P=5.(x+1)-\(\left|x-4\right|\)

P=5.(x+1)-(x-4)

P=5.\(\left[\left(x+1\right)-\left(x-4\right)\right]\)

P=5.(-5)=-25

Q=\(\left|3x-2\right|-3.\left(x+1\right)+1\)

Q=(3x-2)-3.(x+1)+1

Q=\(\left[\left(3x-2\right)-\left(x+1\right)\right]+3+1\)

Q=\(\left(2x-1\right)+3+1\)

Q=\(\left[\left(2x-1\right)+1\right]+3\)

Q=2x+3

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

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

N=x-9

CHÚC BẠN HỌC TỐT!!

a) \(P=5.\left(x+1\right)-\left|x-4\right|\)

\(P=5.\left(x+1\right)-\left(x-4\right)\)

\(P=5x+5-\left(x-4\right)\)

\(P=5x+5-x+4\)

\(P=4x+9.\)

b) \(Q=\left|3x-2\right|-3.\left(x+1\right)+1\)

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

\(Q=\left(3x-2\right)-3x-3+1\)

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

\(Q=-4.\)

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

\(N=x-3+\left(x-5\right)-\left(x+1\right)\)

\(N=x-3+x-5-x-1\)

\(N=x-9.\)

Chúc bạn học tốt!

Ta có: |x + 5| = x + 5 khi x + 5 ≥ 0 hay x ≥ -5.

|x + 5| = -(x + 5) khi x + 5 < 0 hay x < -5.

Vậy :

+ Với x ≥ -5 thì D = 3x + 2 + x + 5 = 4x + 7.

+ Với x < -5 thì D = 3x + 2 – (x + 5) = 3x + 2 – x – 5 = 2x – 3.

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. Viết như thế này khó quan sát quá.

a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\cdot\dfrac{-\left(x-1\right)\left(x+1\right)}{2}\)

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

\(=\dfrac{2\left(x-2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2\cdot\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^2-2x+1\right)\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^3+2x^2-2x^2-4x+x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^3-3x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-x^3+3x-2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)

a: \(B=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2x-1}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{1}{2x-1}=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(2x-1\right)}=\dfrac{-4x}{2x-1}\)

b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\dfrac{-4\cdot3}{2\cdot3-1}=\dfrac{-12}{5}\)

Khi x=-3 thì \(B=\dfrac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\dfrac{12}{-7}=\dfrac{-12}{7}\)