrac{x+y+z}{3} \leq \dfrac{ax+by+cz}{3}$
2, $(a^{n})^2+( b^{n})^2 \geq |a^n.b^n|$
Moi nguoi` lam` chi tie^'t giu'p e nha'. Thanks nhiu`
Đặt $a^n=x,b^n=y$
Thì BĐT trở thành :$x^2+y^2 \geq |xy| \Leftrightarrow x^2-|xy|+y^2 \geq 0$
$ \Leftrightarrow \left( {\left| x \right| - \dfrac{{\left| y \right|}}{2}} \right)^2 +\dfrac{3y^2}{4} \geq 0$(luôn đúng với mọi x,y)
Bài viết đã được chỉnh sửa nội dung bởi dark templar: 08-11-2010 - 22:02
"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.
#3
Đã gửi 08-11-2010 - 21:26
Câu 1
a/BĐT$ \Leftrightarrow \left( {a + b} \right)\left( {x + y} \right) \le 2\left( {ax + by} \right) \Leftrightarrow ay + bx \le ax + by $
$\Leftrightarrow \left( {a - b} \right)\left( {y - x} \right) \le 0 $
(Đúng vì $ a \le b \Leftrightarrow a - b \le 0,x \le y \Leftrightarrow y - x \ge 0 \Rightarrow \left( {a - b} \right)\left( {y - x} \right) \le 0 $)
b/BĐT$ \Leftrightarrow \left( {a + b + c} \right)\left( {x + y + z} \right) \le 3\left( {ax + by + cz} \right)$
Áp dụng câu 1a ta có :
$\dfrac{{16}}{9}\left( {a + b + c} \right)\left( {x + y + z} \right)$
$ = \left[ {\left( {a + b} \right) + \left( {c + \dfrac{{a + b + c}}{3}} \right)} \right]\left[ {\left( {x + y} \right) + \left( {z + \dfrac{{x + y + z}}{3}} \right)} \right] \le 2\left[ {\left( {a + b} \right)\left( {x + y} \right) + \left( {c + \dfrac{{a + b + c}}{3}} \right)\left( {z + \dfrac{{x + y + z}}{3}} \right)} \right] $
$\le 2\left[ {2\left( {ax + by} \right) + 2\left( {cz + \left( {\dfrac{{a + b + c}}{3}} \right)\left( {\dfrac{{x + y + z}}{3}} \right)} \right)} \right] $
$= 4\left( {ax + by + cz} \right) + 4\dfrac{{\left( {a + b + c} \right)\left( {x + y + z} \right)}}{9}$
$\Rightarrow ax + by + cz \ge \dfrac{{\left( {a + b + c} \right)\left( {x + y + z} \right)}}{3}(dpcm)$
a/BĐT$ \Leftrightarrow \left( {a + b} \right)\left( {x + y} \right) \le 2\left( {ax + by} \right) \Leftrightarrow ay + bx \le ax + by $
$\Leftrightarrow \left( {a - b} \right)\left( {y - x} \right) \le 0 $
(Đúng vì $ a \le b \Leftrightarrow a - b \le 0,x \le y \Leftrightarrow y - x \ge 0 \Rightarrow \left( {a - b} \right)\left( {y - x} \right) \le 0 $)
b/BĐT$ \Leftrightarrow \left( {a + b + c} \right)\left( {x + y + z} \right) \le 3\left( {ax + by + cz} \right)$
Áp dụng câu 1a ta có :
$\dfrac{{16}}{9}\left( {a + b + c} \right)\left( {x + y + z} \right)$
$ = \left[ {\left( {a + b} \right) + \left( {c + \dfrac{{a + b + c}}{3}} \right)} \right]\left[ {\left( {x + y} \right) + \left( {z + \dfrac{{x + y + z}}{3}} \right)} \right] \le 2\left[ {\left( {a + b} \right)\left( {x + y} \right) + \left( {c + \dfrac{{a + b + c}}{3}} \right)\left( {z + \dfrac{{x + y + z}}{3}} \right)} \right] $
$\le 2\left[ {2\left( {ax + by} \right) + 2\left( {cz + \left( {\dfrac{{a + b + c}}{3}} \right)\left( {\dfrac{{x + y + z}}{3}} \right)} \right)} \right] $
$= 4\left( {ax + by + cz} \right) + 4\dfrac{{\left( {a + b + c} \right)\left( {x + y + z} \right)}}{9}$
$\Rightarrow ax + by + cz \ge \dfrac{{\left( {a + b + c} \right)\left( {x + y + z} \right)}}{3}(dpcm)$
Bài viết đã được chỉnh sửa nội dung bởi dark templar: 08-11-2010 - 21:28
"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.
#4
Đã gửi 16-11-2010 - 17:53
day la cach viet khac cua trebesep thui!1, Cho a <b c va` x<y z.CM
a, $ \dfrac{a+b}{2} . \dfrac{x+y}{2} \leq \dfrac{ax+by}{2}$
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
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