Thực hiện các phép tính sau: 3 - 4 i 1 + 2 i 1 - 2 i + 4 - 3 i
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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) \(\left(3-5i\right)-\left(4+i\right)=\left(3-4\right)-\left(5i+i\right)=-1-6i\)
a) (3 + 2i)[(2 – i) + (3 – 2i)]
= (3 + 2i)(5 – 3i) = 21 + i
b)(4−3i)+1+i2+i=(4−3i)+(1+i)(2−i)5=(4−3i)(35+15i)=(4+35)−(3−15)i=235−145i(4−3i)+1+i2+i=(4−3i)+(1+i)(2−i)5=(4−3i)(35+15i)=(4+35)−(3−15)i=235−145i
c) (1 + i)2 – (1 - i)2 = 2i – (-2i) = 4i
d) 3+i2+i−4−3i2−i=(3+i)(2−i)5−(4−3i)(2+i)5=7−i5−11−2i5=−45+15i
a) Ta có: (3-2i)(2-3i)=(3.2-2.3)+(-3.3-2.2)i=-13i
b) Ta có: (-1+i)(3+7i)=(-1.3-1.7)+(-1.7+1.3)i=-10-4i
c) Ta có: (5(4+3i)=5.4+5.3i=20+15i
d) Ta có: (-2-5i)4i=(-2.0+5.4)+(2.4-5.0)i=20-8i
\(\dfrac{1}{1-x}\)+\(\dfrac{1}{1+x}\)+\(\dfrac{2}{1+x^2}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=
=\(\dfrac{4}{1-x^4}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{8}{1-x^8}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{16}{1-x^{16}}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{32}{1-x^{32}}\)
(-1 + i)(3 + 7i) = -1.3 + (-1).7i +i.3 + i. 7i = -3 – 7i + 3i – 7 =( -3 – 7)+ ( - 7+3) i = -10 – 4i
a) 2i(3 + i)(2 + 4i) = 2i(2 + 14i) = -28 + 4i
b)
c) 3 + 2i + (6 + i)(5 + i) = 3 + 2i + 29 + 11i = 32 + 13i
d) 4 - 3i + = 4 - 3i + = 4 - 3i +
= (4 + ) - (3 + )i =
3 - 4 i 1 + 2 i 1 - 2 i + 4 - 3 i = 27 5 + 9 5 i