\(a^2+2a+b^2+4b+4c^2-4c+6=0\)

\(\Leftrightarrow\left(a^2+2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

Mà \(\begin{cases}\left(a+1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}\)

\(\Rightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)

\(\Rightarrow\begin{cases}a+1=0\\b+2=0\\2c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=-1\\b=-2\\c=\frac{1}{2}\end{cases}\)

Câu hỏi của Phạm Thị Thùy Linh - Toán lớp 8 - Học toán với OnlineMath

Bạn kiểm tra lại đầu bài đi

Ko có d sao tìm đc:)))))

= (a+1)² +(b+2)² +(2c-1)² =0

=> a = -1

     b = -2

     c = 1/2

đk cần và đủ giỏi toán IQ>100 + chăm

<=>a^2-2a+b^2+4b+4c^2-4c+1+4+1=0

<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0

<=>(a-1)²+(b+2)²+(2c-1)²=0

<=>(a-1)^2=0 hoặc(b+2)^2=0 hoặc (2c-1)^2=0

+,(a-1)^2=0<=>a-1=0<=>a=1

+,(b+2)^2=0<=>b+2=0<=>b=-2

+,(2c-1)^2=0<=>2c-1=0<=>2c=1<=>c=1/2

\(a^2-2a+b^2+4b+4c^2-4c+6=0\)

\(=>\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(=>\left(a^2-2.a.1+1^2\right)+\left(b^2+2.b.2+2^2\right)+\left[\left(2c\right)^2-2.2c.1+1^2\right]=0\)

\(=>\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\left(1\right)\)

Vì : \(\left(a-1\right)^2\ge0\) với mọi a

\(\left(b+2\right)^2\ge0\) với mọi b

\(\left(2c-1\right)^2\ge0\) với mọi c

=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\) với mọi a,b,c

Để (1) thì \(\left(a-1\right)^2=\left(b+2\right)^2=\left(2c-1\right)^2=0=>a=1;b=-2;c=\frac{1}{2}\)

Vậy........

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+1\right)^2+\left(2c-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b+1=0\\2c-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=-1\\c=\frac{1}{2}\end{cases}}\)

 a^2-2a+b^2+4b+4c^2-4c+6=0 
<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0 
<=>(a-1)^2+(b+2)^2+(2c-1)^2=0 
vi (a-1)^2>=0,(b+2)^2>=0,(2c-1)^2>=0 
=>(a-1)^2+(b+2)^2+(2c-1)^2>=0 
dau = xay ra <=>(a-1)^2=0,(b+2)^2=0,(2c-1)^2=0 
<=>a-1=b+2=2c-1=0 
<=>a=2,b=-2,c=1/2 
vay a=2,b=-2,c=1/2

CHÚC BẠN HỌC GIỎI

 a^2-2a+b^2+4b+4c^2-4c+6=0 
<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0 
<=>(a-1)^2+(b+2)^2+(2c-1)^2=0 
vi (a-1)^2>=0,(b+2)^2>=0,(2c-1)^2>=0 
=>(a-1)^2+(b+2)^2+(2c-1)^2>=0 
dau = xay ra <=>(a-1)^2=0,(b+2)^2=0,(2c-1)^2=0 
<=>a-1=b+2=2c-1=0 
<=>a=2,b=-2,c=1/2 
vay a=2,b=-2,c=1/2

CHÚC BẠN HỌC GIỎI

\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0.\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2b-1\right)^2=0\)

Mà \(\left(a-1\right)^2\ge0\forall a\), \(\left(b+2\right)^2\ge0\forall b\),\(\left(2c-1\right)^2\ge0\forall c\)

\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}.\)

  a² - 2a + b² + 4b + 4c² - 4c + 6 = 0

\(\Leftrightarrow\)a² - 2a + 1 + b² + 4b + 4 + 4c² - 4c² + 1 = 0

\(\Leftrightarrow\)( a - 1 )² + ( b + 2 )² + ( 2c - 1 )² = 0

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a-1=0\\b+2=0\\2c-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}\)

Vậy a = 1 , b = -2 , c = \(\frac{1}{2}\)

NGọc Nguyễn suy ra 1 phát luôn là sai đấy mà ko có nhân xét gì à