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a. Vì A thuộc Z
\(\Rightarrow x-2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-3;1;3;7\right\}\)( tm x thuộc Z )
b. Ta có : \(B=\frac{x+2}{x-3}=\frac{x-3+5}{x-3}=1+\frac{5}{x-3}\)
Vì B thuộc Z nên 5 / x - 3 thuộc Z
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;2;4;8\right\}\)( tm x thuộc Z )
c. Ta có : \(C=\frac{x^2-x}{x+1}=\frac{x^2+x-2x+2-2}{x+1}=\frac{x\left(x+1\right)-2x+2-2}{x+1}\)
\(=x-2-\frac{2}{x+1}\)
Vi C thuộc Z nên 2 / x + 1 thuộc Z
\(\Rightarrow x+1\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x\in\left\{-3;-2;0;1\right\}\) ( tm x thuộc Z )
Ta có :
\(A=1+5+5^2+...+5^{32}\)
\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{30}+5^{31}+5^{32}\right)\)
\(A=31+5^3\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)\)
\(A=31+31.5^3+...+31.5^{30}\)
\(A=31\left(1+5^3+...+5^{30}\right)\) chia hết cho 31
Vậy \(A\) chia hết cho 31
\(a)\) Ta có :
\(\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow\)\(ab+ac< ab+bc\)
\(\Leftrightarrow\)\(ac< bc\)
\(\Leftrightarrow\)\(a< b\)
Mà \(a< b\) \(\Rightarrow\) \(\frac{a}{b}< 1\)
Vậy ...
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Ta có : \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2015.5\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{a+c}{c+a}+\frac{b+c}{b+c}=2015.5\)
\(\Leftrightarrow Q+3=2015.5\Rightarrow Q=2015.5-3=10072\)
Ta có : \(\frac{a}{b}=\frac{15}{21}=\frac{135}{189}\)
\(\frac{b}{c}=\frac{9}{12}=\frac{3}{4}=\frac{21}{28}=\frac{189}{252}\)
\(\frac{c}{d}=\frac{9}{11}=\frac{252}{308}\)
\(\Rightarrow a=135\)
\(b=189\)
\(c=252\)
\(d=308\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\), ta được \(a=2k;b=5k;c=7k\)Ta có:
\(\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{4k}{2k+10k-7k}=\frac{4k}{5k}=\frac{4}{5}\)
\(\Rightarrow A=\frac{4}{5}\)
Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\)
=> a = 2k ; b = 5k ; c = 7k . Thay vào A ta được :
\(A=\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{k\left(2-5+7\right)}{k\left(2+2.5-7\right)}=\frac{2-5+7}{2+2.5-7}=\frac{4}{5}\)