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\(a-b=3\Rightarrow a=3+b\) Thay vào B ta được :\(B=\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{\left(3-8\right)+b}{b-5}-\frac{12+3b}{9+b+3}=\frac{b-5}{b-5}-\frac{12+3b}{12+3b}=1-1=0\)
Vậy B = 0
Ta có : \(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)
Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow\left\{\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)
Thay \(a=10k\) và \(b=3k\) vào biểu thức \(A=\frac{3\cdot a-2\cdot b}{a-3\cdot b}\), ta được :
\(A=\frac{3\cdot10k-2\cdot3k}{10k-3\cdot3k}=\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)
Vậy \(A=24\)
Có:
M = \(9+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+...+\frac{2}{8}+\frac{1}{9}\)
= \(9+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)-8\)
= \(1+10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}\right)\)
= \(10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\)
= 10N
=> \(\frac{m}{n}=10\)
\(\frac{x}{2013}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{2013}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{2013}-\left[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\right]=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{2013}-\left[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\right]=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{2013}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{2013}-\frac{3}{8}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2013}=\frac{5}{8}+\frac{3}{8}=1\Rightarrow x=2013\)
Vậy x = 2013
Ta có:
\(A+B=11\left(A-B\right)\)
\(\Rightarrow A+B=11A-11B\)
\(\Rightarrow\) B+11B=11A-A
Suy ra : 12B=10A
\(\Rightarrow\frac{A}{B}=\frac{10}{12}=\frac{6}{5}\)
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
= \(\frac{y+z+1+z+x+2+x+y-3}{x+y+z}\)
= \(\frac{2\left(x+y+z\right)}{x+y+z}\)
= 2
+) \(\frac{1}{x+y+z}=2\)
=> \(x+y+z=\frac{1}{2}=0,5\)
\(\Rightarrow\left\{\begin{matrix}x+y=0,5-z\\x+z=0,5-y\\y+z=0,5-x\end{matrix}\right.\)
=> \(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)
\(\Rightarrow\left\{\begin{matrix}\frac{15-x}{x}=2\\\frac{2,5-y}{y}=2\\\frac{-2,5-z}{z}=2\end{matrix}\right.\)
+ \(\frac{15-x}{x}=2\Rightarrow2x+x=15\Rightarrow3x=15\Rightarrow x=5\)
+ \(\frac{2,5-y}{y}=2\Rightarrow2y+y=2,5\Rightarrow3y=2,5\Rightarrow y=\frac{5}{6}\)
+ \(\frac{-2,5-z}{z}=2\Rightarrow2z+z=\left(-2,5\right)\Rightarrow3z=\left(-2,5\right)\Rightarrow z=\frac{-5}{6}\)
mk nhầm một chỗ nha bn : \(\frac{1,5-x}{x}=2\Rightarrow2x+x=1,5\Rightarrow3x=1,5\Rightarrow x=0,5\)
\(S=\left(\frac{1}{7}\right)^2+\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+...+\left(\frac{10}{7}\right)^2\)
\(=\frac{1^2}{7^2}+\frac{2^2}{7^2}+\frac{3^2}{7^2}+...+\frac{10^2}{7^2}\)
\(=\frac{1^2+2^2+3^2+...+10^2}{7^2}\)
\(=\frac{385}{49}=\frac{55}{7}\)
Vậy S = \(\frac{55}{7}\)
Ta có : 49S= \(1^2+2^2+...+10^2\)
49S= 385
S = \(\frac{385}{49}=\frac{55}{7}.\)
a) Có \(\frac{n}{3n+1}=\frac{2n}{2\left(3n+1\right)}=\frac{2n}{6n+2}< \frac{2n}{6n+1}\)
=) \(\frac{n}{3n+1}< \frac{2n}{6n+1}\)
b) Có B < 1 =) \(B< \frac{10^8+1+9}{10^9+1+9}=\frac{10^8+10}{10^9+10}=\frac{10.\left(10^7+1\right)}{10.\left(10^8+1\right)}=\frac{10^7+1}{10^8+1}=A\)
=) B < A
lấy mik mặt cười ở đâu vậy nhắn tin mik nha mik kết bạn nha!!!!