tính:
\(2\dfrac{3}{13}\) x \(\dfrac{26}{58}\) x 4 x \(2\dfrac{15}{24}\) x \(\dfrac{8}{21}\)
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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: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
a: =13/26*9/18*2/3=2/3*1/4=2/12=1/6
b: =44/22*5/15*3/2=2*1/3*3/2=2*1/2=1
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)
b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)
c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)
c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)
\(\dfrac{21}{24}\cdot\dfrac{2}{11}:\dfrac{9}{8}=\dfrac{21}{24}\cdot\dfrac{2}{11}\cdot\dfrac{8}{9}=\dfrac{16}{99}\)
\(\dfrac{17}{9}\cdot\dfrac{5}{6}:\dfrac{12}{13}=\dfrac{17}{9}\cdot\dfrac{5}{6}\cdot\dfrac{13}{12}=\dfrac{1105}{648}\)
a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)
x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)
x=\(\dfrac{-2}{5}\)
a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)
b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)
\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)
\(=\dfrac{29}{13}\cdot\dfrac{13}{29}\cdot4\cdot\dfrac{21}{8}\cdot\dfrac{8}{21}=4\)