Chủ đề này có 2 trả lời

[Linda Johnson]

Rút gọn:
a)(x+y)^2+(x-y)^2+(x+y)(x-y)-3x^2
b)(x-y+z)^2+(y-z)^2 + 2(x-y+z)(y-z)
Tìm x:
a) (x+3)^2+(x-2)^2=2x^2.
b)7x(x-2)=x-2
c) (3x-5)^2-x(3x-5)=0
d) x^3-25x=0

 

[Linda Johnson]

Giúp mình với các bạn ToT  

[Oo Nguyen Hoang Nguyen oO]

Rút gọn:

a) $(x+y)^2+(x-y)^2+(x+y)(x-y)-3x^2=(x^2+2xy+y^2)+(x^2-2xy+y^2)+(x^2-y^2)-3x^2=y^2$

b) $(x-y+z)^2+(y-z)^2+2(x-y+z)(y-z)=\left [ (x-y+z)+(y-z) \right ]^2=x^2$

Tìm x:

a)$(x+3)^2+(x-2)^2=2x^2\Leftrightarrow (x^2+6x+9)+(x^2-4x+4)-2x^2=0\Leftrightarrow 2x+13=0\Leftrightarrow x=\frac{-13}{2}$

b)$7x(x-2)=x-2\Leftrightarrow 7x(x-2)-(x-2)=0\Leftrightarrow (7x-1)(x-2)=0\Leftrightarrow x=\frac{1}{7}$ hoặc $x=2$

c)$(3x-5)^2-x(3x-5)=0\Leftrightarrow (3x-5)(3x-5-x)=0\Leftrightarrow (3x-5)(2x-5)=0\Leftrightarrow x=\frac{5}{3}$ hoặc $x=\frac{5}{2}$

d)$x^3-25x=0\Leftrightarrow x(x^2-25)=0\Leftrightarrow x(x-5)(x+5)=0\Leftrightarrow x\epsilon \left \{ 0;\pm 5\left. \right \} \right.$