a, thiếu đề 

b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15 

<=> -10 + 5x + 4x - 12 = 10x - 15 

<=> -x = 7 <=> x = 7 

c,-7(5-x)-2(x-10)=15

<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6 

d, -4(x+1) + 89x - 3 = 24 

<=> 85x = 31 <=> x = 31/85 

e, 5(x-30-2(x+6)) = 9 

<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5 

b: =>-10+5x+4x-12=10x-15

=>9x-10x=-15+22

=>-x=7

hay x=-7

c: =>-35+7x-2x+20=15

=>5x-15=15

=>x=6

d: =>-4x-4+72x-24=24

=>68x-32=24

hay x=14/17

-10+5x +4x-12=10x -15

5x +4x -10x=10+12-15

-x =7

x=-7

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

\(\Rightarrow-10+5\chi+4\chi-12=10\chi-15\)

\(\Rightarrow-10-12+15=10\chi-5\chi-4\chi\)

\(\Rightarrow\chi=-7\)

HTDT

bn lm câu 2 hộ mk

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

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

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

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

\(-7.\left(5-x\right)-2.\left(x-10\right)=1\)

           \(-35+7x-2x+20=1\)

                                 \(-15+5x=1\)

                                                \(5x=1+15\)

                                                \(5x=16\)

                                                   \(x=\frac{16}{5}\)

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

        \(-10+5x+4x-12=10x-15\)

                              \(-22+9x=10x-15\)

                                \(9x-10x=-15+22\)

                                            \(-x=7\)

                                                \(x=-7\)

<=> -10+5x+4x-12=10x-15

<=> -22+9x=10x-15

<=> x=-7

D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24

thiếu 1 câu

A= với x = 4

= (x+1)(x+1)(x+1)

= x5−x4+x4+x3−x3+x2−x2+x−1

=x−1=4−1=3

Tương tự với các câu B,C,D

viết đề bài mà k rõ tki lm s mà lm đc bài

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

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

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

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

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

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

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

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

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

4x - 12 - 10 + 5x = 10x - 15

9x - 22               = 10x - 15

9x - 22 + 15       = 10x

9x - 7                = 10x

=> 9x - 7 -10x   = 0

-1x                   = 7

x                     =7 : -1 = -7

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