cảm ơn bn nhìu nha

d

a: \(P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left|3x-2\right|\)

\(=\dfrac{1}{2}-3x+\dfrac{1}{8}-2\left|3x-2\right|\)

\(=-3x+\dfrac{5}{8}-2\left|3x-2\right|\)

Vì x>=2/3 nên 

\(P=-3x+\dfrac{5}{8}-2\left(3x-2\right)\)

\(=-3x+\dfrac{5}{8}-6x+4=-9x+\dfrac{37}{8}\)

b: x<2/3 nên \(P=-3x+\dfrac{5}{8}-2\left(2-3x\right)\)

\(=-3x+\dfrac{5}{8}-4+6x=3x-\dfrac{27}{8}\)

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

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

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

thx

Bài 2: 

3x + 2(5 - x) = 0

<=> 3x + 10 - 2x = 0

<=> x + 10 = 0

<=> x = 0 - 10

<=> x = -10

=> x = -10

Bài 3: 

6(3q + 4q) - 8(5p - q) + (p - q)

= 6.3p + 6.4q - 8.5p - (-8).q + p - q

= 18p + 24q - 40p + 8q + p - q

= (18p - 40p + p) + (24q + 8q - q)

= -21p + 31q

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

a)A=2+x-3x+1

A=3-2x

b)0,4x-4x+12

=12-3,6x