giúp em với em cảm ơn nhiềuuuuu
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TK
2
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
Nguyễn Việt Lâm
Giáo viên
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.
\(\left\{{}\begin{matrix}DD'\perp\left(ABCD\right)\\DD'\in\left(CDD'C'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(CDD'C'\right)\)
26 tháng 9 2021
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)
LN
4
DT
Đỗ Thanh Hải
CTVVIP
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
DT
Đỗ Thanh Hải
CTVVIP
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
NH
1
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.
$(23545-7^5)x:[(8^4-4.10^3)^2-2478]=1$
$6738.x:6738=1$
$x=1$
HP
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
\(P=\dfrac{6}{\sqrt{x}+2}\left(đk:x\ge0\right)=m\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
\(\Rightarrow x\in\left\{0;1;16\right\}\)
\(\Rightarrow m\in\left\{1;2;3\right\}\)
Do n lẻ, đặt \(n=2m+1\)
\(\Rightarrow S=C_{2m+1}^1+C_{2m+1}^2+...+C_{2m+1}^m\)
Áp dụng đẳng thức: \(C_n^k=C_n^{n-k}\)
\(\Rightarrow S=C_{2m+1}^{2m}+C_{2m+1}^{2m-1}+...+C_{2m+1}^{m+1}\)
\(\Rightarrow2S=S+S=C_{2m+1}^1+C_{2m+1}^2+...+C_{2m+1}^{2m}\)
\(=C_{2m+1}^0+C_{2m+1}^1+...+C_{2m+1}^{2m+1}-\left(C_{2m+1}^0+C_{2m+1}^{2m+1}\right)\)
\(=2^{2m+1}-2\)
\(\Rightarrow S=2^{2m}-1\) luôn lẻ (đpcm)