nik lộn

2^x+1. 2^{2009 =} 2²⁰¹⁰

10 - 2x = 25 - 3x

c)

\(4\left(3x-4\right)-2=18\)

<=> \(12x-16-2=18\)
<=> \(12x=36\)

<=> \(x=3\)

Vậy x=3

d)

\(\left(3x-10\right):10=50\)

<=> \(3x-10=500\)

<=> \(3x=510\)

<=> x= \(170\)

Vậy x= 170

f)

\(x-\left[42+\left(-25\right)\right]=-8\)

<=> \(x-17=-8\)

<=> x= \(9\)

Vậy x=9

h)

\(x+5=20-\left(12-7\right)\)

<=> \(x+5=15\)

<=> \(x=10\)

Vậy x= 10

k)

\(\left|x-5\right|=7-\left(-3\right)\)

<=> \(\left|x-5\right|=10\)

* Với \(x>=5\) ; ta được:

\(x-5=10\)

<=> x= 15 (thoả mãn điều kiện )

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=10\)

<=> \(-x+5=10\)

<=> \(-x=5\)

<=> \(x=-5\) (thoả mãn điều kiện)

Vậy x=15 ; x= -5

i)

\(\left|x-5\right|=\left|7\right|\)

<=> \(\left|x-5\right|=7\)

*Với \(x>=5\) ; ta được:

\(x-5=7\)

<=> \(x=12\) (thoả mãn)

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=7\)

<=> \(-x=2\)

<=> \(x=-2\) (thoả mãn)

Vậy x= 12; x= -2

m)

\(2^{x+1}.2^{2009}=2^{2010}\)

<=> \(2^{x+1+2009}=2^{2010}\)

<=> \(2^{x+2010}=2^{2010}\)

=> \(x+2010=2010\)

=> \(x=0\)

Vậy x=0

n)

\(10-2x=25-3x\)

<=>\(x=15\)

Vậy x=15

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

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

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

a) \(3x-10=2x+13\Leftrightarrow x=13+10=23\)

b) \(x+12=-5-x\Leftrightarrow2x=-17\Leftrightarrow x=-\frac{17}{2}\)

c) \(x+5=10-x\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)

d) \(6x+23=2x-12\Leftrightarrow4x=-35\Leftrightarrow x=-\frac{35}{4}\)

e) \(12-x=x+1\Leftrightarrow2x=11\Leftrightarrow x=\frac{11}{2}\)

f) \(14+4x=3x+20\Leftrightarrow x=20-14=6\)

g) \(2\left(x-1\right)+3\left(x-2\right)=x-4\Leftrightarrow2x-2+3x-6=x-4\)

\(\Leftrightarrow5x-8=x-4\Leftrightarrow4x=4\Leftrightarrow x=1\)

h ) \(3\left(4-x\right)-2\left(x-1\right)=x+20\Leftrightarrow12-3x-2x+2=x+20\)

\(\Leftrightarrow14-5x=x+20\Leftrightarrow6x=-6\Leftrightarrow x=-1\)

i ) \(4\left(2x+7\right)-3\left(3x-2\right)=24\Leftrightarrow8x+28-9x+6=24\)

\(\Leftrightarrow34-x=24\Leftrightarrow x=34-24=10\)

k) \(3\left(x-2\right)+2x=10\Leftrightarrow3x-6+2x=5x-6=10\)

\(\Leftrightarrow5x=10+6=16\Leftrightarrow x=\frac{16}{5}\)

Tự KL cho các phần

a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)

\(< =>8x^2+12x-8x^2+x=5x+10\)

\(< =>13x=5x+10< =>8x=10\)

\(< =>x=\frac{10}{8}=\frac{5}{4}\)

b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)

\(< =>9x^2-25-9x^2+x=4\)

\(< =>x=4+29=33\)

c,\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(< =>3-100x+8x^2=8x^2+x-300\)

\(< =>x+100x=3+300\)

\(< =>101x=303< =>x=\frac{303}{101}=3\)

d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)

\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)

\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)

\(< =>\frac{30x-6x-24x+15-4}{20}=5\)

\(< =>\frac{11}{5}=5< =>11=25\)(vo li)

e,  \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(< =>\frac{25x+10}{30}-\frac{80x-10}{30}=\frac{24x+12}{30}-5\)

\(< =>\frac{24x+12+80x-10-25x-10}{30}=5\)

\(< =>79x-8=150\)

\(< =>79x=150+8< =>x=\frac{158}{79}\)

a, 2.x + 7 = 15

   2x         = 8

   x           = 4

b, 25 – 3.(6 – x) = 22

            3.(6-x)    = 3

               6-x      = 1

                  x      = 5

c, [(2x – 11) : 3 + 1].5 = 20

      (2x-11) : 3+1         = 4

       (2x-11):3              = 3

        2x-11                  = 1

        2x                       = 12

        x                         = 6

  e, 2 . 3x = 10 . 312 + 8 . 274

      6x      = 3120 + 2192

      6x      = 5312

        x      = 5312/6

g, x – 12 = (–8) + (–17)

    x  - 12 = -25

    x         = -13

Lần sau tách nhỏ nội dung câu hỏi ra nha em, chứ trả lời thế này biếng lắm '^^ Chị làm chỉ mang tính tham khảo kết quả thôi, còn cụ thể thì em tách từng bước một ra he :>

a, \(2\cdot x+7=15\)

\(\Leftrightarrow2\cdot x=8\)

\(\Leftrightarrow x=4\)

Vậy x = 4.

b, \(25-3\cdot\left(6-x\right)=22\)

\(\Leftrightarrow3\cdot\left(6-x\right)=3\)

\(\Leftrightarrow6-x=1\)

\(\Leftrightarrow x=5\)

Vậy x = 5.

c, \(\left[\left(2x-11\right):3+1\right]\cdot5=20\)

\(\Leftrightarrow\left(2x-11\right):3+1=4\)

\(\Leftrightarrow\left(2x-11\right):3=3\)

\(\Leftrightarrow2x-11=9\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy x = 10.

d, \(\left(25-2x\right)\cdot3:5-32=42\)

\(\Leftrightarrow\)\(\frac{3\cdot\left(25-2x\right)}{5}=74\)

\(\Leftrightarrow3\cdot\left(25-2x\right)=370\)

\(\Leftrightarrow25-2x=\frac{370}{3}\)

\(\Leftrightarrow2x=-\frac{295}{3}\)

\(\Leftrightarrow x\approx49\)

Vậy \(x\approx49\) .

e, \(2\cdot3x=10\cdot312+8\cdot274\)

\(\Leftrightarrow6x=5312\)

\(\Leftrightarrow x=5312:6\approx885\)

Vậy \(x\approx885\) .

g, \(x-12=\left(-8\right)+\left(-17\right)\)

\(\Leftrightarrow x-12=-25\)

\(\Leftrightarrow x=-25+12=-13\)

Vậy x = -13.

h, \(7-2x=18-3x\)

\(\Leftrightarrow-2x+3x=18-7\)

\(\Leftrightarrow x=11\)

Vậy \(x=11\) .

i, \(3\cdot\left(x+5\right)-x-11=24\)

\(\Leftrightarrow3x+15-x-11=24\)

\(\Leftrightarrow2x=24+11-15\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\) .

Thanks mn ^v^