a) 1- 2x - |3 + x| =0

|3 + \(x\)| = 1 - 2\(x\)

Vì |3 + \(x\)| ≥ 0 với mọi \(x\)

nên 1 - 2\(x\) ≥ 0 suy ra 2\(x\) ≤ 1 suy ra \(x\) ≤ 1/2

Với \(x\) ≤ 1/2 ta có:

3 + \(x\) = 1 - 2\(x\) hoặc 3 + \(x\) = 2\(x\) - 1

Th1: 3 + \(x\) = 1 - 2\(x\)

\(2x\) + \(x\) = -3 + 1

3\(x\) = -2

\(x\) = -2/3

3 + \(x\) = 2\(x\) - 1

2\(x\) - \(x\) = 3 + 1

\(x\) = 4

\(x\) = 4

\(x\) = 4 (loại)

Vậy \(x\) = - 2/3

b) 3x - |2x-1| +2 = 2x

|2\(x\) - 1| = 3\(x\) - 2\(x\) + 2

|2\(x\) - 1| = \(x\) + 2

Vì |2\(x\) - 1| ≥ 0 với mọi \(x\) nên \(x\) + 2 ≥ 0

Suy ra: \(x\) ≥ -2

Với \(x\) ≥ - 2 ta có:

2\(x\) - 1 = \(x\) + 2 hoặc 2\(x\) - 1 = - \(x\) - 2

Th1: 2\(x\) - 1 = \(x\) + 2

2\(x\) - \(x\) = 1+ 2

\(x\) = 3 (nhận)

Th2: 2\(x\) - 1 = - \(x\) - 2

2\(x\) + \(x\) = - 2 + 1

3\(x\) = -1

\(x\) = - \(\frac13\)

Vậy \(x\) = 3

Đặt a=2x-1; b=3x+2

=>a+b=2x-1+3x+2=5x+1

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

=>\(a^3+b^3-\left(a+b\right)^3=0\)

=>\(\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3=0\)

=>-3ab(a+b)=0

=>(2x-1)(5x+1)(3x+2)=0

=>\(\left[\begin{array}{l}2x-1=0\\ 5x+1=0\\ 3x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=-\frac15\\ x=-\frac23\end{array}\right.\)

a,x=5 y=2

Xét 2 trường hợp :

TH1 : x > 0

2x - 1 + 2x + 3 = 4

4x +2 = 4

4x = 2

x = 1/2 

TH2 : x < 0

1 - 2x + 3 + 2x = 4

4 - ( 2x - 2x ) = 4

0x = 0

=> loại

Vậy,.......

b/ \(\left|\left|3x-1+9\right|\right|=-\left(-31\right)\)

<=> \(\left|\left|3x+8\right|\right|=31\)

<=> \(\left|3x+8\right|=31\)

<=> \(\orbr{\begin{cases}3x+8=-31\\3x+8=31\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=-39\\3x=23\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-13\\x=\frac{23}{3}\end{cases}}\)

\(M\left(x\right)=-3x^2+6x-4+2x^2-5x+4=-x^2+x\)

Đặt M(x)=0

=>-x(x-1)=0

=>x=0 hoặc x=1

\(M\left(x\right)=-x^2+x=-x\left(x-1\right)\)

Giả sử: \(M\left(x\right)=0\)

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

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

a) \(\frac{x+3}{2x-1}=\frac{2}{5}\)

\(\Leftrightarrow\)\(5.\left(x+3\right)=2.\left(2x-1\right)\)

\(\Leftrightarrow\)\(5x+15=4x-2\)

\(\Leftrightarrow\)\(5x-4x=-2-15\)

\(\Leftrightarrow\)\(x=-17\)

Vậy \(x=-17\)

A)Ta có:\(\frac{x+3}{2x-1}=\frac{2}{5}\)

=>(x+3).5 = (2x-1).2

=>5x+15 = 4x-2

Từ đây ta thực hiện phép chuyển vế

=>5x-4x=-2-15

=>x=-17

B)\(\frac{-x}{4}=\frac{-9}{x}\)

=>\(\frac{x}{-4}=\frac{-9}{x}\)

=>x.x=-9.(-4)

=>x\(^2\)=36

=>x=6 hoặc -6