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Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
Bài 2:
a)|x| < 3
x\(\in\){-2;-1;0;1;2}
b)|x - 4 | < 3
x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }
c) | x + 10 | < 2
x\(\in\){ -2 ; -10 }
Bài 1:
A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99
A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]
A = 1617 + [(-1)+(-1)+...+(-1)]
A = 1617 + (-49)
A = +(1617-49) = A = 1568
B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60
B =
2)
a) \(x\in\left\{2;1;0;-1;-2\right\}\)
b) \(x\in\left\{6;-6;5;-5;4\right\}\)
c) \(x\in\left\{-9;-11;-10\right\}\)
3)
\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
Bài 2:
b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)
\(\Leftrightarrow1⋮d\)
\(\Leftrightarrow d\inƯ\left(1\right)\)
\(\Leftrightarrow d\in\left\{1;-1\right\}\)
\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)
hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)
Bài 1:
a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)
\(=75\cdot\left(-4\right)+603\)
\(=603-300=303\)
Bài 2:
a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ
mà số nguyên tố chẵn duy nhất là 2
nên số lẻ còn lại là 599(thỏa ĐK)
Vậy: Hai số nguyên tố cần tìm là 2 và 599
b,Gọi ƯCLN(21n+4,14n+3)=d
21n+4⋮d ⇒42n+8⋮d
14n+3⋮d ⇒42n+9⋮d
(42n+9)-(42n+8)⋮d
1⋮d ⇒ƯCLN(21n+4,14n+3)=1
Vậy phân số 21n+4/14n+3 là phân số tối giản
