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a)Ta có: a^2 + b^2 + c^2 = ab + bc + ca
<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca
<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0
<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1)
Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c.
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2)
Từ (1) và (2) khẳng định dấu "=" khi:
a - b = 0; b - c = 0 ; c - a = 0 => a=b=c
Vậy a=b=c.
A = (5a – 3b + 8c)(5a – 3b –8c)
= (5a –3b)² – (8c)²
= (25a² – 30ab +9b²) – 64c²
Mà theo đề thì 4c² = a² –b²
Nên ta suy ra:
A = (25a² – 30ab +9b²) – 16(a² –b²)
= 9a² –30ab +25b²
= (3a –5b)²
\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
a) \(a^2+25b^2+17+10b-8a=0\)
\(\Rightarrow a^2-8a+16+25b^2+10b+1=0\)
\(\Rightarrow\left(a-4\right)^2+\left(5b+1\right)^2=0\)
Vì \(\left(a-4\right)^2\ge0\) với mọi a
\(\left(5b+1\right)^2\ge0\) với mọi b
\(\Rightarrow\left(a-4\right)^2+\left(5b+1\right)^2\ge0\) với mọi a,b
Mà \(\left(a-4\right)^2+\left(5b+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-4\right)^2=0\\\left(5b+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-4=0\\5b+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\5b=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=-\dfrac{1}{5}\end{matrix}\right.\)
Ta có : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=\left(5a-3b\right)^2-16.4c^2\)
\(=\left(5a-3b\right)^2-16\left(a^2-b^2\right)\)
\(=25a^2-30ab+9b^2-16a^2+16b^2\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2\left(đpcm\right)\)
biến đổi vế trái
\(\Leftrightarrow\left(5a-3b\right)^2-\left(8c\right)^2\)
\(\Leftrightarrow25a^2-30ab+9b^2-64c^2\)
\(\Leftrightarrow25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(\Leftrightarrow\left(25a^2-16a^2\right)-30ab+\left(9b^2+16b^2\right)\)
\(\Leftrightarrow9a^2-30ab+25b^2\)
\(\Leftrightarrow\left(3a-5b\right)^2\) (điều cần c/m)
\(\frac{\left(2-c\right)\left(b-c\right)}{2a+bc}=\frac{\left(a+b\right)\left(b-c\right)}{a\left(a+b+c\right)+bc}=\frac{\left(a+b\right)\left(b-c\right)}{\left(a+b\right)\left(c+a\right)}=\frac{b-c}{c+a}=\frac{b}{c+a}-\frac{c}{c+a}\)
Tương tự, ta có: \(\frac{\left(2-a\right)\left(c-a\right)}{2b+ca}=\frac{c}{a+b}-\frac{a}{a+b};\frac{\left(2-b\right)\left(a-b\right)}{2c+ab}=\frac{a}{b+c}-\frac{b}{b+c}\)
\(\Rightarrow\)\(VT=\left(\frac{a}{b+c}-\frac{a}{a+b}\right)+\left(\frac{b}{c+a}-\frac{b}{b+c}\right)+\left(\frac{c}{a+b}-\frac{c}{c+a}\right)\)
\(=\frac{a\left(a-c\right)}{\left(a+b\right)\left(b+c\right)}+\frac{b\left(b-a\right)}{\left(b+c\right)\left(c+a\right)}+\frac{c\left(c-b\right)}{\left(c+a\right)\left(a+b\right)}\)
\(=\frac{a\left(a-c\right)\left(c+a\right)+b\left(b-a\right)\left(a+b\right)+c\left(c-b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{3}\)
cái bđt \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\) cô Chi có làm r ib mk gửi link