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

\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)

\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)

\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)

\(-\frac{3}{2}x=-\frac{23}{4}\)

\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)

\(x=\frac{23}{6}\)

b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)

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

\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)

\(x=\frac{4}{5}\)

c) \(\left(2x-1\right).\left(3x+12\right)=0\)

\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)

bài 1

A)37/60

B)13/45

C)1/4

bài 2

a)42/25:y/5=5/6

y=10,08

b)27/y:9/4=3/7

y=28

1 , 10

2 , 8

3 , 5

4 , 2

5 , 1

6 ,

1/ `|x|=10<=> x=\pm 10`

2/ `|x-8|=0<=>x-8=0<=>x=8`

3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`

4/ `|x+1|=3`

$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$

5/ `15-x=16-(14-42)`

`<=>15-x=16+28`

`<=>15-x=44`

`<=>x=-29`

6/ `210-(x-12)=168`

`<=>210-x+12=168`

`<=>222-x=168`

`<=>x=54`

1.

\(\left|x\right|=10\Leftrightarrow x=\pm10\)

2.

\(\left|x-8\right|=0\Leftrightarrow x-8=0\Leftrightarrow x=8\)

3.

\(7+\left|x\right|=12\Leftrightarrow\left|x\right|=5\Leftrightarrow x=\pm5\)

a)\(x\left(x-3\right)-2x+6=0\)

\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)

\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)

\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)

a) \(x\left(x-3\right)-2x+6=0\)

\(x\left(x-3\right)-2\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)

\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)

\(33-39x=0\)

\(3\left(11-13x\right)=0\)

\(11-13x=0\)

\(13x=11\)

\(x=\frac{11}{13}\)

lớp11

Số nguyên âm nhỏ nhất có 2 chữ số là:-99

Ta có:x-7=-99

            x=-99+7

            x=-92

b, x-y=42

Cậu cho đề bài thế thì ai làm đc

96 - 3(x + 1) = 42

<=> 96 - 3x - 3 = 42

<=> 96 - 3 - 42 = 3x

<=> 3x = 51

<=> x = 17

\(96-3\left(x+1\right)=42\)

\(3\left(x+1\right)=96-42=54\)

\(x+1=54:3=18\)

\(x=18-1=17\)

Bài 1: 

Ta có:

\(y-x=25\Rightarrow y=25+x\)

Mà \(7x=4y\Rightarrow7x=4\cdot\left(25+x\right)\)

\(7x=100+4x\)

\(\Rightarrow7x-4x=100\)

\(3x=100\)

\(x=\frac{100}{3}\)

bài 1 :

Ta có: 7x=4y ⇔ x/4=y/7

áp dụng tính chất dãy tỉ số bằng nhau ta có 

x/4=y/7=(y-x)/(7-4)=100/3

⇒x= 4 x 100/3=400/3 ; y = 7 x 100/3=700/3

bài 2 

ta có x/5 = y/6 ⇔ x/20=y/24

         y/8 = z/7 ⇔ y/24=z/21

⇒x/20=y/24=z/21

ADTCDTSBN(bài 1 có)

x/20=y/24=z/21=(x+y)/(20+24)=69/48=23/16

⇒x= 20 x 23/16 = 115/4

   y= 24x 23/16=138/2

   z=21x23/16=483/16

bài 3

x:y:z=3:8:5  ⇔ x/3=y/8=z/5

ADTCDTSBN

x/3=y/8=z/5=(3x+y-4z)/(9+8-10)=14/7=2

⇒x = 2x3 = 6 ; y= 2x8=`16; z=2x5=10

Bài 1 :

a) 7^2x-1 = 343

=> 7^2x-1 = 7³

=> 2x - 1 = 3 => 2x = 4 => x = 2

b) (7x - 11)³ = 2⁵.3² + 200

=> (7x - 11)³ = 32.9 + 200

=> (7x - 11)³ = 488

xem kĩ lại đề này :vvv

c) 174 - (2x - 1)² = 5³

=> (2x - 1)² = 174 - 5³

=> (2x - 1)² = 174 - 125 = 49

=> (2x - 1)² = (\(\pm\)7)²

=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)

Bài 2 :

a) x⁵ = 32 => x⁵ = 2⁵ => x = 2

b) (x + 2)³ = 27

=> (x + 2)³ = 3³

=> x + 2 = 3 => x = 3 - 2 = 1

c) (x - 1)⁴ = 16

=> (x - 1)⁴ = 2⁴

=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)

d) (x - 1)⁸ = (x - 1)⁶

=> (x - 1)⁸ - (x - 1)⁶ = 0

=> (x - 1)⁶ [(x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)

+) x - 1 = 1 => x = 2 ( tm)

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

Vậy x = 1,x = 2,x = 0

     Bài 1:

\(x\) \(\times\) \(\dfrac{15}{16}\) - \(x\) \(\times\) \(\dfrac{4}{16}\) = 2

\(x\) \(\times\) (\(\dfrac{15}{16}\) - \(\dfrac{4}{16}\)) = 2

\(x\) \(\times\) \(\dfrac{11}{16}\) = 2

\(x\) = 2 : \(\dfrac{11}{16}\)

\(x\) = 2 x \(\dfrac{16}{11}\)

\(x\) = \(\dfrac{32}{11}\)

Bài 2: 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{x\left(x+1\right):2}\) = 1 : \(\dfrac{2011}{2012}\)

1 + 2\(\times\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times6}\) + \(\dfrac{1}{2\times10}\) + ... + \(\dfrac{2}{2\times x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\)(\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\) (\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + ... + \(\dfrac{1}{x\times\left(x+1\right)}\)) = 1 + \(\dfrac{1}{2011}\)

1 + 2\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

 1 + 2 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

  1 + 1 - \(\dfrac{2}{x+1}\) = 1 + \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) = 1 + 1 - 1  - \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) =  2 - 1 - \(\dfrac{1}{2011}\)

  \(\dfrac{2}{x+1}\)  = 1 - \(\dfrac{1}{2011}\)

     \(\dfrac{2}{x+1}\) = \(\dfrac{2010}{2011}\)

       \(x\) + 1 = 2 : \(\dfrac{2010}{2011}\)

        \(x\) + 1 = \(\dfrac{2011}{1005}\)

         \(x\) = \(\dfrac{2011}{1005}\) - 1  = \(\dfrac{1006}{1005}\)(loại vì \(\dfrac{1006}{1005}\) không phải là số tự nhiên)

Vậy không có giá trị nào của \(x\) là số tự nhiên thỏa mãn đề bài. 

Bài 3:

\(\dfrac{x}{16}\) \(\times\) (2017 - 1) = 2

\(\dfrac{x}{16}\) \(\times\) 2016 = 2

 \(\dfrac{x}{16}\)             = 2 : 2016

 \(\dfrac{x}{16}\)            = \(\dfrac{1}{1008}\)

   \(x\)             = \(\dfrac{1}{1008}\) x 16

  \(x\) = \(\dfrac{1}{63}\)