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QN
2
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2020
Làm kiểu 11:
Đây là tổng CSN với \(u_1=-1\) ; \(q=-5^2\)
Có tổng cộng \(\frac{50-0}{2}+1=26\) số hạng
Vậy \(S_n=u_1.\frac{q^n-1}{q-1}=-1.\frac{\left(-5^2\right)^{26}-1}{-5^2-1}=\frac{5^{52}-1}{26}\)
Làm kiểu lớp 6:
\(A=5^{50}-5^{48}+5^{46}+...+5^2-1\)
\(5^2A=25A=5^{52}-5^{50}+5^{48}+...+5^4-5^2\)
\(26A=5^{52}-1\Rightarrow A=\frac{5^{52}-1}{26}\)
NN
7 tháng 12 2017
a) \(\left|-7^{50}\right|:7^{48}-\left|6^2\right|\)
\(=7^{50}:7^{48}-6^2\)
\(=7^{50-48}-6^2\)
\(=7^2-6^2\)
\(=49-36\)
\(=13\)
b) \(3x-9=\left|-21\right|\)
\(\Rightarrow3x-9=21\)
\(\Rightarrow3x=21+9\)
\(\Rightarrow3x=30\)
\(\Rightarrow x=30:3\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
c) \(\left|x\right|-3=7\)
\(\Rightarrow\left|x\right|=7+3\)
\(\Rightarrow\left|x\right|=10\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
Vậy \(x=10\) hoặc \(x=-10\)
d) \(\left|x-4\right|-2=\left|-5\right|\)
\(\Rightarrow\left|x-4\right|-2=5\)
\(\Rightarrow\left|x-4\right|=5+2\)
\(\Rightarrow\left|x-4\right|=7\)
Xét trường hợp 1: \(x-4=7\)
\(\Rightarrow x=7+4\)
\(\Rightarrow x=11\)
Xét trường hợp 2: \(x-4=-7\)
\(\Rightarrow x=-7+4\)
\(\Rightarrow x=-\left(7-4\right)\)
\(\Rightarrow x=-3\)
Vậy \(x=11\) hoặc \(x=-3\)
DD
29 tháng 12 2017
a) ∣∣−750∣∣:748−∣∣62∣∣|−750|:748−|62|
=750:748−62=750:748−62
=750−48−62=750−48−62
=72−62=72−62
=49−36=49−36
=13=13
b) 3x−9=|−21|3x−9=|−21|
⇒3x−9=21⇒3x−9=21
⇒3x=21+9⇒3x=21+9
⇒3x=30⇒3x=30
⇒x=30:3⇒x=30:3
⇒x=10⇒x=10
Vậy x=10x=10
c) |x|−3=7|x|−3=7
⇒|x|=7+3⇒|x|=7+3
⇒|x|=10⇒|x|=10
⇒[x=10x=−10⇒[x=10x=−10
Vậy x=10x=10 hoặc x=−10x=−10
d) |x−4|−2=|−5||x−4|−2=|−5|
⇒|x−4|−2=5⇒|x−4|−2=5
⇒|x−4|=5+2⇒|x−4|=5+2
⇒|x−4|=7⇒|x−4|=7
Xét trường hợp 1: x−4=7x−4=7
⇒x=7+4⇒x=7+4
⇒x=11⇒x=11
Xét trường hợp 2: x−4=−7x−4=−7
⇒x=−7+4⇒x=−7+4
⇒x=−(7−4)⇒x=−(7−4)
⇒x=−3⇒x=−3
Vậy x=11x=11 hoặc x=−3
22 tháng 10 2019
a) \(47-\left[\left(45\cdot2^4-5^2\cdot12\right):14\right]\)
\(=47-\left[\left(45\cdot16-25\cdot12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30=17\)
b) \(50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right):2+34\right]\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40=10\)
c) \(10^2-\left[60:\left(5^6:5^4-3\cdot5\right)\right]\)
\(=100-\left[60:\left(5^{6-4}-15\right)\right]\)
\(=100-\left[60:\left(5^2-15\right)\right]\)
\(=100-\left[60:\left(25-15\right)\right]\)
\(=100-\left[60:10\right]\)
\(=100-6=94\)
d) \(50-\left[\left(50-2^3\cdot5\right):2+3\right]\)
\(=50-\left[\left(50-8\cdot5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-\left[5+3\right]\)
\(=50-8=42\)
e) \(10-\left[\left(8^2-48\right)\cdot5+\left(2^3\cdot10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right)\cdot5+\left(8\cdot10+8\right)\right]:28\)
\(=10-\left[16\cdot5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6=4\)
f) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)
\(=8697-\left[3^{7-5}+2\cdot10\right]\)
\(=8697-\left[3^2+20\right]\)
\(=8697-\left[9+20\right]\)
\(=8697-29=8668\)
CHÚC BẠN HỌC TỐT!!!!!!!!!!!
16 tháng 3 2018
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
16 tháng 3 2018
p=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)
=\(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\frac{50}{50}\)
=\(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
=\(50\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
p=50*S
\(\frac{S}{\text{p}}=\frac{1}{50}\)
21 tháng 3 2017
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
H
6 tháng 4 2018
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
........
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
Đpcm
CT
20 tháng 3 2016
S = ( -1 ) + (-1) + ( -1 ) +.......... +(-1) ( có 25 số -1)
S= (-1) . 25
S = -25
20 tháng 3 2016
ta có 1-2+3-4+5-....-48+49-50
(1-2)+(3-4)+(5+6)+....(47-48)+(49-50)
(-1)+(-1)+(-1)+....+(-1) có 50 số -1
=> (-1).50=-50
\(2\times 4+4\times 6+..+48 \times 50\)
\(=2\times (2+2) + 4\times (4+2) + ..+ 48\times(48+2)\)
\(= 2\times 2 + 2\times 2 + 4\times 4 + 4\times 2+..+48 \times 48 + 48 \times 2\)
\(=2\times(2+4+6+...+48) +(2\times2 +4\times 4+...+48\times 48)\)
\(=2\times 600 + 4(1+4+9+..+576)\)
\(= 1200 + 4\times 4900\)
\(=1200 + 19600 = 20 800\)