Phân số nào có giá trị gần nhất với \(\dfrac{1}{2}\)?
A)\(\dfrac{25}{79}\)
B)\(\dfrac{27}{59}\)
C)\(\dfrac{29}{57}\)
D)\(\dfrac{52}{79}\)
E)\(\dfrac{57}{92}\)
K
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18 tháng 6 2021
\(\dfrac{51}{53}+\dfrac{55}{57}+\dfrac{61}{63}+\dfrac{69}{71}+\dfrac{79}{81}+\dfrac{91}{93}\)
\(=\left(\dfrac{52}{53}-\dfrac{1}{53}\right)+\left(\dfrac{56}{57}-\dfrac{1}{57}\right)+\left(\dfrac{62}{63}-\dfrac{1}{63}\right)+\left(\dfrac{70}{71}-\dfrac{1}{71}\right)+\left(\dfrac{80}{81}-\dfrac{1}{81}\right)+\left(\dfrac{92}{93}-\dfrac{1}{93}\right)\)
\(=\left(1-\dfrac{1}{53}-\dfrac{1}{53}\right)+\left(1-\dfrac{1}{57}-\dfrac{1}{57}\right)+\left(1-\dfrac{1}{63}-\dfrac{1}{63}\right)+\left(1-\dfrac{1}{71}-\dfrac{1}{71}\right)+\left(1-\dfrac{1}{81}-\dfrac{1}{81}\right)+\left(1-\dfrac{1}{93}-\dfrac{1}{93}\right)\)
\(=\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)\)
\(=1+1+1+1+1+1\)
\(=6\)
23 tháng 1 2021
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
23 tháng 1 2021
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
KL
3 tháng 8 2017
3. a) Ta có : 13.29 = 377
25.17 = 425
=> \(\dfrac{13}{17}< \dfrac{25}{29}\)
b) Ta có : 59.105 > 56.101
=> \(\dfrac{59}{101}>\dfrac{56}{105}\)
c) Ta có : 14.83 = 1162
20.55 = 1100
=> \(\dfrac{14}{55}>\dfrac{20}{83}\)
d) Ta có : 13.73 = 949
29.57 = 1653
=> \(\dfrac{13}{57}< \dfrac{29}{73}\)
e) Ta có : \(\dfrac{1717}{2121}=\dfrac{17}{21}\)
=> \(\dfrac{17}{21}=\dfrac{1717}{2121}\)
@Đặng Vũ Hoài Anh
KL
3 tháng 8 2017
4. Gọi các phân số cần tìm có dạng \(\dfrac{x}{3}\)
Ta có : \(\dfrac{-1}{2}< \dfrac{x}{3}< \dfrac{1}{2}\)
=> \(\dfrac{-3}{6}< \dfrac{2x}{6}< \dfrac{3}{6}\)
=> -3 < 2x < 3
=> 2x = -2; 0; 2
=> x = -1; 0; 1 (thỏa mãn)
@Đặng Vũ Hoài Anh
MV
3 tháng 8 2017
a,
\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)
Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)
b,
\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)
Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)
c,
\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)
Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)
7 tháng 1
a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)
b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)
c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)
\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)
d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)
\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)
e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)
\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)
\(=\left(3-1\right)^{106}=2^{106}\)
f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)
g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)
\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)
\(=2023^{4\sqrt{2}}\)
ND
Nguyễn Đăng Nhân
CTVHS
11 tháng 9 2023
1,
Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)
\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)
2,
Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)
3,
Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)
\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)
4,
\(\dfrac{48}{47}>1>\dfrac{84}{85}\)
Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)
DQ
26 tháng 8 2018
a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)
\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)
\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)
\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)
\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)
Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)
Nên: \(x+100=0\)
\(x=-100\)
B