Tìm x :
a . | x - 2016 | = x - 2016
b. | x ( x - 2016 ) | = x
c. ( x + 2 )^2 + | 2x + 4 | = 0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(-10x^3+2x^2=0\)
\(\Rightarrow-2x^2\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(5x\left(x-2016\right)-x+2016=0\)
\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)
\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)
a: Ta có: \(-10x^3+2x^2=0\)
\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
a, (x-2016)(2x-1)=0
<=>x=2016 hoặc x=-1/2
b, (x+2)(x+2-x+2)=0
<=>4(x+2)=0
<=>x+2=0
<=>x=-2
a) 2x(x-2016)-x+2016=0
=>2x(x-2016)-(x-2016)=0
=>(x-2016)(2x-1)=0
=>\(\left\{{}\begin{matrix}x-2016=0\\2x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2016\\x=\dfrac{1}{2}\end{matrix}\right.\)
vậy x=2016 hoặc x=\(\dfrac{1}{2}\)
b) (x+2)2-(x-2)(x+2)=0
=>(x+2)[(x+2)-(x-2)]=0
=>(x+2)(x+2-x+2)=0
=>(x+2)4=0
=>x+2=0
=>x=-2
vậy x=-2
Xin lỗi nha. Mk mún giúp lắm nhưng mk mới học lp 5 thui nên đọc đề ko hỉu gì hết đó.
Câu 1:
a)A=|x+1|+2016
Vì |x+1|\(\ge\)0
Suy ra:|x+1|+2016\(\ge\)2016
Dấu = xảy ra khi x+1=0
x=-1
Vậy MinA=2016 khi x=-1
b)B=2017-|2x-\(\frac{1}{3}\)|
Vì -|2x-\(\frac{1}{3}\)|\(\le\)0
Suy ra:2017-|2x-\(\frac{1}{3}\)|\(\le\)2017
Dấu = xảy ra khi \(2x-\frac{1}{3}=0\)
\(2x=\frac{1}{3}\)
\(x=\frac{1}{6}\)
Vậy Max B=2017 khi \(x=\frac{1}{6}\)
c)C=|x+1|+|y+2|+2016
Vì |x+1|\(\ge\)0
|y+2|\(\ge\)0
Suy ra:|x+1|+|y+2|+2016\(\ge\)2016
Dấu = xảy ra khi x+1=0;x=-1
y+2=0;y=-2
Vậy MinC=2016 khi x=-1;y=-1
d)D=-|x+\(\frac{1}{2}\)|-|y-1|+10
=10-|x+\(\frac{1}{2}\)|-|y-1|
Vì -|x+\(\frac{1}{2}\)|\(\le\)0
-|y-1| \(\le\)0
Suy ra: 10-|x+\(\frac{1}{2}\)|-|y-1| \(\le\)10
Dấu = xảy ra khi \(x+\frac{1}{2}=0;x=-\frac{1}{2}\)
y-1=0;y=1
Vậy Max D=10 khi x=\(-\frac{1}{2}\);y=1
Bài 1:
a)Ta thấy: \(\left|x+1\right|\ge0\)
\(\Rightarrow\left|x+1\right|+2016\ge0+2016=2016\)
\(\Rightarrow A\ge2016\)
Dấu = khi x=-1
Vậy MinA=2016 khi x=-1
b)Ta thấy:\(\left|2x-\frac{1}{3}\right|\ge0\)
\(\Rightarrow-\left|2x-\frac{1}{3}\right|\le0\)
\(\Rightarrow2017-\left|2x-\frac{1}{3}\right|\le2017-0=2017\)
\(\Rightarrow B\le2017\)
Dấu = khi x=1/6
Vậy Bmin=2017 khi x=1/6
c)Ta thấy:\(\begin{cases}\left|x+1\right|\\\left|y+2\right|\end{cases}\ge0\)
\(\Rightarrow\left|x+1\right|+\left|y+2\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|y+2\right|+2016\ge0+2016=2016\)
\(\Rightarrow D\ge2016\)
Dấu = khi x=-1 và y=-2
Vậy MinD=2016 khi x=-1 và y=-2
d)Ta thấy:\(\begin{cases}-\left|x+\frac{1}{2}\right|\\-\left|y-1\right|\end{cases}\le0\)
\(\Rightarrow-\left|x+\frac{1}{2}\right|-\left|y-1\right|\le0\)
\(\Rightarrow-\left|x+\frac{1}{2}\right|-\left|y-1\right|+10\le0+10=10\)
\(\Rightarrow D\le10\)
Dấu = khi x=-1/2 và y=1
Vậy MaxD=10 khi x=-1/2 và y=1
a, 3 - 2x = 3 . (5 - x) + 4
3 - 2x = 15 - 3x + 4
-2x + 3x = 15 + 4 - 3
x = 16
b, 4 - (7x + 2017) = 6 . (5 - x) - 2017
4 - 7x - 2017 = 30 - 6x - 2017
-7x + 6x = 30 - 2017 - 4 + 2017
-x = 26
x = -26
c, 15 - x . (x + 1) = 4 - x^2 + 2x
15 - x^2 - x = 4 - x^2 + 2x
-x^2 - x + x^2 - 2x = 4 - 15
-3x = -11
x = 11/3
d, -4 . (x - 5) + 2016 = 3 . (8 - x) - (2x - 2016)
-4x + 20 + 2016 = 24 - 3x - 2x + 2016
-4x + 3x +2x = 24 + 2016 - 20 - 2016
x = 4
đúng 100%
a... . x = 2016
b.....x=0
c.... x = -2
a. x=2016
b. x=2017
c. x= -2