Chọn A.

Cho  2 x - 2 - x = 0 ⇔ 2 2 x - 1 2 x = 0 ⇔ 2 2 x = 1 ⇔ x = 0

Khi đó 

Câu 1: \(I = \int_{0}^{1} \frac{x+1}{x^2+2x+5} \, dx\)

Đặt \(u=x^2+2x+5\)

=>du=(2x+2)dx=2(x+1)dx

=>\(\left(x+1\right)\cdot\left(dx\right)=\frac{du}{2}\)

Khi x=0 thì u=5

Khi x=1 thì u=8

\(I = \int_{5}^{8} \frac{\frac{du}{2}}{u} = \frac{1}{2} \ln\vert{}u\vert{} \bigg\vert{}_{5}^{8} = \frac{1}{2} (\ln 8 - \ln 5) = \frac{1}{2} \ln \left(\frac{8}{5}\right)\)

=>a=2; b=8; c=5

P=a+b=10

Câu 2:

\(2x^2-3x-5\)

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

Đặt \(\frac{3x+4}{2x^2-3x-5}=\frac{3x+4}{\left(2x-5\right)\left(x+1\right)}=\frac{A}{2x-5}+\frac{B}{x+1}\)

=>\(\frac{3x+4}{\left(2x-5\right)\left(x+1\right)}=\frac{A\left(x+1\right)+B\left(2x-5\right)}{\left(2x-5\right)\left(x+1\right)}\)

=>A(x+1)+B(2x-5)=3x+4

=>x(A+2B)+A-5B=3x+4

=>A+2B=3 và A-5B=4

=>A+2B-A+5B=3-4 và A+2B=3

=>7B=-1 và A=3-2B

=>\(B=-\frac17;A=3-2\cdot\frac{-1}{7}=3+\frac27=\frac{23}{7}\)

\(I = \int_{0}^{1} \left( -\frac{1}{7} \cdot \frac{1}{x+1} + \frac{23}{7} \cdot \frac{1}{2x-5} \right) dx\)

\(=\left[-\frac{1}{7}\ln\vert{}x+1\vert{}+\frac{23}{14}\ln\vert{}2x-5\vert{}\right]_0^1\)

\(=\left(-\frac{1}{7}\ln2+\frac{23}{14}\ln\vert{}-3\vert{}\right)-\left(-\frac{1}{7}\ln1+\frac{23}{14}\ln\vert{}-5\vert{}\right)\)

\(=-\frac{1}{7}\ln2+\frac{23}{14}\ln3-\frac{23}{14}\ln5\)

\(=\frac{23}{14}(\ln3-\ln5)-\frac{1}{7}\ln2=\frac{23}{14}\ln\left(\frac{3}{5}\right)-\frac{1}{7}\ln2\)

=>a=3; b=5; c=2

P=a-b+c

=3-5+2

=5-5

=0

Bài 1: Thực hiện phép tính

a)136 - (2 . 52 + 23 . 3)

= 136 - (104 + 69)

= 136 - 173

= -37

b) (-243) + (-12) + (+243) + (-38) + (10)

= [(-243) + (+243)] + (-12) + (-38) + (10)

= 0 + (-40)

= -40

Bài 2 : Tìm x ∈ N, biết:

a) 6 . (x-81) = 54

⇒ x - 81 = 54 : 6

⇒ x - 81 = 9

x = 81 + 9

x = 90

Vậy : x = 90

b) 18 - (x-4) = 32

⇒ x - 4 = 18 - 32

⇒ x - 4 = -14

x = -14 + 4

x = -10

a) 54 – 19i;

b) -15 + i.

Bài 1. 

a) \(\left(3+4i\right)+\left(-1+5i\right)=\left(3-1\right)+\left(4i+5i\right)=2+9i\)

b) \(\left(3-4i\right)-\left(1-5i\right)=\left(3-1\right)-\left(4i-5i\right)=2+i\)

c)\(\left(-3+4i\right)+\left(1-4i\right)=\left(-3+1\right)+\left(4i-4i\right)=-2\)

d) \(\left(3-5i\right)-\left(4+i\right)=\left(3-4\right)-\left(5i+i\right)=-1-6i\)

Bài 2. 

a) \(\left(3+4i\right)\left(-1+5i\right)=3.\left(-1\right)+4i.\left(-1\right)+3.5i+4i.5i\)

\(=-3-4i+15i-20=-23+11i\)

b) \(\left(3-5i\right)-\left(4+i\right)=\left(3-4\right)-\left(5i+i\right)=-1-6i\)

Đáp án D