`c)1/4x+2/5=7/5`

`=>1/4x=7/5-1/5=1`

`=>x=1:1/4=4`

Vậy `x=4` 

`a)2x-2/3=-3/4`

`=>2x=-3/4+2/3=-1/12`

`=>x=-1/24`

Vậy `x=-1/24`

`b)x:3/4+1/4=-2/3`

`=>x:3/4=-2/3-1/4=-11/4`

`=>x=-11/4 xx 3/4=-33/16`

Vậy `x=-33/16`

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

\(\Leftrightarrow2x-3-x+5=x+2-x+1\)

\(\Leftrightarrow x+2=3\)

hay x=1

Vậy: x=1

b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)

\(\Leftrightarrow2x-2-5x-10=-10\)

\(\Leftrightarrow-3x=-10+10+2=2\)

hay \(x=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)

 2x - 3 - x + 5 = x + 2 - x + 1

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

x + 2 = 3

x = 1

\(\left|3x+5\right|=x+1\)

TH1: \(3x+5=x+1\left(x\ge-\dfrac{5}{3}\right)\)

\(\Rightarrow3x-x=1-5\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\left(ktm\right)\)

TH2: \(3x-5=-\left(x+1\right)\left(x< -\dfrac{5}{3}\right)\)

\(\Rightarrow3x-5=-x-1\)

\(\Rightarrow3x+x=-1+5\)

\(\Rightarrow4x=4\)

\(\Rightarrow x=1\)

Vậy không có x thõa mãn

_______

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

\(\Rightarrow2x-3=2x-3\left(x\ge\dfrac{3}{2}\right)\)

\(\Rightarrow0=0\) (luôn đúng)

Nên mọi x đề thỏa mãn khi \(x\ge\dfrac{3}{2}\)

Vậy: ... 

|3x + 5| = x + 1

TH1: x ≥log ) -5/3

(1) ⇒ 3x + 5 = x + 1

3x - x = 1 - 5

2x = -4

x = -2 (loại)

*) TH2: x < -5/3

(1) ⇒ 3x + 5 = -x - 1

3x + x = -1 - 5

4x = -6

x = -3/2 (loại)

Vậy không tìm được x thỏa mãn yêu cầu

--------

|2x - 3| = 2x - 3 (2)

*) TH1: x 3/2

(2) ⇒ 2x - 3 = 2x - 3

0x = 0 (luôn đúng với mọi x ≥ 3/2)

*) TH2: x < 3/2

(2) ⇒ 2x - 3 = 3 - 2x

2x + 2x = 3 + 3

4x = 6

x = 3/2 (loại)

Vậy x ≥  3/2

. 

\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)

Bài 2: 

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

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

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

Vậy a = 2; b = 1; c = 1.

Làm rõ hơn đi bạn

a) |2x-3|+x=21

|2x-3|=21-x

\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)

TH1: 2x-3=21-x

2x-x=21+3

x=24

TH2: 2x-3=-(21-x)

2x-3 = -21+x

2x-x=-21+3

x=-18

Vậy x \(\varepsilon\){-18;24}