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1
A)Z ; Q B)Q C)Q D)Q E)N ; Z ; Q
2
A)> B)< C)< D)<
Bài 1:
a) \(\mathbb{Z}\)
b) \(\mathbb{Q}\)
c) \(\mathbb{Q}\)
d) \(\mathbb{Z}\)
e) \(\mathbb{N}\)
Bài 2:
a) Ta thấy: \(\frac{1}{8}>0; \frac{-3}{8}< 0\) \(\Rightarrow \frac{1}{8}> \frac{-3}{8}\)
b) \(\frac{-3}{7}< 0; 2\frac{1}{2}>0\Rightarrow \frac{-3}{7}< 2\frac{1}{2}\)
c) \(-3,9< 0< 0,1\)
d) \(-2,3< 0< 3,2\)
a) \(-1\in Z\Rightarrow-1\in Q\)
b \(\frac{7}{123}\in Q\)
c) \(3,05\in Q\)
d) \(-\frac{2}{3}\in Q\)
e) \(1035\in N\Rightarrow1035\in Z\Rightarrow1035\in Q\)
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
a)\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)
\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)
\(\dfrac{1}{3}< x< \dfrac{81}{14}\)
Vì\(\dfrac{1}{3}=0,333333333333333333333333...\)
\(\dfrac{81}{14}=5,785714286\)
=>\(x=\left\{1;2;3;4;5\right\}\)
b)\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}\)
\(-\dfrac{1}{12}< x< \dfrac{1}{8}\)
Vì\(-\dfrac{1}{12}=-0.08333333333333333\)
\(\dfrac{1}{8}=0.125\)
=> \(x=\left\{0\right\}\)
a.\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)
\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)
\(\dfrac{1}{3}< x< \dfrac{81}{14}\)
\(0,3333...< x< 5,7857...\)
Vì \(x\in Z\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
Vậy........
b. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(-0,0833...< x< 0,125\)
Vì \(x\in Z\Rightarrow x\in\left\{0\right\}\)
Vậy............
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{-10-6}{-8}=\dfrac{-16}{-8}=2\)\(\Rightarrow\left\{{}\begin{matrix}x=2.2+1=5\\y=2.3+2=8\\z=2.4+3=11\end{matrix}\right.\)
Theo đề bài ta có:
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=k\)
ta có:
\(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=k^3=\dfrac{a}{d}\)
Và \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Ta có đpcm
\(-1\in Z;-1\in Q\\ \dfrac{7}{123}\in Q\\ 3,05\in Q\\ \dfrac{-2}{3}\in Q\\ 1035\in N;1035\in Z;1035\in Q\)
\(a)-1\in Z;-1\in Q\)
\(b)\dfrac{7}{123}\in Q\)
\(c)3,05\in Q\)
\(d)\dfrac{-2}{3}\in Q\)
\(e)1035\in N;1035\in Z;1035\in Q\)