Thực hiện các phép tính sau :
a) \(\left(3+2i\right)\left(2-i\right)+\left(3-2i\right)\)
b) \(\left(4-3i\right)+\dfrac{1+i}{2+i}\)
c) \(\left(1+i\right)^2-\left(1-i\right)^2\)
d) \(\dfrac{3+i}{2+i}-\dfrac{4-3i}{2-i}\)
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
Thực hiện các phép tính :
a) \(\left(2+3i\right)^2-\left(2-3i\right)^2\)
b) \(\dfrac{\left(1+i\right)^5}{\left(1-i\right)^3}\)
Thực hiện các phép tính sau :
a) \(\dfrac{\left(2+i\right)+\left(1+i\right)\left(4-3i\right)}{3+2i}\)
b) \(\dfrac{\left(3-4i\right)\left(1+2i\right)}{1-2i}+4-3i\)
BÀI 1 : THỰC HIỆN PHÉP TÍNH
a, \(\left(1+\sqrt{3}-\sqrt[2]{2}\right)\times\left(1+\sqrt{3}+\sqrt[2]{2}\right)\)
b, \(\left(\dfrac{3}{2}\times\sqrt{6}+2\times\sqrt{\dfrac{2}{3}}-4\times\sqrt{\dfrac{3}{2}}\right)\times\left(3\times\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
BÀI 2 : rút gọn
B = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-2}}\)
Câu 1: Cho bt: A= \(\left(\dfrac{1}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{1}{\sqrt{1-x^2}}+1\right)\)
a) Tìm x để A có nghĩa
b) Rút gọn
c) Tính A với x =\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
Câu 2: Cho bt B= \(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\)
a) Rút gọn
b) CM B\(\ge\)0
c) So sánh B với \(\sqrt{B}\)
Thực hiện các phép tính sau :
a) \(2i\left(3+i\right)\left(2+4i\right)\)
b) \(\dfrac{\left(1+i\right)^2\left(2i\right)^3}{-2+i}\)
c) \(3+2i+\left(6+i\right)\left(5+i\right)\)
d) \(4-3i+\dfrac{5+4i}{3+6i}\)
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 =
Bài 1: Rút gọn: A= \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
Bài 2: Rút gọn: B=\(\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
Bài 3: Rút gọn D=\(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+b}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)
Bài 1:
\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)
\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7-x}{x-7}\)
\(=\dfrac{-\left(x-7\right)}{x-7}\)
\(=-1\)
A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)
A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)
A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)
Bài 2:
\(B=\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
\(=\left(\dfrac{3}{x+1}+\dfrac{3\left(x^2+2x+1\right)-x^2}{x\cdot\left(x^2+2x+1\right)}\cdot\dfrac{x+1}{2x^2+3x}\right)\cdot\dfrac{x^2+x}{1+3x}\)
\(=\left(\dfrac{3}{x+1}+\dfrac{3x^2+6x+3-x^2}{x\left(x+1\right)^2}\cdot\dfrac{x+1}{2x^2+3x}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)
\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)}\cdot\dfrac{1}{2x^2+3x}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)
\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)
\(=\dfrac{3x\cdot\left(2x^2+3x\right)+2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\cdot\dfrac{x\left(x+1\right)}{1+3x}\)
\(=\dfrac{6x^3+9x^2+2x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)
\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)
\(=\dfrac{6x^3+11x^2+6x+3}{\left(2x^2+3x\right)\left(1+3x\right)}\)
\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+6x^3+3x+9x^2}\)
\(=\dfrac{6x^3+11x^2+6x+3}{11x^2+6x^3+3x}\)
Bài 1: A= \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a) RÚt gọn A
b) tính A khi \(a^2\) -3 =0
Bài 2:B= \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a) Rút gọn B
b) C/m rằng: B>0 với mọi x>0 , x khác 1
Bài 3:C = \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{2}{a-1}\right)\)
Rút gọn C
Bài 3:
\(C=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a-1}{\sqrt{a}+3}\)
\(=\dfrac{\left(a-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
1. Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\left(a+b\right)^2=10+2ab\\\left(a+b\right)\left(a-\dfrac{2}{ab}\right)=\dfrac{4}{3}\end{matrix}\right.\)
2.Giải phương trình:
\(\sqrt{\sqrt{2}-1-x}+\sqrt[4]{x}=\dfrac{1}{\sqrt[4]{2}}\)
