Bài làm

( 3 + 6 + x ) . 9 = ( 2 - x ) . 2

27 + 54 + 9x = 4 - 2x

9x + 2x = 4 - 27 - 54

11x = -77

x = -7

Vậy x = -7

b) ( 6 - x ) . 3 = ( 9 + x ) . 3

18 - 3x = 27 + 3x

-3x - 3x = 27 - 18

-6x = 9

x = 9/-6

x = -3/2

Vậy x = -3/2

a,\(\text{(3+6+x) . 9 = (2-x).2}\)

\(27+54+9x=4-2x\)

\(9x+2x=4-54-27\)

\(11x=-77\)

\(\Rightarrow x=-7\)

b, \(\text{(6-x) . 3 = (9+x) . 3}\)

\(18-3x=27+3x\)

\(-3x-3x=27-18\)

\(-6x=9\)

\(\Rightarrow x=\frac{9}{-6}\)

học tốt

\(\left(3+6+x\right)9=\left(2-x\right)2\)

\(27+54+9x=4-2x\)

\(27+54+9x-4+2x=0\)

\(77+11x=0\)

\(11x=-77\)

\(x=-7\)

\(\left(6-x\right)3=\left(9+x\right)3\)

\(18-3x=27+3x\)

\(18-3x-27-3x=0\)

\(-9-6x=0\)

\(6x=-9\)

\(x=-\frac{3}{2}\)

\(\Leftrightarrow x^2-36-x^2+12x-9=9\)

\(\Leftrightarrow12x=54\)

hay x=9/2

Ta có

( x   –   6 ) ( x   +   6 )   –   ( x   +   3 ) 2   =   9     ⇔   x 2   –   36   –   ( x 2   +   6 x   +   9 )   =   9     ⇔   x 2   –   36   –   x 2   –   6 x   –   9   –   9   =   0

ó - 6x – 54 = 0 ó 6x = -54 ó x = -9

Vậy x = -9

Đáp án cần chọn là: A

a) 6 . x - 15 = 15

6 . x = 15 + 15

6 . x = 30

x = 30 : 6

x = 5

b) 2 . x + 3 = 9

2 . x = 9 - 3

2 . x = 6

x = 6 : 2

x = 3

c) 17 - 6 : x = 15

6 : x = 17 - 15

6 : x = 2

x = 6 : 2

x = 3

d) x : 3 + 12 = 14

x : 3 = 14 - 12

x : 3 = 2

x = 2 . 3

x = 6

a) 6 . x -15 = 15 

⇒ 6 .x = 15 + (-15)

⇒ 6 .x = 0

⇔ x = 0

Vậy x = 0.

b) 2.x + 3 = 9 

⇒2.x = 9 -3 

⇒  2.x = 6

⇔ x = 3

Vậy x = 3.

a) x - 3 = -6

x=(-6)+3

x=-3

b) x - (-5)=4

x=4+5

x=9

c)x - (-9) = 4-(-9)

x - (-9) =13

x =13-9

x = 4

d) 4-x=-3-(-6)

4-x=3

x=4-1

x=3

a) -3

b) 9

c)4

d)3

d. (x - 3)(x² + 3x + 9) + x(x + 2)(2 - x) = 1

<=> x³ - 9 + (x² + 2x)(2 - x) = 1

<=> x³ - 9 + 2x² - x³ + 4x - 2x² = 1

<=> 4x = 10

<=> x = \(\dfrac{10}{4}=\dfrac{5}{2}\)

d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1

\(<=> x^3-27-x(x^2-4)=1\)

\(<=> x^3-27-x^3-4x=1<=>-4x=28<=> x=-7\)

=> ptrình có tập nghiệm S={-7}

e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19

\(<=> x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+19=0\)

\(<=>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)

\(<=>12x=15<=>x=12/15 \)

=> ptrình có tập nghiệm S={12/15}

a) \(x+\dfrac{1}{2}=\dfrac{3}{5}\)

    \(x=\dfrac{3}{5}-\dfrac{1}{2}\) 

    \(x=\dfrac{6}{10}-\dfrac{5}{10}\)

    \(x=\dfrac{1}{10}\)

b) \(x\times\dfrac{6}{7}=\dfrac{5}{21}\)

    \(x=\dfrac{5}{21}:\dfrac{6}{7}\)

    \(x=\dfrac{5}{21}\times\dfrac{7}{6}\)

    \(x=\dfrac{5}{18}\)

c) \(x:\dfrac{4}{3}=\dfrac{8}{9}\)

    \(x=\dfrac{8}{9}\times\dfrac{4}{3}\)

   \(x=\dfrac{32}{27}\)

a: x=3/5-1/2=6/10-5/10=1/10

b: x=5/21:6/7=5/21x7/5=35/105=1/3

c: x=8/9x4/3=32/27

\(a.x+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{10}\)

\(b.x\times\dfrac{6}{7}=\dfrac{5}{21}\)

\(x=\dfrac{5}{21}:\dfrac{6}{7}=\dfrac{5}{18}\)

\(c.x:\dfrac{4}{3}=\dfrac{8}{9}\)

\(x=\dfrac{8}{9}\times\dfrac{4}{3}=\dfrac{32}{27}\)

Bài này cho sai đề nghiệm cực lẻ 

x^3 -9x^2 +27x -27 -(x^3 -27) +6(x^2 +2x+1) +3x^2 =-33

x^3 -9x^2 +27x -27 -x^3 +27 +6x^2 + 12x+ 6 +3x^2 =-33

39x+6=-33

39x=-39

x=-1

Vậy x=-1

a)4/9 : x/6 = 2/3

           x/6 = 4/9 : 2/3

          x/6  =   2/3 = 4/6

vậy x = 4

b)x/5 x 3/7 = 9/35

   x/5          = 9/35 : 3/7

   x/5          = 3/5

vậy x = 3