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21 tháng 11 2021

\(a^3b-ab^3=ab\left(a^2-b^2\right)=ab\left(a^2-ab+ab-b^2\right)=ab\left(a-b\right)\left(a+b\right)\)

Với a hoặc b chẵn \(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮2\)

Với a và b lẻ \(\Leftrightarrow\left(a-b\right)⋮2\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮2\)

Vậy \(ab\left(a-b\right)\left(a+b\right)⋮2,\forall a,b\left(1\right)\)

Với a hoặc b chia hết cho 3 thì \(ab\left(a-b\right)\left(a+b\right)⋮3\)

Với \(a=3k+1;b=3q+1\Leftrightarrow\left(a-b\right)=3\left(k-q\right)⋮3\)

\(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮3\)

Với \(a=3k+1;b=3q+2\Leftrightarrow\left(a+b\right)=\left(3k+1+3q+2\right)=3\left(k+q+1\right)⋮3\)

\(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮3\)

Mà a,b có vai trò tương đương nên \(ab\left(a-b\right)\left(a+b\right)⋮3,\forall a,b\left(2\right)\)

\(\left(1\right)\left(2\right)\Leftrightarrowđpcm\)

21 tháng 11 2021

Ta có : a3b -ab3 
=a3b -ab -ab3 +ab
=ab (a2 -1) -ab (b2 -1) 
=ab (a-1)(a+1) -ab (b-1)(b+1)
Vì a (a-1)(a+1) là 3 số tự nhiên liên tiếp nên chia hết cho 6 .Tương tự b (b-1)(b+1) cũng chia hết cho 6
=> a3b -ab3 chia hết cho 6 (đpcm )

 


 

NV
23 tháng 8 2021

71.

\(\left\{{}\begin{matrix}BB'\perp\left(ABCD\right)\\BB'\in\left(ABB'A'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(ABB'A'\right)\)

74.

\(\left\{{}\begin{matrix}DD'\perp\left(ABCD\right)\\DD'\in\left(CDD'C'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(CDD'C'\right)\)

undefined

26 tháng 9 2021

\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đk:x>0,x\ne1\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)

15 tháng 4 2021

1 Japanese is too difficult for me to learn

2 If he felt well, he wouldn't go to bed early

3 I wish I hadn't gone out in the rain

4 They had me carry te box upstairs

5 Your sister is too young to enjoy this film

6 THey had this letter posted by me

7 The morning was too cold for them to go out

8 If the tickets were cheap enough, we would go there

9 The pupils collected these old clothes

10 The coffee is so excellent so I can drink it

15 tháng 4 2021

11 Her house was sold 2 years ago

12 Some tickets was bought there by him

13 Waste paper was recycled to save money and labour

14 The TV set was so heavy that the girl couldn't move it

15 I will have someone type the report for you

16 There was so much fog that the driver couldn't see far

17 THe day was fine enoigh for them to enjoy sunbathing on the beach

18 I wish your were here now

19 Susan offeref me to lend her some money

20 They asked me is I could fill in the form

AH
Akai Haruma
Giáo viên
19 tháng 8 2023

Lời giải:

a.

$(5x-6)(1999^2+2.1999+1)=4.10^3$

$(5x-6)(1999+1)^2=(4.10^3)^2=4000^2$
$(5x-6).2000^2=4000^2$

$5x-6=\frac{4000^2}{2000^2}=2^2=4$

$5x=10$

$x=10:5=2$

b.

$(23545-7^5)x:[(8^4-4.10^3)^2-2478]=1$

$6738.x:6738=1$

$x=1$

3 tháng 9 2021

ĐK: \(x\ge0\)

TH1: \(m\le0\Rightarrow\) phương trình vô nghiệm.

TH2: \(m>0\)

\(pt\Leftrightarrow\sqrt{x}+2=\dfrac{6}{m}\)

\(\Leftrightarrow\sqrt{x}=\dfrac{6-2m}{m}\)

Phương trình có nghiệm khi: \(\dfrac{6-2m}{m}\ge0\Leftrightarrow6-2m\ge0\Leftrightarrow m\le3\).

\(\Rightarrow0< m\le3\)

Mà \(m\in Z\Rightarrow m\in\left\{1;2;3\right\}\).

3 tháng 9 2021

\(P=\dfrac{6}{\sqrt{x}+2}\left(đk:x\ge0\right)=m\in Z\)

\(\Rightarrow\sqrt{x}+2\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

\(\Rightarrow x\in\left\{0;1;16\right\}\)

\(\Rightarrow m\in\left\{1;2;3\right\}\)