a, 3 - 2 | 5x - 4 | = -11

2|5x - 4| = 14

|5x - 4| = 7

Th1: 5x -4 =7

5x = 11

x= 11/5

Th2:

5x -4 =-7

5x = -3

x= -3/5

a) => 2/5x-4/=14

   =>   /5x-4/=7

  => 5x-4=7 hoac 5x-4=-7

       x=11/5         x=-3/5

h/ Với mọi x, y ta có :

\(\left\{{}\begin{matrix}\left|x-0,5\right|\ge0\\\left|x+y-17\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|x-0,5\right|+\left|x+y-17\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-0,5\right|=0\\\left|x+y-17\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-0,5=0\\x+y-17=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=16,5\end{matrix}\right.\)

Vaayj...

m/ \(\left(5-x\right)+\left(3x-\frac{1}{4}\right)>0\)

\(\Leftrightarrow5-x+3x-\frac{1}{4}>0\)

\(\Leftrightarrow2x-4,75>0\)

\(\Leftrightarrow x>2,375\)

Vậy...

q/ \(5^{3x-1}=625\)

\(\Leftrightarrow5^{3x-1}=5^4\)

\(\Leftrightarrow3x-1=4\Leftrightarrow x=\frac{5}{3}\)

Vậy..

(3x+2)(x-1)=0

vi.(3x+2)(x-1)=0

suy ra3x+2=0 hoacx-1=0

với3x+2=0

3x=-2

x=-2/3

vớix-1=0

x=1

3x^2 - 3x + 2x - 2 = 0

3x^2 - x - 2 = 0

3x^2 - 3x + 2x -2 = 0

3x(x - 1) + 2(x - 1) = 0

(x - 1) * (3x + 2) =0

x - 1 = 0 hoặc 3x + 2 =0

x = 1      hoặc x = -2/3

a) Ta có:

f(0) = -2.0³ + 3.0² - 0 + 5 = 0 + 0 - 0 + 5 = 5

g(-1) = 2.(-1)³ - 2.(-1)² + (-1) - 9 = -2 - 2 - 1 - 9 = -14

b) f(x) + g(x) = (-2x³ + 3x² - x + 5) + (2x³ - 2x² + x - 9)

                   = -2x³ + 3x² - x + 5 + 2x³ - 2x² + x - 9

                  = (-2x³ + 2x³) + (3x² - 2x²) - (x - x) + (5 - 9)

                 = x² - 4

f(x) - g(x) = (-2x³ + 3x² - x + 5) - (2x³ - 2x² + x - 9)

               = -2x³ + 3x² - x + 5 - 2x³ + 2x² - x + 9

              = -(2x³ + 2x³) + (3x² + 2x²) - (x + x) + (5 + 9)

             = -4x³ + 5x² - 2x + 14

a) 1- 2x - |3 + x| =0

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

Vì |3 + \(x\)| ≥ 0 với mọi \(x\)

nên 1 - 2\(x\) ≥ 0 suy ra 2\(x\) ≤ 1 suy ra \(x\) ≤ 1/2

Với \(x\) ≤ 1/2 ta có:

3 + \(x\) = 1 - 2\(x\) hoặc 3 + \(x\) = 2\(x\) - 1

Th1: 3 + \(x\) = 1 - 2\(x\)

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

3\(x\) = -2

\(x\) = -2/3

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

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

\(x\) = 4

\(x\) = 4

\(x\) = 4 (loại)

Vậy \(x\) = - 2/3

b) 3x - |2x-1| +2 = 2x

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

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

Vì |2\(x\) - 1| ≥ 0 với mọi \(x\) nên \(x\) + 2 ≥ 0

Suy ra: \(x\) ≥ -2

Với \(x\) ≥ - 2 ta có:

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

Th1: 2\(x\) - 1 = \(x\) + 2

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

\(x\) = 3 (nhận)

Th2: 2\(x\) - 1 = - \(x\) - 2

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

3\(x\) = -1

\(x\) = - \(\frac13\)

Vậy \(x\) = 3

ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

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

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

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

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