Question

a) cho a²+b²+c²=ab+bc+ca. chứng minh a=b=c

b) cho 2( x²+t²)+(y+t)(y-t) = 2x(y+t). chứng minh x=y=t

c) cho a+b+c=0; ab+bc+ac=0. chứng minh A= (a-1)²⁰⁰³+b²⁰⁰⁴+(c+1)²⁰⁰⁵

^{Giúp mk vs các bạn ơi, mk cần gấp}

Answer 1

a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)

Answer 2

Ta có : a² + b² + c² = ab + bc + ca

=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

= (a² - 2ab + b²) +  (b² - 2bc + c²) + (c² - 2ca + a²) = 0

=> (a - b)² + (b - c)² + (c - a)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)

b) Ta có :  2(x² + t²) + (y + t)(y - t) = 2x(y + t)

=> 2x² + 2t² + y² - t² = 2xy + 2t

=> 2x² + t² + y² = 2xt + 2xy

=> 2x² + t² + y² - 2xt - 2xy = 0

=> (x² - 2xy + y²) + (x² + t² - 2xt)  = 0

=> (x - y)² + (x - t)² = 0

=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)

c) Ta có a + b + c = 0 

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> a² + b² + c² + 2(ab + bc + ca) = 0

=> a² + b² + c² = 0

=> a = b = c = 0

Khi đó A = (0 - 1)²⁰⁰³ + 0²⁰⁰⁴ + (0 + 1)²⁰⁰⁵

= - 1 + 0 + 1 = 0

Vậy A = 0

Answer 3

b) Ta có: \(2\left(x^2+t^2\right)+\left(y+t\right)\left(y-t\right)=2x\left(y+t\right)\)

\(\Leftrightarrow2x^2+2t^2+y^2-t^2-2xy-2xt=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xt+t^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-t\right)^2=0\)

Tương tự phần a => \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-t\right)^2=0\end{cases}}\Rightarrow x=y=t\)

Answer 4

c) -.-

Ta có: \(\hept{\begin{cases}a+b+c=0\\ab+bc+ca=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b+c\right)^2=0\\2\left(ab+bc+ca\right)=0\end{cases}}\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=0\)

Dấu "=" xảy ra khi: \(a=b=c=0\)

=> \(A=\left(0-1\right)^{2003}+0^{2004}+\left(0+1\right)^{2005}=-1+0+1=0\)

Answer 5

a) \(a^2+b^2+c^2=ab+bc+ca< =>2a^2+2b^2+2c^2-2ab-2bc-2ca\)

\(< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>a=b=c\)

b) \(2\left(x^2+t^2\right)+\left(y-t\right)\left(y+t\right)=2x\left(y+t\right)< =>2x^2+2t^2+y^2-t^2-2xy-2xt=0\)

\(< =>x^2-2xy+y^2+x^2-2xt+t^2=0< =>\left(x-y\right)^2+\left(x-t\right)^2=0< =>x=y=t\)

c) Ta có \(a+b+c=0< =>a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(< =>a^2+b^2+c^2=0< =>a=b=c=0\)

Khi đó \(A=\left(0-1\right)^{2003}+0^{2004}+\left(0+1\right)^{2005}=-1+1=0\)