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25 tháng 10 2015

2/17/21

1/5 +1/7/8=2/3/40

25 tháng 10 2015

a/

=3/21+14/21+42/21=3+14+42/21=59/21

b/

=4/5+15/8

=32/40+45/40=107/40

15 tháng 5 2023

a: =7/4*4/7=1

b: =5/2*2/9*5/3=50/54=25/27

17 tháng 4 2023

\(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)

\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)

\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)

\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)

\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6}{30}-\dfrac{5}{30}=\dfrac{1}{30}\)

\(\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{7}{42}-\dfrac{6}{42}=\dfrac{1}{42}\)

`@mt`

5 tháng 7 2023

1: =1/8*9/4=9/32

2: =8/27*243/32=9/4

3: =(5/4*4/5)^5*(4/5)^2=16/25

4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)

5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)

6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)

8: =(0,2*5)^4*5^2=25

10: =-0,5^5*2^10

=-0,5^5*2^5*2^5

=-32

13: =(0,5*2)^2*2^2=4

 

6 tháng 7 2023

mình cảm ơn ạ

10 tháng 8 2021

Xóa nhanh đi ko CTV xóa giờ

10 tháng 8 2021

xóa j

35 : 7 = 5

16 tháng 12 2021

= 5

20 tháng 10 2021

1) \(-6x^4+4x^3-2x^2\)

2) \(=x^2+4x-21-x^2-4x+5=-16\)

3) \(=6x^2-4x-x^2-4x-4=5x^2-8x-4\)

4) \(=2x^3-4x^2-8x-3x^2+6x+12=2x^3-7x^2-2x+12\)

20 tháng 10 2021

giúp mình với

 

 

11 tháng 5

1: \(\left(\sqrt{15}-2\sqrt3\right)^2+12\sqrt5\)

\(=15+12-2\cdot\sqrt{15}\cdot2\sqrt3+12\sqrt5\)

\(=27-4\sqrt{45}+12\sqrt5=27\)

2: \(3\sqrt2\left(4-\sqrt2\right)+3\left(1-2\sqrt2\right)^2\)

\(=12\sqrt2-6+3\left(9-4\sqrt2\right)\)

\(=12\sqrt2-6+27-12\sqrt2=21\)

3: \(\frac12\left(\sqrt6+\sqrt5\right)^2-\frac14\cdot\sqrt{120}-\sqrt{\frac{15}{2}}\)

\(=\frac12\left(11+2\sqrt{30}\right)-\frac14\cdot2\sqrt{30}-\sqrt{\frac{30}{4}}\)

\(=\frac{11}{2}+\sqrt{30}-\frac12\sqrt{30}-\frac12\sqrt{30}=\frac{11}{2}\)

4: \(\left(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7}\right)^2\)

\(=4-\sqrt7+4+\sqrt7-2\cdot\sqrt{\left(4-\sqrt7\right)\left(4+\sqrt7\right)}\)

\(=8-2\cdot\sqrt{16-7}=8-2\cdot\sqrt9=8-2\cdot3=8-6=2\)

5: \(\left(\sqrt{\sqrt{14}+\sqrt5}+\sqrt{\sqrt{14}-\sqrt5}\right)^2\)

\(=\sqrt{14}+\sqrt5+\sqrt{14}-\sqrt5+2\cdot\sqrt{\left(\sqrt{14}+\sqrt5\right)\left(\sqrt{14}-\sqrt5\right)}\)

\(=2\sqrt{14}+2\cdot\sqrt{14-5}=2\sqrt{14}+2\cdot\sqrt9=2\sqrt{14}+6\)

6: \(\left(\sqrt3+1\right)^3-\left(\sqrt3-1\right)^3\)

\(=\left(3\sqrt3+3\cdot3\cdot1+3\cdot\sqrt3\cdot1^2+1\right)-\left(3\sqrt3-3\cdot3\cdot1+3\sqrt3-1\right)\)

\(=\left(6\sqrt3+10\right)-\left(6\sqrt3-10\right)=20\)

7: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left\lbrack\left(\sqrt2\right)^3+3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2+1^3\right\rbrack-\left\lbrack\left(\sqrt2\right)^3-3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2-1^3\right\rbrack\)

\(=\left(2\sqrt2+6+3\sqrt2+1\right)-\left(2\sqrt2-6+3\sqrt2-1\right)=5\sqrt2+7-5\sqrt2+7=14\)

8: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+2\cdot\sqrt{360}}\)

\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}-\sqrt{45+2\cdot3\sqrt5\cdot2\sqrt2+8}\)

\(=\sqrt{\left(2\sqrt2-\sqrt5\right)^2}-\sqrt{\left(3\sqrt5+2\sqrt2\right)^2}\)

\(=2\sqrt2-\sqrt5-3\sqrt5-2\sqrt2=-4\sqrt5\)

9: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)\)

\(=\frac12\left(\sqrt5-1+\sqrt5+1\right)=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

27 tháng 7 2021

1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)

\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)

\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)

\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)

11 tháng 5

1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)

\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)

\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)

2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)

\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)

\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)

3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)

\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)

4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)

\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)

\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)

6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)

\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)

=-3-9

=-12

5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)

7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)

\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)

\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)

8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)

\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)

9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)

\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)

\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)

10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)

\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)

\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)

11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)


`@` `\text {Ans}`

`\downarrow`

`1.`

\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)

`=`\(\left(-4xy\right)\left(2xy^2\right)+\left(-4xy\right)\left(-3x^2y\right)\)

`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)

`=`\(-8x^2y^3+12x^3y^2\)

`2.`

\(\left(-5x\right)\left(3x^3+7x^2-x\right)\)

`=`\(\left(-5x\right)\left(3x^3\right)+\left(-5x\right)\left(7x^2\right)+\left(-5x\right)\left(-x\right)\)

`=`\(-15x^4-35x^3+5x^2\)

`3.`

\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)

`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)

`=`\(12x^2+15x-8x-10-12x^2+6x\)

`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)

`=`\(13x-10\)

`4.`

\(2x^2\left(x^2-7x+9\right)\)

`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)

`=`\(2x^4-14x^3+18x^2\)

`5.`

\(\left(3x-5\right)\left(x^2-5x+7\right)\)

`=`\(3x\left(x^2-5x+7\right)-5\left(x^2-5x+7\right)\)

`=`\(3x^3-15x^2+21x-5x^2+25x-35\)

`=`\(3x^3-20x^2+46x-35\)

C xem lại bài cuối ạ.