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a){720:[41-(7.2\(^2\)- X)] }:(2\(^3\).5)=1
{720:[41-(7.4-X)] }:(8.5)=1
{720:[41-(28-X)] }:40=1
41-(28-X):40=1.720
41-(28-X):40=720
(28-X):40=720+41
(28-X):40=716
28-X =716.40
28-X=28640
X=28-28640
X=(-28612)
b)420+65.4=(x+175):5+30
420+260=(x+175):5+30
280=(x+175):5+30
(x+175):5=280-30
(x+175):5=250
x+175=250.5
x+175=1250
x=1250-175
x=1075
c)(32.15):2=(x+70):14-40
480:2=(x+70):14-40
240=(x+70):14-40
(x+70):14=240+40
(x+70):14=280
x+70=280.14
x+70=3920
x=3920-70
x=3850
\(\left(0,25-30\%.x\right).\frac{1}{3}-\frac{1}{4}=-5\frac{1}{6}\)
\(< =>\left(\frac{1}{4}-\frac{3x}{10}\right).\frac{1}{3}-\frac{1}{4}=-\frac{30}{6}\)
\(< =>\left(\frac{1}{4}-\frac{3x}{10}\right).\frac{1}{3}=-5+\frac{1}{4}=-\frac{19}{4}\)
\(< =>\frac{1}{4}-\frac{3x}{10}=\frac{-19}{4}.\frac{3}{1}=-\frac{57}{4}\)
\(< =>\frac{-3x}{10}=\frac{-57}{4}-\frac{1}{4}=-\frac{58}{4}\)
\(< =>-12x=-580\)
\(< =>12x=580< =>x=\frac{580}{12}=\frac{145}{3}\)
\(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=-5\frac{1}{6}\)
\(\left(\frac{1}{4}-\frac{3}{10}x\right)\frac{1}{3}-\frac{1}{4}=\frac{-31}{6}\)
\(\frac{1}{4}-\frac{3}{10}x=\left(\frac{-31}{6}+\frac{1}{4}\right)\div\frac{1}{3}\)
\(\frac{1}{4}-\frac{3}{10}x=\frac{-59}{4}\)
\(\frac{3}{10}x=\frac{1}{4}-\frac{-59}{4}\)
\(\frac{3}{10}x=15\Leftrightarrow x=50\)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(1,\Leftrightarrow7x=42\Leftrightarrow x=6\\ 2,\Leftrightarrow36-4x=4\Leftrightarrow4x=32\Leftrightarrow x=8\\ 3,\Leftrightarrow3x=35\Leftrightarrow x=\dfrac{35}{3}\\ 4,\Leftrightarrow x-12=144\Leftrightarrow x=156\\ 5,\Leftrightarrow x-14=16\Leftrightarrow x=30\\ 6,\Leftrightarrow3x-24=\dfrac{148}{73}\Leftrightarrow3x=\dfrac{1900}{73}\Leftrightarrow x=\dfrac{1900}{219}\\ 7,\Leftrightarrow33+x=45\Leftrightarrow x=12\\ 8,Sai.đề\\ 9,\Leftrightarrow\left(x+9\right):2=39\Leftrightarrow x+9=78\Leftrightarrow x=69\\ 11,\Leftrightarrow2\left(x+7\right)=38\Leftrightarrow x+7=19\Leftrightarrow x=12\\ 13,\Leftrightarrow2\left(x-51\right)=66\Leftrightarrow x-51=33\Leftrightarrow x=84\)
\(15,\Leftrightarrow x-19=9\Leftrightarrow x=28\\ 17,\Leftrightarrow\left(x-3\right):2=48\Leftrightarrow x-3=96\Leftrightarrow x=99\\ 19,\Leftrightarrow0x=46\Leftrightarrow x\in\varnothing\\ 8,Sai.đề\\ 14,\Leftrightarrow2x=209\Leftrightarrow x=\dfrac{209}{2}\\ 16,\Leftrightarrow2x+6=157\Leftrightarrow2x=151\Leftrightarrow x=\dfrac{151}{2}\\ 18,\Leftrightarrow5\left(x+4\right)=100\Leftrightarrow x+4=20\Leftrightarrow x=16\\ 20,\Leftrightarrow\left(3x-5\right)^3=27=3^3\Leftrightarrow3x-5=3\Leftrightarrow x=\dfrac{8}{3}\)
