\(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)
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VM
26 tháng 10 2019
a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)
\(\Rightarrow x=\frac{1}{81}\)
Vậy \(x=\frac{1}{81}.\)
b) \(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)
\(\Rightarrow\frac{297}{164}=x:\frac{75}{90}\)
\(\Rightarrow x=\frac{297}{164}.\frac{75}{90}\)
\(\Rightarrow x=\frac{495}{328}\)
Vậy \(x=\frac{495}{328}.\)
c) \(x+\left|-\frac{1}{2}\right|=3\frac{1}{3}-4\frac{1}{2}\)
\(\Rightarrow x+\frac{1}{2}=\frac{10}{3}-\frac{9}{2}\)
\(\Rightarrow x+\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x=\left(-\frac{7}{6}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!
KS
1
5 tháng 4 2017
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
M
19 tháng 7 2017
a, 2/1/3:1/3=7/9:x b, x:1/3=12/99:15/90
7 = 7/9 : x x :1/3= 8/11
x = 7/9:7 x = 8/11 * 1/3
x = 1/9 x = 8/33
c, 0,15:x=3/1/3:2,25 d, 3/4:0,75=x:75/90
0,15:x =40/27 1 =x:75/90
x = 0,15:40/27 x = 1*75/90
x = 81/800 x = 75/90
e, x/-15=-60/x f, -2/x=-x/8
=>x*x=-15*(-60) => (-2)*8=x*-x
=>x2=900 => -16 = -x2
=>x2=302 hoặc x2=(-30)2 => 16=x2
=> x=30 hoặc x= -30 => x2=42 hoặc x2=(-4)2
=> x=4 hoặc x=-4
19 tháng 7 2017
a)\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(\frac{7}{3}\times3=\frac{7}{9}:x\)
\(7=\frac{7}{9}:x\)
\(x=\frac{7}{9}:7\)
\(x=\frac{7}{9}\times\frac{1}{7}\)
\(x=\frac{1}{9}\)
25 tháng 2 2019
\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)
\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)
\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)
\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)
29 tháng 6 2022
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
\(\frac{3}{4} :\frac{41}{99}=x:\frac{75}{90}\)
\(\frac{297}{164}\) \(=x:\frac{75}{90}\)
\(x=\)\(\frac{297}{164}.\)\(\frac{75}{90}\)
\(x=\frac{495}{328}\)