\(\left(5^{19}\cdot5^{14}\right):5^{32}\)

\(=5^{19+14}:5^{32}\)

\(=5^{33}:5^{32}\)

\(=5^{33-32}\)

\(=5\)

a) 59 . 73 - 30² +27 . 59

=59x(73+27)-30²

=59x100-30²

=5900-900

=5000

b) 1560 : [5 . 79 -( 125 + 5. 49) + 5. 21]

=1560:[5.79-(125+245)+5.21)

=1560:[5.(79+21)-(125+245)

=1560:[500-370]

=1560:130

=12

13: \(\left(-15\right)+8+7\)

\(=\left(-15\right)+\left(8+7\right)\)

=-15+15

=0

14: \(\left(-8\right)+2+6\)

\(=\left(-8\right)+\left(2+6\right)\)

=-8+8

=0

15: \(\left(-1\right)+3-2\)

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

=-3+3

=0

16: \(25-8-7\)

\(=25-\left(8+7\right)\)

=25-15

=10

17: \(8-2-6\)

\(=8-\left(2+6\right)\)

=8-8

=0

18: \(\left(-12\right)-3+15\)

\(=\left(-12-3\right)+15\)

=-15+15

=0

\(a,2x^2+6x=2x\left(x+3\right)\\ b,x^2+2xy+y^2-9z^2\\ =\left(x^2+2xy+y^2\right)-\left(3z\right)^2\\ =\left(x+y\right)^2-\left(3z\right)^2\\ =\left(x+y-3z\right)\left(x+y+3z\right)\\ b,x^3-2x^2+x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ d,x^2-2x-15=x^2-5x+3x-15\\ =x\left(x-5\right)+3\left(x-5\right)\\ =\left(x+3\right)\left(x-5\right)\)

a) 2x² + 6x

=2x (x+3)

b) x² + 2xy +y²- 9z²

(x² + 2xy +y2)- 9z²

(x + y)² - (3z)²

(x+y-3z) (x+y+3z) 

Chủ yếu em phân tích sao cho có nhân tử chung ra là được

a) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}=\sqrt{5}-\sqrt{16.3}+5\sqrt{9.3}-\sqrt{9.5}\)

\(=\sqrt{5}-4\sqrt{3}+15\sqrt{3}-3\sqrt{5}=11\sqrt{3}-2\sqrt{5}\)

b) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}=2\sqrt{3}+\sqrt{16.3}-\sqrt{25.3}-\sqrt{81.3}\)

\(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)

c) \(3\sqrt{50}-2\sqrt{75}-4\dfrac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\dfrac{1}{3}}\)

\(=3\sqrt{25.2}-2\sqrt{25.3}-4\sqrt{\dfrac{54}{3}}-\sqrt{9.\dfrac{1}{3}}=15\sqrt{2}-10\sqrt{3}-4\sqrt{18}-\sqrt{3}\)

\(=15\sqrt{2}-11\sqrt{3}-4\sqrt{9.2}=15\sqrt{2}-11\sqrt{3}-12\sqrt{2}=3\sqrt{2}-11\sqrt{3}\)

mấy câu dưới bạn làm tương tự thôi

d) Ta có: \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{2}+\dfrac{12}{\sqrt{6}}\)

\(=\sqrt{2}-\sqrt{2}+2\sqrt{6}-8\sqrt{6}\)

\(=-6\sqrt{6}\)

e) Ta có: \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{27}+5\sqrt{1\dfrac{1}{3}}\)

\(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-3\sqrt{3}+5\cdot\dfrac{2}{\sqrt{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-3\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)

\(=-11\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)

\(=\dfrac{-33+10\sqrt{3}}{3}\)

f) Ta có: \(\sqrt{150}+\sqrt{96}+\dfrac{9}{2}\cdot\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)

\(=5\sqrt{6}+4\sqrt{6}+\dfrac{9}{2}\cdot\sqrt{\dfrac{8}{3}}-\sqrt{6}\)

\(=8\sqrt{6}+\dfrac{9}{2}\cdot\dfrac{2\sqrt{2}}{\sqrt{3}}\)

\(=8\sqrt{6}+3\sqrt{6}\)

\(=11\sqrt{6}\)

n)\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}-\sqrt{\dfrac{1}{6}}\)

=\(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-\sqrt{\dfrac{1}{6}}\)

=\(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}-\sqrt{\dfrac{1}{6}}\)

=\(3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{\dfrac{1}{6}}\)

=\(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{\dfrac{1}{6}}\)

=\(\sqrt{6}-11-\sqrt{\dfrac{1}{6}}\)

=\(\dfrac{5-11\sqrt{6}}{\sqrt{6}}\)

h)\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

=\(\left|\sqrt{3}-3\right|+\sqrt{3-2\sqrt{3}+1}\)

=\(3-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

=\(3-\sqrt{3}+\left|\sqrt{3}-1\right|\)

=\(3-\sqrt{3}+\sqrt{3}-1\)

=2