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[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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ĐĂNG BÀI NGAY
Bài 1: Cho
{ x , y , z ∈ R y ≠ z ( x − y ) ( x − z ) = 1 \left\{\begin{matrix} x,y,z\in R \\ y\neq z\\ (x-y)(x-z)=1 \end{matrix}\right. ⎩ ⎪ ⎨ ⎪ ⎧ x , y , z ∈ R y = z ( x − y ) ( x − z ) = 1 . Tìm GTNN của
P = 1 ( x − y ) 2 + 1 ( y − z ) 2 + 1 ( z − x ) 2 P=\frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(z-x)^2} P = ( x − y ) 2 1 + ( y − z ) 2 1 + ( z − x ) 2 1 Bài 2: Cho
{ a , b , c > 0 a b + b c + c a + a b c = 4 \left\{\begin{matrix} a,b,c>0\\ ab+bc+ca+abc=4 \end{matrix}\right. { a , b , c > 0 a b + b c + c a + a b c = 4 . Tìm GTLN của
Q = 1 2 ( a 2 + b 2 ) + 4 + 1 2 ( b 2 + c 2 ) + 4 + 1 2 ( c 2 + a 2 ) + 4 Q=\frac{1}{\sqrt{2(a^2+b^2)}+4}+\frac{1}{\sqrt{2(b^2+c^2)}+4}+\frac{1}{\sqrt{2(c^2+a^2)}+4} Q = 2 ( a 2 + b 2 ) + 4 1 + 2 ( b 2 + c 2 ) + 4 1 + 2 ( c 2 + a 2 ) + 4 1
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Bài 1: Cho
{ x , y , z ∈ R y ≠ z ( x − y ) ( x − z ) = 1 \left\{\begin{matrix} x,y,z\in R \\ y\neq z\\ (x-y)(x-z)=1 \end{matrix}\right. ⎩ ⎪ ⎨ ⎪ ⎧ x , y , z ∈ R y = z ( x − y ) ( x − z ) = 1 . Tìm GTNN của
P = 1 ( x − y ) 2 + 1 ( y − z ) 2 + 1 ( z − x ) 2 P=\frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(z-x)^2} P = ( x − y ) 2 1 + ( y − z ) 2 1 + ( z − x ) 2 1 Bài 2: Cho
{ a , b , c > 0 a b + b c + c a + a b c = 4 \left\{\begin{matrix} a,b,c>0\\ ab+bc+ca+abc=4 \end{matrix}\right. { a , b , c > 0 a b + b c + c a + a b c = 4 . Tìm GTLN của
Q = 1 2 ( a 2 + b 2 ) + 4 + 1 2 ( b 2 + c 2 ) + 4 + 1 2 ( c 2 + a 2 ) + 4 Q=\frac{1}{\sqrt{2(a^2+b^2)}+4}+\frac{1}{\sqrt{2(b^2+c^2)}+4}+\frac{1}{\sqrt{2(c^2+a^2)}+4} Q = 2 ( a 2 + b 2 ) + 4 1 + 2 ( b 2 + c 2 ) + 4 1 + 2 ( c 2 + a 2 ) + 4 1
Bài 1:
P =
1 ( x − y ) 2 + 1 ( y − z ) 2 + 1 ( z − x ) 2 = ( x − y ) 2 + ( z − x ) 2 + 1 ( y − z ) 2 = [ ( y − x ) + ( x − z ) ] 2 + 1 [ ( y − x ) + ( x − z ) ] 2 + 2 > = 2 + 2 = 4 \frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(z-x)^2}=(x-y)^2+(z-x)^2+\frac{1}{(y-z)^2}=[(y-x)+(x-z)]^2+\frac{1}{[(y-x)+(x-z)]^2}+2>= 2+2=4 ( x − y ) 2 1 + ( y − z ) 2 1 + ( z − x ) 2 1 = ( x − y ) 2 + ( z − x ) 2 + ( y − z ) 2 1 = [ ( y − x ) + ( x − z ) ] 2 + [ ( y − x ) + ( x − z ) ] 2 1 + 2 > = 2 + 2 = 4
Bài 2:
a b + b c + c a + a b c = 4 → 1 a + 2 + 1 b + 2 + 1 c + 2 = 1 ab+bc+ca+abc=4\rightarrow \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1 a b + b c + c a + a b c = 4 → a + 2 1 + b + 2 1 + c + 2 1 = 1
Xét:
1 2 ( a 2 + b 2 ) + 4 ≤ 1 ( a + 2 ) + ( b + 2 ) ≤ 1 4 ( 1 a + 2 + 1 b + 2 ) \frac{1}{\sqrt{2(a^2+b^2)}+4}\leq \frac{1}{(a+2)+(b+2)}\leq \frac{1}{4}(\frac{1}{a+2}+\frac{1}{b+2}) 2 ( a 2 + b 2 ) + 4 1 ≤ ( a + 2 ) + ( b + 2 ) 1 ≤ 4 1 ( a + 2 1 + b + 2 1 )
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
