x3 +2x - 3 = 0
HELP ME!!!
help me (2x-3)*(6-2x)=0
TH1 :2x-3=0
=>2x=3
=>x\(\frac{3}{2}\)
TH2: 6-2x=0
=>2x=6
=>x=3
em moi hoc lop 5 thoi nen thong cam nhe tri nhan bao
Cho x1+x2+x3+....+x2009+x2010+x2011=0 và x1+x2+x3=x4+x5+x6=....=x2008+x2009+x2010=1
P/S: Help me... Thanks very much☺☺☺
Ta có:
x1 + x2 + x3 + ... + x2008 + x2009 + x2010
= (x1 + x2 + x3) + ... + (x2008 + x2009 + x2010)
= 1 + 1 + 1 + ... + 1(670 số 1)
= 670
\(\Rightarrow\) x1 + x2 + x3 + ... + x2009 + x2010 + x2011 = 670 + x2011 = 0
\(\Rightarrow\) x2011 = -670
Tìm x :
a) x^2 - 2x + 3 = 0
b) 2x^2 - x - 3 = 0
Help me !!!
Bài : Cho đa thức M = x3 + x2y - 2x2 - xy - y2 + 3y + x + 2017. Tính giá trị của đa thức M biết x + y - 2 = 0.
Help me !
M = x3 + x2y - 2x2 - xy - y2 + 3y + x + 2017
M = (x3 + x2y - 2x2) - (xy + y2 - 2y) + (x + y - 2) + 2019
M = x2. (x + y - 2) - y(x + y - 2) + (x + y - 2) + 2019 = 2019
\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)
\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)
\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)
\(M=x^2.0-y.0+0+2019\)
\(M=0-0+0+2019\)
\(M=2019\)
help me
(x-3).(x^4+2x^2+1)=0. Tìm x?
vì \(x^4+2x^2+1=\left(x^2+1\right)^2\) mà \(x^2\ge0\Rightarrow x^2+1>0\Rightarrow\left(x^2+1\right)^2>0\)với mọi x.Nên x-3=0 .Từ đó suy ra x=3
(2x-3).(x+1/4)=0
plz help me!!
(2x-3)(x+\(\frac{1}{4}\))=0
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-1}{4}\end{cases}}}\)
Bài làm
Ta có: \(\left(2x-3\right)\left(x+\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{4}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}}\)
Vậy x = 3/2 hoặc x = -1/4
\(\left(2x-3\right)\left(x+\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{4}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)
P=
\(\frac{\text{2x 3 x4+ 2 x3x 4 x2018 +2 x3 x4 x2019 -2x3 x4 x2020}}{\text{5 x6 x7+ 5x 6 x7 x2018 +5 x6 x7 x2019- 5x 6 x7 x2020}}\)
HELP ME !!!
Ta có \(\frac{2.3.4+2.3.4.2018+2.3.4.2019-2.3.4.2020}{5.6.7+5.6.7.2018+5.6.7.2019-5.6.7.2020}\)
\(=\frac{2.3.4\left(1+2018+2019-2020\right)}{5.6.7.\left(1+2018+2019-2020\right)}=\frac{2.3.4}{5.6.7}=\frac{4}{35}\)
TÌM x , biêt ,
a, ( 2x + 1 ) + ( 3 - x ) = 0
Help me !!!!
(2x+1)+(3-x)=0
=>2x+1=-3+x
=>2x+1-x=-3
=>x+1=-3
=>x=-3-1=-4
Vậy x=-4
a)(2x+1)+(3-x)=0
2x+1+3-x=0
x+4=0
x=-4
vậy x=-4