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Ta có :
\(A=\dfrac{34}{7.13}+\dfrac{51}{13.22}+\dfrac{85}{22.37}+\dfrac{68}{37.49}\)
\(\dfrac{A}{17}=\dfrac{2}{7.13}+\dfrac{3}{13.22}+\dfrac{5}{22.37}+\dfrac{4}{37.49}\)
\(A.\dfrac{3}{17}=\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}\)
\(A.\dfrac{3}{17}=\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\)
\(A.\dfrac{3}{17}=\dfrac{1}{7}-\dfrac{1}{49}\)
\(A.\dfrac{3}{17}=\dfrac{6}{49}\)
\(\Rightarrow A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{34}{49}\)
\(B=\dfrac{39}{7.16}+\dfrac{65}{16.31}+\dfrac{52}{31.43}+\dfrac{26}{37.49}\)
\(\dfrac{B}{13}=\dfrac{3}{7.16}+\dfrac{5}{16.31}+\dfrac{4}{31.43}+\dfrac{2}{37.49}\)
\(B.\dfrac{3}{13}=\dfrac{9}{7.16}+\dfrac{15}{16.31}+\dfrac{12}{31.43}+\dfrac{6}{43.49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)
\(\Rightarrow B=\dfrac{6}{49}:\dfrac{3}{13}=\dfrac{26}{49}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{34}{49}:\dfrac{26}{49}=\dfrac{17}{13}\)
Chúc bn học tốt!!!!!!!!!
Link này bạn: Câu hỏi của Quỳnh Anh Shuy - Toán lớp 7 | Học trực tuyến
1. Ta có : (\(\dfrac{-3}{8}\))3 < 0
(\(\dfrac{8}{243}\))3 > 0
=> (\(\dfrac{-3}{8}\))3 < (\(\dfrac{8}{243}\))3
@Cuber Việt
\(\left(\dfrac{-3}{8}\right)^3< 0< \left(\dfrac{8}{243}\right)^3\)
Vậy \(\left(\dfrac{-3}{8}\right)^3< \left(\dfrac{8}{243}\right)^3\)
\(A=\dfrac{34}{7\cdot13}+\dfrac{51}{13\cdot22}+\dfrac{85}{22\cdot37}+\dfrac{68}{37\cdot49}\\ =\dfrac{17}{3}\cdot\dfrac{6}{7\cdot13}+\dfrac{17}{3}\cdot\dfrac{9}{13\cdot22}+\dfrac{17}{3}\cdot\dfrac{15}{22\cdot37}+\dfrac{17}{3}\cdot\dfrac{12}{37\cdot49}\\ =\dfrac{17}{3}\cdot\left(\dfrac{6}{7\cdot13}+\dfrac{9}{13\cdot22}+\dfrac{15}{22\cdot37}+\dfrac{12}{37\cdot49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\dfrac{6}{49}\\ =\dfrac{34}{49}\)
a: -51/136=-3/8=-27/72
-60/108=-5/9=-40/72
26/-156=-1/6=-12/72
b: -14/35=-2/5
-26/65=-2/5
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)
=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)
=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)
=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
\(M=\dfrac{26}{7.11}+\dfrac{65}{11.21}+\dfrac{39}{21.27}+\dfrac{143}{27.49}\)
\(M=\dfrac{13.2}{7.11}+\dfrac{13.5}{11.21}+\dfrac{13.3}{21.27}+\dfrac{13.11}{27.49}\)
\(M=13.\dfrac{2}{7.11}+13.\dfrac{5}{11.21}+13.\dfrac{3}{21.27}+13.\dfrac{11}{27.49}\)
\(M=13.\left(\dfrac{2}{7.11}+\dfrac{5}{11.21}+\dfrac{3}{21.27}+\dfrac{11}{27.49}\right)\)
\(M=13.\dfrac{1}{2}\left(\dfrac{4}{7.11}+\dfrac{10}{11.21}+\dfrac{6}{21.27}+\dfrac{22}{27.49}\right)\)
\(M=\dfrac{13}{2}\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\)
\(M=\dfrac{13}{2}.\dfrac{6}{49}=\dfrac{39}{49}\)
N sai đề rồi chuyển 35 thành 85 tính tương tự xong lấy \(P=\dfrac{M}{N}+\dfrac{34}{29}\)
P=2
M=26/7x11+65/11x21+39/21x27+143/37x49
M=26/77+65/231+13/189+143/1813
M=0,766705481
N=51/7x16+17/16x19+35/19x34+85/34x49
N=51/112+17/304+35/646+5/98
N=0,6164781702
P=0,766705481/0,6164781702+29/34
P=2,096627519