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a) \(\dfrac{x}{4}=\dfrac{16}{128}\Rightarrow x=\dfrac{4.16}{128}=\dfrac{1}{2}=0,5\)
b)\(1\dfrac{5}{6}=\dfrac{-x}{5}\\ \Leftrightarrow\dfrac{11}{6}=\dfrac{-x}{5}\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\Rightarrow x=-\dfrac{55}{6}\)
c) \(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\Rightarrow x=\dfrac{8.\left(-3,5\right)}{4,25}=-\dfrac{112}{17}\)
a,X/4=16/28
=»x/4=4/7
=4/7.4
=7
b,1/5/6=-x/5
-x/5=11/6
X=11/6.5
X=55/6
\(a,\dfrac{x}{4}=\dfrac{16}{128}\\ \Leftrightarrow128x=4.16\\ \Leftrightarrow128x=64\\ \Leftrightarrow x=\dfrac{1}{2}\)
Vậy...
\(b,1\dfrac{5}{6}=\dfrac{-x}{5}\\ \Leftrightarrow-6x=25\\ \Leftrightarrow x=-\dfrac{25}{6}\)
Vậy...
\(c,4,25:8=-3,5:x\\ \Leftrightarrow4,25x=-3,5.8\\ \Leftrightarrow x=-28:4,25\\ \Leftrightarrow x=-\dfrac{112}{17}\)
Vậy...
Câu a :
\(\dfrac{x}{4}=\dfrac{16}{128}\) \(\Leftrightarrow x.128=4.16\) \(\Rightarrow x=\dfrac{1}{2}\)
Câu b :
\(1\dfrac{5}{6}=-\dfrac{x}{5}\) \(\Leftrightarrow11.5=-x.6\) \(\) \(\Rightarrow-x=-\dfrac{55}{6}\)
Câu c :
\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\) \(\Leftrightarrow4,25.x=8.-3,5\) \(\Rightarrow x=-\dfrac{112}{17}\)
a,Ta có : x.128=16.4
x.128=64
Suy ra : x=1/2
b,\(1\dfrac{5}{6}=\dfrac{11}{6}\)
Ta có ; 11.5=-x.6
-x.6=55
Suy ra : x=-55/6
c,Ta có : 4,25.x=-3,5.8
4,25.x=-28
Suy ra : x=-112/17
a)\(\dfrac{x}{4}=\dfrac{16}{128}\)
\(\Leftrightarrow\dfrac{x}{4}=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1.4}{8}=\dfrac{1}{2}\)
b)\(1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Leftrightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)
\(\Rightarrow-x=\dfrac{11.5}{6}=\dfrac{-55}{6}\)
c)\(4,25:8=-3,5:x\)
\(\Leftrightarrow\dfrac{17}{32}=\dfrac{-3,5}{x}\)
\(\Rightarrow x=\dfrac{\left(-3,5\right).32}{17}=-\dfrac{112}{17}\)
Tìm x
a)\(\dfrac{x}{4}\)=\(\dfrac{16}{128}\)
\(\Rightarrow\)x.128=16.4
x.128=64
x=64:128
x=\(\dfrac{1}{2}\)
Vậy x=\(\dfrac{1}{2}\)
b)\(1\dfrac{5}{6}\)=\(\dfrac{-x}{5}\)
\(\dfrac{11}{6}\)=\(\dfrac{-x}{5}\)
-x:5=\(\dfrac{11}{6}\)
-x=\(\dfrac{11}{6}\).5
-x=\(\dfrac{55}{6}\)
\(\Rightarrow\)x=\(\dfrac{-55}{6}\)
Vậy x=\(\dfrac{-55}{6}\)
c)4,25:8 = -3,5:x
-3,5:x=\(\dfrac{17}{4}\)
x=-3,5:\(\dfrac{17}{4}\)
x=\(\dfrac{14}{17}\)
Vậy x=\(\dfrac{14}{17}\)
a) \(\left|x\right|=3,5\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
b) \(\left|1-x\right|+0,73=3\)
\(\Rightarrow\left|1-x\right|=3-0,73\)
\(\Rightarrow\left|1-x\right|=2,27\)
\(\Rightarrow\left[{}\begin{matrix}1-x=2,27\\1-x=-2,27\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,27\\x=3,27\end{matrix}\right.\)
c) \(\dfrac{-5}{8}+x=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{4}{9}-\left(\dfrac{-5}{8}\right)\)
\(\Rightarrow x=\dfrac{77}{72}\)
1)
a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)
\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)
\(\Rightarrow x=-\dfrac{55}{6}\)
b) 4,25 : 8 = -3,5 : x
\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(x=\dfrac{-3,5.8}{4,25}\)
\(x=\dfrac{-28}{4,25}\)
2.
\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)
\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
Ta có thể lặp đc các tỉ lệ thức sau:
\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)
\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)
\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)
a) \(\frac{x}{4}=\frac{16}{128}\)
\(x.128=4.16\)
\(x.128=64\)
\(x=64:128\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\frac{11}{6}=\frac{-x}{5}\)
\(11.5=\left(-x\right).6\)
\(55=\left(-x\right).6\)
\(-x=55:6\)
\(-x=\frac{55}{6}\)
\(\Rightarrow x=-\frac{55}{6}\)
Vậy \(x=-\frac{55}{6}\)
c) \(4,25:8=-3,5:x\)
\(\frac{17}{32}=-3,5:x\)
\(x=\frac{-35}{10}:\frac{17}{32}\)
\(x=-6\frac{10}{17}\)
Vậy \(x=-6\frac{10}{17}\)
\(a.\)
\(\frac{x}{4}=\frac{16}{128}\)
\(\Rightarrow x=\frac{16.4}{128}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy : \(x=\frac{1}{2}\)
\(b.\)
\(\frac{11}{6}=\frac{-x}{5}\)
\(\Rightarrow x=\frac{11.5}{-6}\)
\(\Rightarrow x=-\frac{55}{6}\)
Vậy : \(x=-\frac{55}{6}\)
\(c.\)
\(\frac{4,25}{8}=\frac{-3,5}{x}\)
\(\Rightarrow x=\frac{-3,5.8}{4,25}\)
\(\Rightarrow x=-6\frac{10}{7}\)
Vậy : \(x=-6\frac{10}{7}\)
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
\(a.\)
\(\dfrac{x}{4}=\dfrac{16}{128}\)
\(\Rightarrow x=\dfrac{16.4}{128}=\dfrac{1}{2}\)
\(b.\)
\(1\dfrac{5}{6}=-\dfrac{x}{5}\)
\(\Rightarrow\dfrac{11}{6}=-\dfrac{x}{5}\)
\(\Rightarrow x=\dfrac{11.\left(-5\right)}{6}=-\dfrac{55}{6}\)
\(c.\)
\(4,25:8=-3,5:x\)
\(\Rightarrow\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(\Rightarrow x=\dfrac{-3,5.8}{4,25}=-\dfrac{112}{17}\)
\(a)\dfrac{x}{4}=\dfrac{16}{128}\)
\(\Rightarrow\dfrac{16}{128}.4=\dfrac{1}{2}\)
\(b)1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Rightarrow x=\dfrac{11}{6}.\left(-5\right)=\dfrac{-55}{6}\)
\(c)4,25:8=-3,5:x\)
\(\Rightarrow\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(\Rightarrow x=\dfrac{-3,5.8}{4,25}=\dfrac{-112}{17}\)
Chúc bạn học tốt!