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Ta có: \(x-y-z=0\Rightarrow x-z=y,z-y=x,y-x=-z\)
\(B=\left(1-\frac{z}{x}\right)\cdot\left(1-\frac{x}{y}\right)\cdot\left(1-\frac{y}{z}\right)\)
\(\Rightarrow B=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z-y}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=\frac{-xyz}{xyz}=-1\)
x - y - z = 0
=> x = y + z
y = x - z
-z = x - y
Thay vào B ta được :
\(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1-\frac{y}{z}\right)\)
\(=\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1-\frac{x-z}{z}\right)\)
\(=\left(\frac{-y}{x}\right)\left(\frac{z}{y}\right)\left(\frac{-x}{z}\right)\)
\(=\frac{-yz\left(-x\right)}{xyz}\)
\(=\frac{xyz}{xyz}=1\)
Mình k dám chắc nhá
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
câu 1: Câu hỏi của Vương Ái Như - Toán lớp 7 - Học toán với OnlineMath
câu 2:
Ta có: \(8^7-2^{18}=2^{21}-2^{18}=2^{17}.\left(2^4-2\right)=2^{17}.14⋮14\)
câu 3:
\(4x=7y=3x\Rightarrow\frac{4x}{84}=\frac{7y}{84}=\frac{3z}{84}\Rightarrow\frac{x}{21}=\frac{y}{12}=\frac{z}{28}=\frac{x+y+z}{21+12+28}=\frac{61}{61}=1\)
\(\Rightarrow x=21,y=12,z=28\)
câu 4:
\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)
\(\Rightarrow a=5.12=60,b=9.5=45,c=8.5=40\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
bạn biết đề trước sao ?
ko!!!!!!!!Thi roi ma!!!!!!!!!!