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.

nguyễn văn dương gì vậy bạn???

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) ( x + 1 )( y + 2 ) = 0

\(\Rightarrow\) x + 1 = 0 hoặc y + 2 = 0

+) x + 1 = 0 \(\Rightarrow\) x = -1

+) y + 2 = 0 \(\Rightarrow\) y = -2

Vậy x = -1; y = -2

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

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