\(^{x^2-2x+1=0}\)
\(=>\left(x-1\right)^2=0\)
\(=>x-1=0\)
\(=>x=1\)
T nha
\(x^2-2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
x2 - 2x + 1 = 0
<=>x(x-2)+1=0
=>x(x-2)=-1=1.(-1)
Vì x>(x-2)
=>x=1
\(^{x^2-2x+1=0}\)
\(=>\left(x-1\right)^2=0\)
\(=>x-1=0\)
\(=>x=1\)
T nha
\(x^2-2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
x2 - 2x + 1 = 0
<=>x(x-2)+1=0
=>x(x-2)=-1=1.(-1)
Vì x>(x-2)
=>x=1
giup mik voi
tim x,y
2x^2(1-y)+y(y^2+xy-2x)=0
tim x biet
a) 9x^2+6x+1=0
b) 25x^2=4
c) 8-125x^3=0
d) (2x+1)-10(2x+1)(x+2)+25(x+2)^2=0
GIÚP MIK VS NHÉ
tim x,y,z biết
a)(2x-3)^2=0
b) 25x^2-10x+1=0
c) 6x+9=-x^2
d) 169-y^2=0
e) 2x^2-2xy-4x+5=0
giup mik vs
tôi đi hok
nhanh nhé
mai mik kiểm tra rùi giúp mik vs pls
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
Tim min
A= x^2+x-2
B=x^2-x
C=1/4x^2-x+7
D=1/2x^2+3x+1
E=(x-1)(x^2+x+1)-x(x-1)(x+1)+x^2
Tim max
A= -2(x-1)^2+(x+3)
B=-x^2+4x-1
C=-2x^2+x
D=(x-3)(2-x)-3(x+5)(x+7)
E=-3x^2+4x-1
AI HELP MIK DAU TIEN MIK SẼ HAU TẠ
tl nhanh cho mik nhá, mik đag cần gấp. Tks nhiều :)))
1, Rút gọn biểu thức sau:
a, (2x-3)(x+2)-2(x+1)^2
b,(x-2)^2+2(x-2)(2x+2)+4(x+1)^2
c,(x^2-2x+4)(x+2)-(x-1)^3+3(x-1)(x+1)
d,(x+y+z-t)^2-(-x-y-z+t)^2
e,(x+1)^3-3(x-2)(x+1)-(x^2+x+1)(x+1)
2,PTĐTTNT
a,x^2-9-x^2(x^2-9)
b, 9x^2-25y^2-6x+10y
c,x^2-8x-4y^2+16
tìm x biết :
a,(2x-3)^2 =(x+ 5)^2
b,x^2(x-1) -4x^2 +8x -4 =0
c, (x-4)^2 -36 =0
giúp mik nha mik đang gấp