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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2=a^2-2ab+b^2+2ab\)
\(=\left(a-b\right)^2+2ab\)
\(=m^2+2n\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{1}{y-1}-\frac{y}{1-y^2}\left(y\ne\pm1\right)\)
\(\Leftrightarrow A=\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}=\frac{y+1}{\left(y-1\right)\left(y+1\right)}+\frac{y}{\left(y-1\right)\left(y+1\right)}=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\)
Thay y=2 (tm) vao A ta co:
\(A=\frac{2\cdot2+1}{\left(2-1\right)\left(2+1\right)}=\frac{5}{3}\)
Vay \(A=\frac{5}{3}\)voi y=2
b) Ta co: \(\hept{\begin{cases}A=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\left(y\ne\pm1\right)\\B=\frac{y^2-y}{2y+1}=\frac{y\left(y-1\right)}{2y+1}\left(y\ne\frac{-1}{2}\right)\end{cases}}\)
\(\Rightarrow M=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}\cdot\frac{y\left(y-1\right)}{2y+1}=\frac{\left(2y+1\right)\cdot y\cdot\left(y-1\right)}{\left(y-1\right)\left(y+1\right)\left(2y+1\right)}=\frac{y}{y+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= (a+b)(a2-ab+b2) + 3ab((a+b)2-2ab) + 6a2b2(a+b)
Thay a+b = 1 vài biểu tức trên ta có:
a2-ab+b2+ 3ab(1-2ab)+6a2b2=a2-ab+b2+3ab-6a2b2+6a2b2
= a2 + 2ab + b2
= (a+b)2
= 1
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (2a + 2b +2c - 3c)^2 + (2b + 2c +2a - 3a)^2 + (2c + 2a +2b -3b)^2
Đặt a + b + c = x thì
A = (2x - 3c)^2 + (2x - 3a)^2 + (2x - 3b)^2
=4x^2 - 12cx + 9c^2 + 4x^2 - 12ax + 9x^2 + 4x^2 - 12bx + 9b^2
=12x^2 - 12x(a + b + c) + 9(a^2 + b^2 + c^2)
=12x^2 - 12x^2 + 9(a^2 + b^2 + c^2) =9(a^2 + b^2 + c^2) =9m
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: Cho \(a^2+b^2+c^2=m\)
Tính: \(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)
Giải:
Ta có: \(\left(x+y-z\right)^2=\left(x+y\right)^2-2\left(x+y\right).z+z^2=x^2+y^2+z^2+2xy-2xz-2yz\)
Ứng dụng vào bài trên:
\(A=\left[\left(2a\right)^2+\left(2b\right)^2+c^2+2\left(2a\right)\left(2b\right)-2\left(2a\right)c-2\left(2b\right)c\right]\)
\(+\left[\left(2b\right)^2+\left(2c\right)^2+a^2+2\left(2b\right)\left(2c\right)-2\left(2b\right)a-2\left(2c\right)a\right]\)
\(+\left[\left(2c\right)^2+\left(2a\right)^2+b^2+2\left(2c\right)\left(2a\right)-2\left(2c\right)b-2\left(2a\right)b\right]\)
\(=4a^2+4b^2+c^2+8ab-4ac-4bc\)
\(+4b^2+4c^2+a^2+8bc-4ba-4ca\)
\(+4c^2+4a^2+b^2+8ca-4cb-4ab\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
\(=9m\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)
\(=\left(4a^2+4b^2+c^2+8ab-4ac+4bc\right)+\left(4b^2+4c^2+a^2+8bc-4ba-4ac\right)\)\(+\left(4c^2+4a^2+b^2+8ac-4cb-4ab\right)\)
\(=9a^2+9b^2+9c^2\)
\(=9\left(a^2+b^2+c^2\right)\)
\(=9m\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (2a + 2b - c)2 + ( 2b + 2c -a)2 + (2c + 2a -b) 2
= [ (2a + 2b)2 - 2 . 2 ( a+b) .c + c2 ] + [ ( 2b + 2c )2 -2.2(b+c) .a + a2 ] + [(2c + 2a)2 - 2.2(a+c)b + b2 ]
= (4a2+ 8ab+4b2- 4ac - 4bc +c2) + ( 4b2+ 8bc+ 4c2- 4ab - 4ac+a2) +
(4c2+8ac+4a2- 4ab - 4bc+b2)
=9a2 +9b2 +9c2
Thay a2 +b2 +c2 =m vào trên , tá dược :
A= 9m
a, Ta có :
\(a^2+b^2=a^2-2ab+b^2+2ab\) \(=\left(a-b\right)^2+2ab=m^2+2n\)
b, Ta có :
\(\left(a+b\right)^2=a^2+2ab+b^2\) \(=a^2-2ab+b^2+4ab=\left(a-b\right)^2+4ab\) \(=m^2+4n\)