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10 tháng 8 2023

A với D là gì ấy? Gõ muốn lủng cái tay 

31 tháng 7 2016

Ta có: ab= 10a + b

          ba=10b + a

=> ab + ba = 10a + b+ 10b + a = 11a + 11b Chia hết cho 11

abc -cba= 100a + 10b + c - 100c -10b -a = ( 100a -a ) + (10b - 10b) + ( 100c - c ) = 99a - 99c chia hết cho 99

25 tháng 9 2017

a, |x+1| +2x \(=\)-2

\(\Leftrightarrow\) |x+1|\(=\)-2-2x (*)

TH1: Nếu x+1\(\ge0\)\(\Leftrightarrow x\ge-1\)thì \(\left|x+1\right|=x+1\)

Thay vào (*) ta có:

\(x+1=-2-2x\)

\(\Leftrightarrow x+2x=-2-1\)

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)(TMĐK)

TH2: Nếu \(x+1< 0\Leftrightarrow x< -1\Rightarrow\left|x+1\right|=-\left(x+1\right)\)

Thay vào (*), ta có:

\(-x-1=\)\(-2-2x\)

\(\Leftrightarrow-x+2x=-2+1\)

\(\Leftrightarrow x=-1\)(kTMĐK)

Vậy S\(=\){-1}

b, \(x^2-6x+9=1\)

\(\Leftrightarrow x^2-2.3.x+3^2\)

\(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\left(x-3\right)^2-1^2=0\)

\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy S\(=\){4;2}

26 tháng 9 2017

a) |x + 1| + 2x = 2

\(\Rightarrow\) |3x + 1| = 2

\(\Rightarrow\) |3x| = 1

\(\Rightarrow\) |x| = 1 : 3 = \(\dfrac{1}{3}\)

\(\Rightarrow\) x = \(\dfrac{1}{3}\) hoặc \(-\dfrac{1}{3}\)

a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\\left(-2x-2\right)^2-\left(x+1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\3\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=-1\)

b: \(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(-2x-2\right)^2-\left(x+1\right)^2=0\\-2x-2>=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\left(x+1\right)^2=0\\x< =-1\end{matrix}\right.\Leftrightarrow x=-1\)

b: \(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow x\in\left\{4;2\right\}\)

27 tháng 9 2016

a) \(\left|x+1\right|+2x=-2\)

\(\Leftrightarrow\left|x+1\right|+2\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

b) \(x^2-6x+9=1\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow x^2-2x-4x+8=0\)

\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=4\end{array}\right.\)

 

28 tháng 11 2016

\(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=x\left(x+1\right)\left(4x^2-1\right)\)

\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

Dạng 3 :

a) 3x - 10 = 2x + 13

=> 3x - 2x = 13 - 10

=> x = 3

b) x + 12 = -5 - x

=> x + x = -5 - 12

=> 2x = -17

=> x = -8,5

c) x + 5 = 10 - x 

=> x + x = 10 - 5

=> 2x = 5

=> x = 2,5

d) 6x + 23 = 2x - 12

=> 2x - 6x = 23 + 12

=> -4x = 35

=> x = -8,75

e) 12 - x = x + 1

=> x + x = 12 - 1

=> 2x = 11

=> x = 5,5

f) 14 + 4x = 3x + 20

=> 4x - 3x = 20 - 14

=> x = 6