ta có:

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !

Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html

Mình cảm ơn trước nhaa

Bài 1:

\(a.5^5-5^4+5^3\)

\(=5^3.5^2-5^3.5+5^3.1\)

\(=5^3\left(5^2-5+1\right)\)

\(=5^3.21\)

\(=5^3.3.7⋮7\)

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Bài 2:

\(a.32< 2^n< 128\)

\(\Rightarrow2^5< 2^n< 2^7\)

\(\Rightarrow n=2\)

\(b.9.27\le3^n\le243\)

\(\Rightarrow3^2.3^3\le3^n\le3^5\)

\(\Rightarrow3^5\le3^n\le3^5\)

\(\Rightarrow n=5\)

Giải:

1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\dfrac{55}{24}\)

\(=\dfrac{-19}{8}\)

2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)

\(=\dfrac{5}{12}\)

3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)

\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)

\(=-\dfrac{67}{120}\)

4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)

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

\(=-\dfrac{43}{30}\)

5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)

\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)

\(=\dfrac{3}{20}\)

6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)

\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)

\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)

\(=8+\dfrac{5}{8}\)

\(=\dfrac{69}{8}\)

7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+6+1\right)\)

\(=-\dfrac{1}{4}.20=-5\)

8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)

\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)

\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)

\(=-7\left(6+1\right)\)

\(=-7.7=-49\)

Vậy ...

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

Đây là những bài cơ bản mà bạn!

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này

Bài 1:

a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)

\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)

\(\Leftrightarrow-5=0\)(vl)

Vậy: \(x\in\varnothing\)

b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)

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

hay x=1

Vậy: x=1

c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)

\(\Leftrightarrow2x-72=0\)

\(\Leftrightarrow2\left(x-36\right)=0\)

mà 2>0

nên x-36=0

hay x=36

Vậy: x=36

d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)

\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)

\(\Leftrightarrow120x+36=56-64x\)

\(\Leftrightarrow120x+36-56+64x=0\)

\(\Leftrightarrow184x-20=0\)

\(\Leftrightarrow184x=20\)

hay \(x=\frac{5}{46}\)

Vậy: \(x=\frac{5}{46}\)

e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)

\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)

\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)

\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)

\(\Leftrightarrow-23x+29=0\)

\(\Leftrightarrow-23x=-29\)

hay \(x=\frac{29}{23}\)

Vậy: \(x=\frac{29}{23}\)

f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)

\(\Leftrightarrow2x+8-10x-50-25=0\)

\(\Leftrightarrow-8x-67=0\)

\(\Leftrightarrow-8x=67\)

hay \(x=\frac{-67}{8}\)

Vậy: \(x=\frac{-67}{8}\)

g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)

\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)

\(\Leftrightarrow10-5x-8x-8+12x-30=0\)

\(\Leftrightarrow-x-28=0\)

\(\Leftrightarrow-x=28\)

hay x=-28

Vậy: x=-28

h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)

\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)

\(\Leftrightarrow0x=0\)

Vậy: \(x\in R\)

Bài 2:

a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)

b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)

c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)

\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: Tập nghiệm S={-3}

d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)

\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)

\(\Leftrightarrow12-7x=0\)

\(\Leftrightarrow7x=12\)

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

Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)

e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x

\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow31x=1\)

hay \(x=\frac{1}{31}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)

\( a)\dfrac{{x - 3}}{5} = 6 - \dfrac{{1 - 2x}}{2}\\ \Leftrightarrow 2\left( {x - 3} \right) = 60 - 5\left( {1 - 2x} \right)\\ \Leftrightarrow 2x - 6 = 60 - 5 + 10x\\ \Leftrightarrow 8x = - 61\\ \Leftrightarrow x = - \dfrac{{61}}{8}\\ b)\dfrac{{3x - 2}}{6} - 5 = \dfrac{{3 - 2\left( {x + 7} \right)}}{4}\\ \Leftrightarrow 2\left( {3x - 2} \right) - 60 = 3\left( { - 11 - 2x} \right)\\ \Leftrightarrow 6x - 4 - 60 = - 33 - 6x\\ \Leftrightarrow 12x = 31\\ \Leftrightarrow x = \dfrac{{31}}{{12}} \)

\(a.\frac{x-3}{5}=6-\frac{1-2x}{2}\\\Leftrightarrow \frac{2\left(x-3\right)}{10}=\frac{60}{10}-\frac{5\left(1-2x\right)}{10}\\ \Leftrightarrow2\left(x-3\right)=60-5\left(1-2x\right)\\\Leftrightarrow 2x-6=60-5+10x\\\Leftrightarrow 2x-10x=6+60-5\\\Leftrightarrow -8x=61\\ \Leftrightarrow x=-\frac{61}{8}\)

Vậy nghiệm của phương trình trên là \(-\frac{61}{8}\)