Tìm x
2x-3/5=x-2/4
(2x-3 là tử 5 là mẫu, x-2 là tử 4 mẫu)
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\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)
31−53+75−97+119−1311+1513+1311−119+97−75+53−31
=(13−13)−(35−35)+(57−57)−(79−79)+(911−911)−(1113−1113)+1315=(31−31)−(53−53)+(75−75)−(97−97)+(119−119)−(1311−1311)+1513
=1315=1513
`# \text {Ryo}`
So sánh ạ?
\(2^{225}\text{ và }3^{150}\)
Ta có:
\(2^{225}=2^{3\cdot75}=\left(2^3\right)^{75}=8^{75}\\ 3^{150}=3^{2\cdot75}=\left(3^2\right)^{75}=9^{75}\)
Vì `9 > 8 \Rightarrow`\(9^{75}>8^{75}\)
\(\Rightarrow2^{225}< 3^{150}\)
2²²⁵ = (2³)⁷⁵ = 8⁷⁵
3¹⁵⁰ = (3²)⁷⁵ = 9⁷⁵
Do 8 < 9 nên 8⁷⁵ < 9⁷⁵
Vậy 2²²⁵ < 3¹⁵⁰
a) 17/20 = 323/380
18/19 = 360/380
Do 323 < 360 nên 323/380 < 360/380
Vậy 17/20 < 18/19
b) 19/18 = 1 + 1/18
2023/2022 = 1 + 1/2022
Do 18 < 2022 nên 1/18 > 1/2022
1 + 1/18 > 1 + 1/2022
Vậy 19/18 > 2023/2022
c) 13/17 = 455/595
135/175 = 27/35 = 459/595
Do 455 < 459 nên 455/595 < 459/595
Vậy 13/17 < 135/175
d) 53/63 = 11236/13356
535/636 = 11235/13356
Do 11236 > 11235 nên 11236/13356 > 11235/13356
Vậy 53/63 > 535/636
e) 13/15 = 65/75
22/25 = 66/75
Do 65 < 66 nên 65/75 < 66/75
Vậy 13/15 < 22/25
8⁵ = (2³)⁵ = 2¹⁵ = 2.2¹⁴
3.4⁷ = 3.(2²)⁷ = 3.2¹⁴
Do 2 < 3 nên 2.2¹⁴ < 3.2¹⁴
Vậy 8⁵ < 3.4⁷
\(8^5=2^{15}=2.2^{14}\)
\(3.4^7=3.2^{14}>2.2^{14}\)
\(\Rightarrow8^5< 3.4^7\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
`# \text {Kaizu DN}`
`a)`
`(3x + 6) + (7x - 14) = 0?`
\(\Rightarrow3x+6+7x-14=0\\ \Rightarrow\left(3x+7x\right)+\left(6-14\right)=0\\ \Rightarrow10x-8=0\\ \Rightarrow10x=8\Rightarrow x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{4}{5}\)
Vậy, \(x=\dfrac{4}{5}\)
`b)`
`17y + 35 + 4x + 17 = 42`
\(\Rightarrow\left(17y+17\right)+\left(35+4x\right)=42\\ \Rightarrow17\left(y+1\right)+\left(35+4x\right)=42\)
Bạn xem lại đề ;-;.
a) \(\dfrac{5}{3a-1}=1\)
\(\Rightarrow3a-1=5\)
\(\Rightarrow3a=6\)
\(\Rightarrow a=\dfrac{6}{3}=2\)
b) \(\dfrac{5}{3a-1}=-5\)
\(\Rightarrow3a-1=5:\left(-5\right)=-1\)
\(\Rightarrow3a=-1+1=0\)
\(\Rightarrow a=0:3=0\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
\(\dfrac{2x-3}{5}=\dfrac{x-2}{4}\)
\(\Rightarrow\left(2x-3\right)\cdot4=\left(x-2\right)\cdot5\)
\(\Rightarrow8x-12=5x-10\)
\(\Rightarrow8x-5x=-10+12\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)
#kễnh
\(\dfrac{2x-3}{5}=\dfrac{x-2}{4}\)
\(\Leftrightarrow4\left(2x-3\right)=5\left(x-2\right)\)
\(\Leftrightarrow8x-12=5x-10\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)