\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)

= x^2(X-1) - 4(x^2-2x+1)

=x^2(x-1)-4(x-1)^2

=(x-1)(x^2-4x+4)

=(x-1)(x-2)^2

x²+x- y²+y

=x²-y²+ x+y

=(x-y)(x+y)+(x+y)

=(x-y+1)(x+y)

h nhé

cảm ơn

x²+x-y²+y

=x²-y²+x+y

=(x-y)(x+y)+(x+y)

=(x+y)(x-y+1)

=> 2      [   \(P\left(x\right)+Q\left(x\right)+G\left(x\right)\)]= \(x^3+6x^2+5x-4+2x^2+5x^2-x-3-x^3+3x^2-6x+5=16x^2-2x-12\)

=>\(P\left(x\right)+Q\left(x\right)+G\left(x\right)=8x^2-x-6\)

=> \(G\left(x\right)=2x^2-x^3-6x-2\), \(P\left(x\right)=x^2-3\), \(Q\left(x\right)=x^3+5x^2+5x-1\)

đúng cái đi

Bài làm :

\(a,2.x-3=185\)

\(2.x=185+3\)

\(2.x=188\)

\(x=94\)

\(b,357-\left(2.x-23\right)=122\)

\(2.x-23=357-122\)

\(2.x-23=235\)

\(2.x=258\)

\(x=129\)

\(c,152+\left(x+231\right):2=358\)

\(\left(x+231\right):2=358-152\)

\(\left(x+231\right):2=206\)

\(x+231=412\)

\(x=181\)

Học tốt nhé

x^2 +5x^2-7x+15x-21=9

6x^2 +8x -12 =0

2(x^2 +4x -6) =0

x1= -2+căn 10

x2= -2-căn 10

      x^2+(x+3)(5x-7)=9

<=>x^2+5x^2+8x-21-9=0

<=>6x^2+8x-30=0

<=>2(3x^2+4x-15)=0

<=>3x^2+4x-15=0

<=>3x^2-5x+9x-15=0

<=>x(3x-5)+3(3x-5)=0

<=>(x+3)(3x-5)=0

<=>x+3=0 hoặc 3x-5=0

+)x+3=0<=>x=-3

+)3x-5=0<=>3x=5<=>x=5/3

A=|x - 2009| + |x - 2010| + |x - 2011| 

*TH1: Xét x ≤ 2009 ; khi đó 

. A = 2009 - x + 2010 - x + 2011 -x 

. A = 6030 - 3x 

có x ≤ 2009 --> -x ≥ -2009 --> -3x ≥ -6027 --> 6030 - 3x ≥ 3 

Dấu " = " <=> x = 2009 

--> Amin = 3 <=> x = 2009 

*TH2 : Xét 2009 < x ≤ 2010 ; ta có 

. A = x - 2009 + 2010 - x + 2011 - x 

. A = 2012 - x 

có x ≤ 2010 --> -x ≥ -2010 --> 2012 - x ≥ 2 

--> Amin = 2 <=> x = 2010 

*TH3 : Xét 2010 < x < 2011 ; ta có : 

. A = x - 2009 + x - 2010 + 2011 - x 

. A = x - 8 > 2010 - 8 = 2002 --> không có min 

*TH4 : Xét x ≥ 2011 ; ta có : 

. A = x - 2009 + x - 2010 + x - 2011 

. A = 3x - 6030 ≥ 3.1011 - 6030 = 3 

Dấu " = " <=> xảy ra <=> x = 2011 

--> Amin = 3 <=> x = 2011 

** Kết hợp các trường hợp trên lại ta có : 

Amin = 2 <=> x = 2010 

oái,sai rồi

\(-5\left(2-x\right)+4\left(x-3\right)=10x-15.\)

\(\Leftrightarrow-10+5x+4x-12=10x-15\)

\(\Leftrightarrow5x+4x-10x=10+12-15\)

\(\Leftrightarrow-x=7\)

\(\Rightarrow x=-7\)

         \(-5\left(2-x\right)+4\left(x-3\right)=10x-15\)

\(\Leftrightarrow-10+5x+4x-12-10x+15=0\)

\(\Leftrightarrow-x-7=0\)

\(\Leftrightarrow-x=7\)

\(\Leftrightarrow x=-7\)

~HOK TỐT~

(x^2-6x+8)(x^2-8x+15)+1

=(x^2-4x-2x+8)(x^2-5x-3x+15)+1

=(x(x-4)-2(x-4))(x(x-5)-3(x-5))+1

=(x-4)(x-2)(x-5)(x-3)+1

=(x-2)(x-5)(x-3)(x-4)+1

=(x^2-7x+10)(x^2-7x+12)+1

Gọi a=x^2-7x+11, ta có

(a-1)(a+1)+1

= a² - 1 + 1

= a²

= (x² - 7x + 11)²