a, \(231-\left(x-6\right)=1339:13\)

\(\Leftrightarrow231-\left(x-6\right)=103\)

\(\Leftrightarrow x-6=128\)

\(\Leftrightarrow x=134\)

Vậy .....

b, \(70-5\left(2x-3\right)=45\)

\(\Leftrightarrow5\left(2x-3\right)=25\)

\(\Leftrightarrow2x-3=5\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)

Vậy ....

c, \(156-\left(x+61\right)=82\)

\(\Leftrightarrow x+61=74\)

\(\Leftrightarrow x=13\)

Vậy ....

d, \(6\left(5x+35\right)=330\)

\(\Leftrightarrow5x+35=55\)

\(\Leftrightarrow5x=20\)

\(\Leftrightarrow x=4\)

Vậy ......

e, \(936-\left(4x+24\right)=72\)

\(\Leftrightarrow4x+24=864\)

\(\Leftrightarrow4x=840\)

\(\Leftrightarrow x=210\)

Vậy ...

231-(x-6)=1339:13

237-x=103

=>x=237-103=134

Vậy x=134

70-5(2x-3)=45

70-10x+15=45

=>-10x=45-15-70=-40

=>x=4

Vậy x=4

Các caau sau tương tự nhé bn

a, 231 - ( x - 6 ) = 1339 : 13

231 - ( x -6 ) = 103

x - 6 = 231 - 103 = 128

x = 128 + 6 = 134

b, 70 - 5 (2x -3 ) = 45

5 ( 2x -3 ) = 70 - 45 = 25

( 2x -3 ) = 25 : 5 = 5

2x = 5 + 3 = 8

x = 8 : 2 = 4

c, 156 - ( x + 61 ) = 82

x + 61 = 156 - 82 =74

x = 74 - 61

x = 13

d, 6 ( 5x + 35 ) = 330

5x + 35 = 330 : 6 = 55

5x = 55 - 35 = 20

x = 20 : 5 = 4

e, 936-(4x+ 24 ) = 72

4x + 24 = 936 - 72 = 864

4x = 864 - 24 = 840

x = 840 : 4 = 210

Chuc ban hoc tot nhe

Không có đâu bạn

ko hiểu lớp 9 lại đi làm những hệ như thế này làm j 

ý kiến cá nhân thôi

a) \(70-5\left(2x-3\right)=45\)

\(70-10x+15=45\)

\(-10x=45-70-15\)

\(-10x=-40\)

\(x=4\)

vay \(x=4\)

 b) \(156-\left(x+61\right)=82\)

\(x+61=156-82\)

\(x+61=74\)

\(x=74-61\)

\(x=13\)

vay \(x=13\)

c) \(6\left(5x+35\right)=330\)

\(30x+210=330\)

\(30x=330-210\)

\(30x=120\)

\(x=4\)

vay \(x=4\)

d) \(936-\left(4x+24\right)=72\)

\(4x+24=936-72\)

\(4x+24=864\)

\(4x=864-24\)

\(4x=840\)

\(x=210\)

vay \(x=210\)

\(70-5.\left(2x-3\right)\)\(=45\)

\(5.\left(2x-3\right)\)\(=\)\(70-45\)

\(5.\left(2x-3\right)\)\(=25\)

\(2x-3=25:5\)

\(2x-3=5\)

\(2x=8\)

\(x=8:2\)

\(x=4\)

\(156-\left(x+61\right)\)\(=82\)

\(x+61=156-82\)

\(x+61=74\)

\(x=74-61\)

\(x=13\)

\(6.\left(5x+35\right)\)\(=330\)

\(5x+35=330:6\)

\(5x+35=55\)

\(5x=55-35\)

\(5x=20\)

\(x=20:5\)

\(x=4\)

\(936-\left(4x+24\right)\)\(=72\)

\(4x+24=936-72\)

\(4x+24=864\)

\(4x=864-24\)

\(4x=840\)

\(x=840:4\)

\(x=210\)

Tìm x :

a) 70 - 5 . ( 2x - 3 ) = 45

           5 . ( 2x - 3 ) = 70 - 45

           5 . ( 2x - 3 ) = 25

                    2x - 3 = 25 : 5

                    2x - 3 = 5

                         2x = 5 + 3

                         2x = 8

                           x = 8 : 2

                           x = 4

b)    156 - ( x + 61 ) = 82

                     x + 61 = 156 - 82

                     x + 61 =  74

                             x = 74 - 61

                             x = 13

c) 6 . ( 5x + 35 ) = 330

             5x + 35 = 330 : 6

             5x + 35 = 55

                     5x = 55 - 35

                     5x = 20

                       x = 20 : 5 

                       x = 4

d) 936 - ( 4x + 24 ) = 72

                  4x + 24 = 936 - 72

                  4x + 24 = 364

                          4x = 364 - 24

                          4x = 340

                            x = 340 : 4

                            x = 85

a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

=>\(x^3-x^2-4x^2+4x+4x-4=0\)

=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2: \(x^3+3x^2=x+6\)

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

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

=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

=>\(2x+3\sqrt{x}-1=0\)

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

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

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

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b.

\(x^3+3x^2-x-6=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

\(2x+3\sqrt{x}+1=0\)

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

\(x^4+4x^2+1=3x^3+3x\)

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

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

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

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

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

a, 1000

b,17460

a) = (325 + 75) + ( 70 + 530 )

= 400 + 600

= 1000

b) = 15 x ( 76 + 24)

= 15 x 100

= 1500

a) = (325 + 75) + ( 70 + 530 )

= 400 + 600

= 1000

b) = 15 x ( 76 + 24)

= 15 x 100

= 1500

a) 70 - 5.(2x-3)=45

5.(2x-3)=70-45

10x-15 =25

10x     =25 +15

10x     =40

x        =40:10

x        = 4

b)6.(5x+35) =330

5x +35 = 330 : 6

5x +35 =55

5x = 55-35

5x = 20

x = 20 : 5

x = 4

c)936-(4.x+24)=72

4x+24 =936 - 72

4x+24 =864

4x = 864 - 24

4x = 840

x = 840 :4

x = 210

d)(x+1)+(x+3)+(x+5)

x+1+x+3+x+5=30

(x+x+x)+(1+3+5) = 30

3x + 9 =30

3x = 30-9

3x =21

x = 21 : 3

x = 7

e)Vì số lượng x bằng  số số hạng nên số lượng x là :

(100 - 2) : 2 + 1 = 50 (số x)

Tổng 2 +4+6+8+...+100 là :

(100+2) x 50 : 2 = 2550

Ta có :

50 x + 2550 = 2650

50x = 2650 -2550

50x = 100

x = 100 : 50

x = 2

Tick nghen !

a)   70 - 5 ( 2x-3 ) = 45

              5( 2x-3 ) = 25

                  2x - 3 = 25 : 5

                 2x - 3  = 5

                       2x = 8 

                         x = 8 : 2

                        x = 4

b)   6 ( 5x + 35 ) = 330

    5x + 35 = 330 : 6

    5x + 35 = 55

            5x = 55 - 35

            5x = 20

              x = 20 : 5

              x = 4 

c)  936 - ( 4x + 24 ) = 72

     4x + 24 = 936 - 72

     4x + 24 = 864

             4x = 864 - 24

             4x = 840

              x = 840 : 4

              x = 210

 d)   5( 3x + 34 ) = 515

             3x + 34 = 515 : 5

             3x + 34 = 103

                     3x = 103 -  34

                     3x = 69

                       x = 69 : 3

                      x = 23

\(a,70-5\left(2x-3\right)=45\)

\(\Leftrightarrow10x-15=70-45\)

\(\Leftrightarrow10x=25+15\)

\(\Leftrightarrow x=\frac{40}{10}=4\)

Vậy x=4

\(b,6\left(5x+35\right)=330\)

\(\Leftrightarrow5x+35=55\)

\(\Leftrightarrow5x=20\)

\(\Leftrightarrow x=4\)

vậy x=4

\(c,936-\left(4x+24\right)=72\)

\(\Leftrightarrow4x+24=936-72\)

\(\Leftrightarrow4x=864-24\)

\(\Leftrightarrow x=840:4=210\)

Vậy x=210

\(d,5\left(3x+34\right)=515\)

\(\Leftrightarrow3x+34=515:5\)

\(\Leftrightarrow3x=103-34\)

\(\Leftrightarrow x=69:3=23\)

Vậy x=23

nói chung là lên gg mà sợt;vvv

Lời giải:
$\frac{x}{10}=\frac{y}{5}; \frac{y}{2}=\frac{z}{3}$

$\frac{x}{2}=y; \frac{y}{2}=\frac{z}{3}$

$\Rightarrow \frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. Áp dụng tính chất dãy tỉ số bằng nhau:

$ \frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

$=\frac{2x}{8}=\frac{3y}{6}=\frac{4z}{12}$

$=\frac{2x-3y+4z}{8-6+12}=\frac{330}{14}=\frac{165}{7}$

$\Rightarrow x=\frac{165}{7}.4=\frac{660}{7}; y=\frac{165}{7}.2=\frac{330}{7}; z=\frac{165}{7}.3=\frac{495}{7}$

$