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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
\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
Bài 1:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
\(\Rightarrowđpcm\)
Bài 2:
Ta có: \(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=25a^2-30ab+9b^2-64c^2\)
\(=25a^2-30ab+9b^2-16a^2+16b^2\left(a^2-b^2=4c^2\right)\)
\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2=VP\)
\(\Rightarrowđpcm\)
\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)\(\Leftrightarrow\)\(a=b=c\)
=> đpcm
a)Ta có:a^2+b^2+c^2-ab-bc-ca=0
<=>2(a^2+b^2+c^2-ab-bc-ca)=2.0
<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0
<=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0
<=>(a-b)^2+(b-c)^2+(c-a)^2=0
Vì (a-b)^2>=0 với mọi a,b
(b-c)^2>=0 với mọi b,c
(c-a)^2>=0 với mọi c,a
Do đó:(a-b)^2+(b-c)^2+(c-a)^2>=0
Dấu "=" xảy ra <=>a-b=0<=>a=b
b-c=0<=>b=c
c-a=0<=>c=a
hay a=b=c(đpcm)
Ta có :(5a-3b+8c)(5a-3b-8c)=((5a-3b)+8c)((5a-3b)-8c)=(5a-3b)^2-(8c)^2=25a^2-30ab+9b^2-64c^2
=25a^2-30ab+9b^2-16.4c^2=25a^2-30ab+9b^2-16.4c^2(*)
Thay a^2-b^2=4c^2 vào(*)=>25a^2-30ab+9b^2-16(a^2-b^2)=25a^2-30ab+9b^2-16a^2+16b^2
=9a^2-30ab+25b^2=(3a)^2-3a.2.5b+(5b)^2=(3a-5b)^2
=>đpcm
bài 1: ta có : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
\(\Leftrightarrow\left(5a-3b\right)^2-\left(8c\right)^2=\left(3a-5b\right)^2\) \(\Leftrightarrow\left(5a-3b\right)^2-\left(3a-5b\right)^2=\left(8c\right)^2\)
\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=\left(8c\right)^2\)
\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=64c^2\) \(\Leftrightarrow16\left(a+b\right)\left(a-b\right)=64c^2\)
\(\Leftrightarrow a^2-b^2=4c^2\left(đpcm\right)\)
bài 2 : bài này yc CM j bn ?? ?
bài 3 : a) ta có : \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\left(đpcm\right)\)
b) ta có : \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=a^2+b^2+c^2\) \(\Rightarrow\) giống câu a
c) ta có : \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\) \(\Rightarrow\) giống câu a2. >=