Áp dụng các hằng đẳng thức đáng nhớ để tính :
a) \(\left(2+i\sqrt{3}\right)^2\)
b) \(\left(1+2i\right)^3\)
c) \(\left(3-i\sqrt{2}\right)^3\)
d) \(\left(2-i\right)^3\)
Áp dụng bằng hằng đẳng thức đáng nhớ để thực hiện phép chia :
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
b) \(\left(125x^3+1\right):\left(5x+1\right)\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
Bài giải:
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
a) (x2 +2xy + y2 ) : (x +y)
= (x +y)2 : (x +y)
= x + y
b) (125x3 + 1) : (5x + 1)
= (5x + 1)(25x2 - 5x + 1) : (5x + 1)
= 25x2 - 5x + 1
c) \(\left(x^2-2xy+y^2\right)\left(y-x\right)\)
= \(\left(x-y\right)^2:\left(y-x\right)\)
= \(x-y\)
Bài 1: Rút gọn căn bậc 2 theo hằng đẳng thức 1:
a)\(\sqrt{\left(23-15\sqrt{3}\right)^2}\)
b) \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)
c) \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)
d)\(\sqrt{\left(16-6\sqrt{7}\right)^2}\)
f)\(\sqrt{\left(22-8\sqrt{3}\right)^2}\)
g) \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)
h) \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)
i)\(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
(mink đag cần gấp)
Bài 1:
a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)
\(=\left|23-15\sqrt{3}\right|\)
\(=\left|\sqrt{529}-\sqrt{675}\right|\)
\(=\sqrt{675}-\sqrt{529}\)
\(=15\sqrt{3}-23\)
b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)
\(=\left|2-2\sqrt{3}\right|\)
\(=2\sqrt{3}-2\)
c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)
\(=\left|15-4\sqrt{3}\right|\)
\(=15-4\sqrt{3}\)
d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)
\(=\left|16-6\sqrt{7}\right|\)
\(=\left|\sqrt{256}-\sqrt{252}\right|\)
\(=16-6\sqrt{7}\)
f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)
\(=\left|22-8\sqrt{3}\right|\)
\(=\left|\sqrt{484}-\sqrt{192}\right|\)
\(=22-8\sqrt{3}\)
g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)
\(=\left|9-4\sqrt{2}\right|\)
\(=9-4\sqrt{2}\)
h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)
\(=\left|13-4\sqrt{3}\right|\)
\(=13-4\sqrt{3}\)
i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
\(=\left|7-3\sqrt{3}\right|\)
\(=7-3\sqrt{3}\)
Tính giá trị biểu thức (Nhân thêm số căn vào biểu thức để làm xuất hiện hằng đẳng thức \(\left(a\pm\sqrt{b}\right)^2\) hoặc \(\left(\sqrt{a}\pm\sqrt{b}\right)^2\) rồi phá căn)
a. \(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
b. \(\dfrac{\sqrt{3}+1}{2}.\sqrt{8-2\sqrt{3}}\)
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
Giải các phương trình sau trên tập số phức :
a) \(3x^2+\left(2+2i\sqrt{2}\right)x-\dfrac{\left(1+i\right)^3}{1-i}=i\sqrt{8}x\)
b) \(\left(1-ix\right)^2+\left(3+2i\right)x-5=0\)
tính: \(\left(2a-b\right)^2-2\times\left(2a-b\right)\times\left(a+b\right)+\left(a+b\right)^2\)
ÁP DỤNG HẰNG ĐẲNG THỨC ĐÁNG NHỚ
(2a-b)2 - 2 x ( 2a-b) x (a+b) + (a + b)2
= [(2a-b) - (a+b)]2
Áp dụng các hằng đẳng thức đáng nhớ để tính:
a) ( 2 + i 3 ) 2 ;
b) ( 1 + 2 i ) 3 ;
c) ( 3 - i 2 ) 2 ;
d) ( 2 - i ) 3 .
a) 1 + 4i 3 ;
b) – 11 – 2i;
c) 7 − 6i 2 ;
d) 2 – 11i.
Thực hiện các phép tính :
a) \(\left(2+3i\right)\left(3-i\right)+\left(2-3i\right)\left(3+i\right)\)
b) \(\dfrac{2+i\sqrt{2}}{1-i\sqrt{2}}+\dfrac{1+i\sqrt{2}}{2-i\sqrt{2}}\)
c) \(\dfrac{\left(1+i\right)\left(2+i\right)}{2-i}+\dfrac{\left(1+i\right)\left(2-i\right)}{2+i}\)
Bài 1:Tính
1.\(\sqrt{12,5}\cdot\sqrt{0,2}\cdot\sqrt{0,1}\)
2.\(\sqrt{48,4}\cdot\sqrt{5}\cdot\sqrt{0,5}\)
Bài 2:Khai triển các hằng đẳng thức sau:
a,\(\left(\sqrt{7}+\sqrt{3}\right)^2\)
b,\(\left(\sqrt{11}-\sqrt{5}\right)^2\)
c,\(\left(\sqrt{x}+\sqrt{y}\right)^2\)
d,\(\left(\sqrt{13}+\sqrt{7}\right)^2\)
e,\(\left(\sqrt{a}-\sqrt{b}\right)^2\)
f,\(\left(\sqrt{3}-1\right)^2\)
B1:
1. \(\sqrt{12.5}\cdot\sqrt{0.2}\cdot\sqrt{0.1}\) \(=\sqrt{12.5\cdot0.2\cdot0.1}\) \(=\sqrt{0.25}=0.5\)
2.\(\sqrt{48.4}\cdot\sqrt{5}\cdot\sqrt{0.5}\) = \(\sqrt{48.4\cdot5\cdot0.5}\) =\(\sqrt{121}=11\)
B2:
a, \(\left(\sqrt{7}+\sqrt{3}\right)^2=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3=7+2\cdot\sqrt{21}+3\)\(=10+2\sqrt{21}\)
b,\(\left(\sqrt{11}-\sqrt{5}\right)^2=11-2\sqrt{55}+5=16-2\sqrt{55}\)
c,\(\left(\sqrt{x}+\sqrt{y}\right) ^2=x+2\sqrt{xy}+y\)
d.\(\left(\sqrt{13}+\sqrt{7}\right)^2=13+2\sqrt{7}+7=20+2\sqrt{7}\)
e,\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)
f,\(\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\)
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 =