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2003 / 2001 = 1 + 2/2001
1999/1997 = 1 + 2/1997
vì 2/ 2001 < 2/1997
nên 1 + 2/2001 < 1 + 2/1997
hay 2003 < 1999/1997
b, = 5/9 x 1/4 + 4/9 x 1/4
= 1/4 x ( 5/9 + 4/9 )
= 1/4 x 1
= 1/4
* Ý a mk k nhớ cách làm ^^, xl *
\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)
\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)
\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)
a )
\(4\dfrac{5}{9}:2\dfrac{5}{8}-7=-\dfrac{995}{189}=-5\) ( đã làm tròn )
\(\left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\) \(=\dfrac{2}{5}=0\) ( đã làm tròn )
Ta có :
\(-5< x< 0\)
\(\Rightarrow x=\left\{-4;-3;-2;-1\right\}\)
b )
\(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)=-\dfrac{20}{21}=-1\) ( đã làm tròn )
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{7}{12}=1\) ( đã làm tròn )
Ta có : \(-1< x+0,6< 1\)
\(\Rightarrow x=\left\{-1;0\right\}\)
a) \(4\dfrac{5}{9}:2\dfrac{5}{8}-7< x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>4\dfrac{5}{9}:2\dfrac{5}{8}-7\\x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{227}{36}\\x< -\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow-\dfrac{2}{5}>x>-\dfrac{227}{36}\)
b) \(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)< x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{7}>-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)\\x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{32}{21}\\x< \dfrac{1}{84}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{84}>x>-\dfrac{32}{21}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
a) ta có : \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) = \(\dfrac{x}{16}=\dfrac{y}{24}\) ( 1)
\(\dfrac{y}{8}=\dfrac{z}{5}\) = \(\dfrac{y}{24}=\dfrac{z}{15}\) (2)
từ (1) và (2) , ta có : \(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}\)
mà x - y + z = 35
theo tính chất của dãy tỉ số bằng nhau , ta có :
\(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x-y+z}{16-24+15}=\dfrac{35}{7}=5\)
do đó : \(\dfrac{x}{16}=5\) => x = 5. 16 = 80
\(\dfrac{y}{24}=5\) => y = 5.24 = 120
\(\dfrac{z}{15}=5\) => z = 5.15 = 75
vậy x = 80
y = 120
z = 75
1.
=2/5 x 12/3 + 2/5 x 15/3 + 2/5 x 1
= 2/5 x (12/3 + 15/3 + 1)
=2/5 x 1
=2/5
2.a=1;2
Theo đề: \(2x+y=0\Leftrightarrow y=-2x\) \(\left(1\right)\)
Ta có:
\(\dfrac{3-x}{y-4}=\dfrac{2}{5}\)
\(\Leftrightarrow5\left(3-x\right)=2\left(y-4\right)\)
\(\Leftrightarrow15-5x=2y-8\)
\(\Leftrightarrow15+8=2y+5x\)
\(\Leftrightarrow5x+2y=23\) \(\left(2\right)\)
Thế (1) vào (2), suy ra:
\(5x+2.\left(-2x\right)=23\)
\(\Leftrightarrow5x-4x=23\)
\(\Leftrightarrow x=23\)
\(\Rightarrow y=-2.23=-46\)
= nhau nha
\(2/3*4/5=8/15\)
\(4/5*2/3=8/15\)
\(=>2/3*4/5=4/5*2/3\) nhé