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a: =-3/4-1/4+2/7+5/7+2023/2024
=-1+1+2023/2024=2023/2024
b: 2/3x=2/7
=>x=2/7:2/3=3/7
c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
=>x=3/5:2/3=3/5*3/2=9/10
a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)
\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)
\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)
b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)
\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)
\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)
=-3-1+1
=-3
c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)
\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)
Lời giải:
\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)
\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)
\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)
\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)
\(B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...-\dfrac{1}{2022}+\dfrac{1}{2023}\\ \Rightarrow B=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)
\(\Rightarrow B=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\\ \Rightarrow2^2B=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)
\(\Rightarrow4B-B=\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\\ \Rightarrow3B=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)
\(\Rightarrow3B=1-\dfrac{3}{2^{2024}}\\ \Rightarrow B=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)
\(\Rightarrow B=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\\ B=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)
\(A=1-2+3-4+5-6+7-8+...+99-100\)
\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(A=\left(-1\right).50\)
\(A=-50\)
\(B=1+3-5-7+9+11-...-397-399\)
\(B=1-2+2-2+2-...+2-2-399\)
\(B=1-399\)
\(B=-398\)
\(C=1-2-3+4+5-6-7+...+97-98-99+100\)
\(C=-1+1-1+1-...-1+1\)
\(C=0\)
\(D=2^{2024}-2^{2023}-...-1\)
\(D=2^{2024}-\left(2^0+2^1+2^2+...2^{2023}\right)\)
\(D=2^{2024}-\left(\dfrac{2^{2024}-1}{2-1}\right)\)
\(D=2^{2024}-\left(2^{2024}-1\right)\)
\(D=2^{2024}-2^{2024}+1\)
\(D=1\)
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +...+ 99 - 100
A = (1 - 2) + ( 3 - 4) + ( 5- 6) +....+(99 - 100)
Xét dãy số 1; 3; 5;...;99
Dãy số trên là dãy số cách đều có khoảng cách là: 3 - 1 = 2
Dãy số trên có số số hạng là: (99 - 1) : 2 + 1 = 50 (số)
Vậy tổng A có 50 nhóm, mỗi nhóm có giá trị là: 1- 2 = -1
A = - 1\(\times\)50 = -50
b,
B = 1 + 3 - 5 - 7 + 9 + 11-...- 397 - 399
B = ( 1 + 3 - 5 - 7) + ( 9 + 11 - 13 - 15) + ...+( 393 + 395 - 397 - 399)
B = -8 + (-8) +...+ (-8)
Xét dãy số 1; 9; ...;393
Dãy số trên là dãy số cách đều có khoảng cách là: 9-1 = 8
Dãy số trên có số số hạng là: ( 393 - 1): 8 + 1 = 50 (số hạng)
Tổng B có 50 nhóm mỗi nhóm có giá trị là -8
B = -8 \(\times\) 50 = - 400
c,
C = 1 - 2 - 3 + 4 + 5 - 6 +...+ 97 - 98 - 99 +100
C = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7+ 8) +...+ ( 97 - 98 - 99 + 100)
C = 0 + 0 + 0 +...+0
C = 0
d, D = 22024 - 22023- ... +2 - 1
2D = 22005- 22004 + 22003+...- 2
2D + D = 22005 - 1
3D = 22005 - 1
D = (22005 - 1): 3
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$
$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$
$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$
$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$
$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$
A=23.41+24.51+....+22023.20241
=23.4+24.5+...+22023.2024=3.42+4.52+...+2023.20242
=2(4−33.4+5−44.5+...+2024−20232023.2024)=2(3.44−3+4.55−4+...+2023.20242024−2023)
=2(13−14+14−15+....+12023−12024)=2(31−41+41−51+....+20231−20241)
=2(13−12024)=20213036=2(31−20241)=30362021