a) x=5397/20

b)x=5/3

a) (15 x 19 - x - 0,15) : 0,25 = 15 : 0,15

    (285 - x - 0,15) : 0,25 = 60

    (285 - x - 0,25) = 60 : 0,25

    ( 285 - x - 0,25) = 15

      285 - x = 15 + 0,25

      285 - x = 15,25

              x = 285 - 15,25 

              x = 269,75

b) \(2\frac{2}{3}-x=\frac{1}{3}.\frac{6}{2}\)

\(\frac{8}{3}-x=1\)

\(x=\frac{8}{3}-1\)

\(x=\frac{5}{3}\)

Chứng minh rằng (9^2n+1994^93)chia heets cho 5

a) \(x\cdot\frac{1}{2}+x\cdot\frac{1}{4}+x\cdot\frac{1}{8}=\frac{21}{24}\)

\(x\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=\frac{7}{8}\)

\(x\cdot\frac{7}{8}=\frac{7}{8}\)

\(\Rightarrow x=\frac{7}{8}\div\frac{7}{8}=1\)

b) \(\left(x+4\right)+\left(x+9\right)+\left(x+14\right)+.....+\left(x+44\right)+\left(x+49\right)=1430\)

\(\left(x+x+x+....+x+x\right)+\left(4+9+14+...+44+49\right)=1430\)

\(10x+265=1430\)

\(10x=1430-265\)

\(10x=1165\)

\(\Rightarrow x=\frac{1165}{10}=116,5\)

c) \(x\cdot0,25-0,5=1\)

\(x\cdot0,25=1+0,5\)

\(x\cdot0,25=1,5\)

\(\Rightarrow x=1,5\div0,25=6\)

\(\left(\frac{7}{3}\times x-0.6\right)\div3\frac{2}{5}=1\)

\(\left(\frac{7}{3}\times x-\frac{3}{5}\right)\div\frac{17}{5}=1\)

\(\frac{7}{3}\times x-\frac{3}{5}=1\times\frac{17}{5}\)

\(\frac{7}{3}\times x=\frac{17}{5}+\frac{3}{5}\)

\(\frac{7}{3}\times x=4\)

\(x=4\div\frac{7}{3}\)

\(x=4\times\frac{3}{7}\)

\(x=\frac{12}{7}\)

\(x=1\frac{5}{7}\)

   3/3 - 1/4.x=0,25
=> 1-1/4x=0,25
        1/4x=1-0,25
        1/4x=0,75
             x=0,75:1/4
             x=3

Vậy x=3
 

5.[3/4-2/3x]-13/4=-2{(5/4)²-5/2.1/4+(1/16)²

5.[3/4-2/3x]-13/4=2{25/16-5/8+1/256}

........................=2{400/256-160/250+1/256}

........................=2.241/256

........................=241/128

5.[3/4-2/3x]=241/128+13/4

5.[3/4-2/3x]=241/128+416/128

[3/4-2/3x]=657/128:5

[3/4-2/3x]=657/640

vậy 3/4-2/3x=657/640 và 3/4-2/3x=-657/640

th1:3/4-2/3x va th2:3/4-2/3x=-657/640

còn lại bạn chỉ cần tính kết quả mà dấu[..] là giá tri tuyệt đối đó

\(\begin{array}{l}a)x + 0,25 = \frac{1}{2}\\x = \frac{1}{2} - 0,25\\x = \frac{1}{2} - \frac{1}{4}\\x = \frac{2}{4} - \frac{1}{4}\\x = \frac{1}{4}\end{array}\)

Vậy \(x = \frac{1}{4}\)

\(\begin{array}{l}b)x - \left( { - \frac{5}{7}} \right) = \frac{9}{{14}}\\x = \frac{9}{{14}} + \left( { - \frac{5}{7}} \right)\\x = \frac{9}{{14}} + \left( { - \frac{{10}}{{14}}} \right)\\x = \frac{{ - 1}}{{14}}\end{array}\)

Vậy \(x = \frac{{ - 1}}{{14}}\)

\(\left(0,25-30\%.x\right).\frac{1}{3}-\frac{1}{4}=-5\frac{1}{6}\)

\(< =>\left(\frac{1}{4}-\frac{3x}{10}\right).\frac{1}{3}-\frac{1}{4}=-\frac{30}{6}\)

\(< =>\left(\frac{1}{4}-\frac{3x}{10}\right).\frac{1}{3}=-5+\frac{1}{4}=-\frac{19}{4}\)

\(< =>\frac{1}{4}-\frac{3x}{10}=\frac{-19}{4}.\frac{3}{1}=-\frac{57}{4}\)

\(< =>\frac{-3x}{10}=\frac{-57}{4}-\frac{1}{4}=-\frac{58}{4}\)

\(< =>-12x=-580\)

\(< =>12x=580< =>x=\frac{580}{12}=\frac{145}{3}\)

\(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=-5\frac{1}{6}\)

\(\left(\frac{1}{4}-\frac{3}{10}x\right)\frac{1}{3}-\frac{1}{4}=\frac{-31}{6}\)

\(\frac{1}{4}-\frac{3}{10}x=\left(\frac{-31}{6}+\frac{1}{4}\right)\div\frac{1}{3}\)

\(\frac{1}{4}-\frac{3}{10}x=\frac{-59}{4}\)

\(\frac{3}{10}x=\frac{1}{4}-\frac{-59}{4}\)

\(\frac{3}{10}x=15\Leftrightarrow x=50\)

a, \(7\left(x-1\right)+2x\left(1-x\right)=0\)

\(\Rightarrow\left(7-2x\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}\)

b, \(0,25-\left|3,5-x\right|=0\)

\(\Rightarrow\left|3,5-x\right|=2,5\)

\(\Rightarrow\orbr{\begin{cases}3,5-x=2,5\\3,5-x=-2,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

c, \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}\)

a) 7.(x-1) + 2x.(1-x) = 0

7.(x-1) - 2x.(x-1) = 0

(x-1).(7-2x) = 0

=> (x-1) = 0 => x = 1

7-2x = 0 => 2x = 7 => x = 7/2

KL:...

b) 0,25 - | 3,5-x| = 0

=> |3,5 - x| = 0,25

TH1: 3,5 - x = 0,25

x = 3,25

TH2: 3,5 - x = -0,25

x = 3,75

phần c bn dựa vào phần b mak lm nha

a) 7.(x - 1) + 2x.(1 - x) = 0

<=> -2x² + 9x - 7 = 0

<=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}\)

b) 0,25 - |3,5 - x| = 0

<=> - |3,5 - x| = 0 - 0,25

<=> - |3,5 - x| = -0,25 

<=> |3,5 - x| = 0,25

Ta xét 2 th:

Th1: |x| = 3,5 - 0,25

<=> |x| = \(\frac{13}{4}\)

Th2: |x| = 3,5 + 0,25

<=> x = \(\frac{15}{4}\)

=> \(\orbr{\begin{cases}x=\frac{13}{4}\\\frac{15}{4}\end{cases}}\)

c) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

<=> \(\left|x+\frac{3}{4}\right|=0+\frac{1}{2}\)

<=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

Ta xét 2 th:

Th1: \(x+\frac{3}{4}=\frac{1}{2}\)

<=> \(x=\frac{1}{2}-\frac{5}{4}\)

<=> \(x=-\frac{3}{4}\)

Th2: \(x+\frac{3}{4}=\frac{1}{2}\)

<=> \(x=\frac{1}{2}-\frac{3}{4}\)

<=> \(x=-\frac{1}{4}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{1}{4}\end{cases}}\)