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16 tháng 4 2017

8(3x-2)-10x=2(4-7x)+15

\(\Leftrightarrow24x-16-10x=8-14x+15\)

\(\Leftrightarrow14x-16=-14x+23\)

\(\Leftrightarrow14x+14x=23+16\)

\(\Leftrightarrow28x=39\)

\(\Leftrightarrow x=\frac{39}{28}\)

16 tháng 4 2017

8(3x-2)-10x = 2(4-7x)+15

24x - 16 - 10x = 8 - 14x + 15

14x - 16 = 23 - 14x

14x + 14x = 23 + 16

28x = 39

x = 39:28

x = 39/28

11 tháng 4 2019

a. Rút gọn đa thức và sắp xếp theo thứ tự giảm dần của biến..

\(A\left(x\right)=13x^4+3x^2+15x+7x^2-10x^4-7x-6-8x+15\)

\(=\left(13x^4-10x^4\right)+\left(3x^2+7x^2\right)+\left(15x-7x-8x\right)+\left(15-6\right)\)

\(=3x^4+10x^2+9.\)

\(B\left(x\right)=5x^4+10-5x^2-18+3x-10x^2-3x-4x^4\)

\(=\left(5x^4-4x^4\right)+\left(-5x^2-3x^2\right)+\left(3x-3x\right)+\left(10-18\right)\)

\(=x^4-8x^2-8\)

b. Tính M = A(x) + B(x) ; N = A(x) - B(x)

\(M=A\left(x\right)+B\left(x\right)=\left(3x^4+10x^2+9\right)+\left(x^4-8x^2-8\right)\)

\(=\left(3x^4+x^4\right)+\left(10x^2-8x^2\right)+\left(10-8\right)\)

\(=4x^4+2x^2+2\)

\(N=A\left(x\right)-B\left(x\right)=\left(3x^4+10x^2+9\right)-\left(x^4-8x^2-8\right)\)

\(=3x^4+10x^2+9-x^4+8x^2+8\)

\(=\left(3x^4-x^4\right)+\left(10x^2+8x^2\right)+\left(9+8\right)\)

\(=2x^4+18x^2+17\)

27 tháng 11 2018

\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)

\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)

\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)

\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)

\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)

b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1

26 tháng 7 2023

a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)

23 tháng 11 2018

a.x2+5x+4

=>x2+4x+x+4

=>x(x+4)+x+4

=>(x+4)(x+1)

b. x2-7x=6

=>x2-x-6x=6

=>x(x-1)-6(x-1)

=>(x-1)(x-6)

c.x2-6x+8

=>x2-2x-4x+8

=>x(x-2)-4(x-2)

=>(x-2)(x-4)

d.x2-10x+21

=>x2-3x-7x+21

=>x(x-3)-7(x-3)

=>(x-3)(x-7)

e.Bạn xem lại í e đi xem có đúng đề không?

g.x2+2x-15

=>x2+5x-3x-15

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

=>(x-5)(x-3)

h.x2-3x-28

=>x2+4x-7x-28

=>x(x+4)-7(x+4)

=>(x+4)(x-7)

Nhớ tick cho mình nhé

23 tháng 11 2018

1, \(x^2+5x+4\)

\(=x^2+x+4x+4\)

\(=x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right).\left(x+4\right)\)

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

\(=x^2-x-6x+6\)

\(=x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(x+1\right)\left(x-6\right)\)

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

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

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

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

4, \(x^2-10x+21\)

\(=x^2-3x-7x+21\)

\(=x\left(x-3\right)-7\left(x-3\right)\)

\(=\left(x-3\right)\left(x-7\right)\)

5, \(x^2-2x+8\)

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

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

\(=x\left(x+2\right)+4\left(x+2\right)\)

\(=\left(x+2\right)\left(x+4\right)\)

6, \(x^2+2x-15\)

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

\(=x\left(x+5\right)-3\left(x+5\right)\)

\(=\left(x+5\right)\left(x-3\right)\)

7, \(x^2-3x-28\)

\(=x^2+4x-7x-28\)

\(=x\left(x+4\right)-7\left(x+4\right)\)

\(=\left(x+4\right)\left(x-7\right)\)

\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)

=>\(x^2+x+1=4x\)

=>\(x^2-3x+1=0\)

\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)

\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)

\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)

\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)

\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)

\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)