|3x-2|=x+2

\(\orbr{\begin{cases}3x-2=x+2\\3x-2=-x-2\end{cases}}\)

\(\orbr{\begin{cases}3x-x=2+2\\3x+x=-2+2\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\4x=0\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x\in\varnothing\end{cases}}\)

VẬy x = 2

3x-2=x+2    hoặc        3x-2 =-x-2

2x=4                            4x=0

x=2                                 x=0

       K!

\(a)\)

\(\left(x^2-2x-1\right)^2\)

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

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

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

\(c)\)

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

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

\(=x^7+x^3+x^5+x+x^6+x^2+x^4+1\)

\(=x^7+x^6+x^5+x^4+x^3+x^2+x+1\)

\(e)\)

\(\left(m^2+2m-3\right)^2\)

\(=m^4+4m^2+9+2.m^2.2m-2.2m.3-2.m^2.3\)

\(=m^4+4m^3-2m^2-12m+9\)

\(d)\)

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

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

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

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

\(=3^8-1\)

\(=6560\)

b16

a, -(2x)^2+(3y)^2=(3y)^2-(2x)^2=(3y-2x)(3y+2x)

-4x² + 9y² = ( 3y - 2x )( 3y + 2x )

( x + 1 )³ - ( 2 - x )³ = ( 2x - 1 )( x² - x + 7 )

8 + ( 4x - 3 )³ = ( 4x - 1 )( 16x² - 32x + 19 )

81 - ( 9 - x² )² = -x²( x² - 18 )

( x + y + z + t )( x + y - z - t ) = ( x + y )² - ( z + t )² = x² + y² - z² - t² + 2xy - 2zt

( x - y + z - t )( x - y - z + t ) = ( x - y )² - ( z - t )² = x² + y² - z² - t² - 2xy + 2zt

mấy ý kia khai triển lâu vl nên thôi:(

a,x-1          b, x^2-x-1

c,x^2-2x-4     d,x+2

e, 1           g, 16x

h, x^2+x+1        i, 1

a, \(\left(x+y+x\right)^2-2\left(x+y+x\right)\left(y+z\right)+\left(y+z\right)^2\)

\(=\left(x+y+x-y-z\right)^2=\left(2x-z\right)^2\)

b, \(\left(x+y+x\right)^2-\left(y+z\right)^2=\left(2x+y\right)^2-\left(y+z\right)^2\)

\(=\left(2x+y-y-z\right)\left(2x+y+y+z\right)=\left(2x-z\right)\left(2x+2y+z\right)\)

c, \(\left(x+3\right)^2+4\left(x+3\right)+4=\left(x+3+2\right)^2=\left(x+5\right)^2\)

d, \(25+10\left(x+1\right)+\left(x+1\right)^2=\left(5+x+1\right)^2=\left(x+6\right)^2\)

e, \(\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2=\left(x+2+x-2\right)^2=\left(2x\right)^2\)

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

\(=\left(x-3-x-3\right)^2=\left(-6\right)^2=6^2\)

b10

a, 25^2 -15^2= (25-15)(25+15)= 10. 40=400

b,205^2-95^2=(205-95)(205+95)=33000

c,36^2-14^2=(36-14)(36+14)=1100

d, 950^2-850^2=(950-850)(950+850)=180000

\(a)\)

\(1-5x\le x^2-4\)

\(\Leftrightarrow x^2-4+5x-1\ge0\)

\(\Leftrightarrow x^2+5x-5\ge0\)

\(\Leftrightarrow x\le\frac{\left(-5-\sqrt{45}\right)}{2}\)hoặc \(x\ge\frac{\left(-5+\sqrt{45}\right)}{2}\)

\(c)\)

\(3x^2-6x+7\)

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

\(=3\left(x-1\right)^2+4>0\)(Vô lý)

=> Bất phương trình vô nghiệm

\(d)\)

\(\frac{4-x}{x-9}>2\)

\(\Leftrightarrow\frac{\left(4-x\right)}{\left(x-9\right)}-2>0\)

\(\Leftrightarrow\frac{\left(-3x+22\right)}{x-9}>0\)

\(\Leftrightarrow\frac{22}{3}< x< 9\)

Bổ sung b)

2/7x-4 >1 
<=> 2/( 7x - 4) - 1 > 0
<=> [ 2 - ( 7x -4)]/( 7x - 4) > 0
<=> ( 6-7x)/( 7x -4) > 0
<=> ( 7x - 6).( 7x - 4) < 0
<=> 4/7 < x < 6/7