[ 2314 - ( 346 + 2x ) ] + 124 = 2036

[ 2314 - ( 346 + 2x ) ] = 2036 - 124

[ 2314 - ( 346 + 2x ) ] = 1912

346 + 2x = 2314 - 1912

316 + 2x = 402

2x = 402 - 316

2x = 86

x = 86 : 2 

x = 43

[2314-(346+2x)]+124=2036

[2314-(346+2x)]=2036-134

[2314-(346+2x)]=1902

346+2x=2314-1902

346+2x=412

2x=412-346

2x=66

x=66/2

x=33

hok tốt

11: Ta có: \(\left(x+3\right)^3=125\)

\(\Leftrightarrow x+3=5\)

hay x=2

12: Ta có: \(\left(2x\right)^4=16\)

\(\Leftrightarrow x^4=1\)

hay \(x\in\left\{1;-1\right\}\)

còn câu 13 -> 20 nữa ạ.

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)

<=> \(\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=0\)

<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

<=> \(x-2163=0\)(vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)

<=> \(x=2163\)

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)

<=> \(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}-3=0\)

<=> \(\left(\frac{x-28-124}{2011}-1\right)+\left(\frac{x-124-2011}{28}-1\right)+\left(\frac{x-2011-28}{124}-1\right)=0\)

<=> \(\frac{x-28-124-2011}{2011}+\frac{x-124-2011-28}{28}+\frac{x-2011-28-124}{124}=0\)

<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

Vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)

=> \(x-2163=0\)

=> \(x=2163\)

 câu này bó quá ... ko thể giải được

Bài làm

(4x - 124) - 316 = 144

 4x - 124            = 144 + 316

 4x - 124            = 460

 4x                     = 460 - 124 

 4x                     = 336

   x                     = 336 : 4

   x                     = 84

Vậy x = 84

B1: 

a, 25-x=15 

x=25-15  =10

b,9+2.x=3^7 : 3^4

9+2.x = 3^3

2.x=27-9

2.x=18

x=18:2=9

2:

a: 15x-9x+2x=72

=>17x-9x=72

=>8x=72

=>x=9

b:Sửa đề: 96-7*(x+1)=12^4:12^3

=>96-7(x+1)=12

=>7*(x+1)=84

=>x+1=12

=>x=11

c: 2^x*4=128

=>2^x=32

=>x=5

1:

c: 41-(2x-5)=18

=>2x-5=41-18=23

=>2x=28

=>x=13

d: \(2^{x-2}=11\)

mà x nguyên

nên \(x\in\varnothing\)

a) 4x+6=3x-4

<=>4x-3x=-4-6

<=>x=-10

a)Tìm x:

4x+6=3x-4

4x-3x=-4-6

       x=-10

Vậy x=-10

c)Tính nhanh:

2315.(-2314+1)-2314.(1-2315)

=2315.(-2314)+2315-2314-2314.2315

=2315.(-2314+1-2314)-2314

=2315.(-4627)-2314

=-10711505-2314

=-10713819

/2x-5/+x=2
2x-5+x=2
3x=-3
x=-1
 

nếu 2x-5 lớn hơn hoặc bằng 0 thì x lớn hơn hoặc bằng 3 

khi đó /2x-5/+x

=>2x+x-5=2

=>3x-5=2

=>3x=5+2=7 (loại vì 7 không chia hết cho 3) n

nếu 2x-5 bé hơn 0 thì x bé hơn 3 

khi đó /2x-5/+x=2

=>5-2x+x=2

=>5-x=3 (loại vì theo điều kiện x bé hơn 3)

vậy không có x thỏa mãn

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}\)\(=3\)

\(\Leftrightarrow\frac{x-152}{2011}+\frac{x-2135}{28}+\frac{x-2039}{124}=3\)\(\Leftrightarrow\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=3-1-1-1\)

\(\Leftrightarrow\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

\(\Leftrightarrow\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

        Vì \(\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)>0\)

  \(\Rightarrow x-2163=0\)

\(\Leftrightarrow x=2163\)

          Vậy \(x=2163\)

Bài làm:

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

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\left(x^2-x+1\right)-\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\left(x^2-x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)\left(2x-5\right)=0\)

GPT này ra ta được: \(x\in\left\{-1;0;1;\frac{5}{2}\right\}\)

\(\left(2x-5\right)\left(x^3+1\right)-\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow2x^4+2x-5x^3-5-\left(2x^2+2x-5x-5\right)=0\)

\(\Leftrightarrow2x^4+2x-5x^3-5-2x^2-2x+5x+5=0\)

\(\Leftrightarrow2x^4+5x-5x^3-2x^2=0\)

\(\Leftrightarrow x\left(2x^3+5-5x^2-2x\right)=0\)

\(\Leftrightarrow x=0;\frac{5}{2};\pm1\)