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Tìm x:
a) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=-\dfrac{7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(-1\dfrac{1}{5}+x=\dfrac{-9}{10}\)
\(\Rightarrow x=\dfrac{3}{10}\)
b) \(\dfrac{5}{7}+\dfrac{2}{3}x=\dfrac{3}{10}\)
\(\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-87}{140}\)
c) \(\dfrac{-22}{15}x+\dfrac{1}{3}=\left|\dfrac{-2}{3}+\dfrac{1}{5}\right|\)
\(-\dfrac{22}{15}x+\dfrac{1}{3}=\dfrac{7}{15}\)
\(\dfrac{-22}{15}x=\dfrac{4}{15}-\dfrac{1}{3}\)
\(\dfrac{-22}{15}x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-1}{11}\)
a) \(\dfrac{x}{27}=\dfrac{-2}{3,6}\)
=> x. 3,6 = 27. (-2)
=> x.3,6 = -54
x = (-54) : 3,6
x = -15
b) -0,52 : x = -9,36 : 16,38
- 0,52 : x = \(\dfrac{-4}{7}\)
x = \(\dfrac{-4}{7}\) . ( -0,52)
x = \(\dfrac{52}{175}\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
Đặt P(x)=0
\(\Leftrightarrow x\left(x^4+7x^3-9x^2-2x-\dfrac{1}{4}\right)=0\)
=>x=0
Đặt Q(x)=0
\(\Leftrightarrow-5x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}=0\)
hay \(x\in\varnothing\)
I , tìm x :
a, \(\left|x\right|=1,21\)
Ta có : \(\left|x\right|=\left|1,21\right|\rightarrow\left|x\right|=\pm1,21\)
b, \(\dfrac{11}{12}-\left(\dfrac{2}{5}-x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{1}{4}\) => \(x=\dfrac{2}{5}-\dfrac{1}{4}\)
=> \(x=\dfrac{3}{20}\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}\div x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}\div x=\dfrac{-7}{20}\) => \(x=\dfrac{1}{4}\div\dfrac{-7}{20}\)
=> \(x=\dfrac{-5}{7}\)
d,\(3^x=81\)
Ta có 81= \(3^4\)
Vì : \(3^x=3^4\Rightarrow x=4\)
e,\(\dfrac{1}{2}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\left|x\right|-\dfrac{5}{6}=\dfrac{8}{3}:\dfrac{1}{2}\)
=> \(\left|x\right|-\dfrac{5}{2}=\dfrac{16}{3}\) => \(\left|x\right|=\dfrac{16}{3}+\dfrac{5}{2}\)
=> \(\left|x\right|=\dfrac{47}{6}\)
Vì \(\left|x\right|=\left|\dfrac{47}{6}\right|\Rightarrow x=\pm\dfrac{47}{6}\)
f, \(2^{x-3}=4\)
\(2^{x-3}=2^2\)
=> \(x-3=2\)
=> \(x=5\)
a, Ta có \(\left|x\right|=1,21\)
\(\Rightarrow\left[{}\begin{matrix}x=1,21\\x=-1,21\end{matrix}\right.\)
Vậy \(x\in\left\{1,21;-1,21\right\}\)
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
\(x\in Z\Rightarrow2x-17\ne0;Goi:d=\left(x-8,2x-17\right)\Rightarrow\left\{{}\begin{matrix}x-8⋮d\\2x-17⋮d\end{matrix}\right.\Leftrightarrow2\left(x-8\right)-\left(2x-17\right)⋮d\Leftrightarrow1⋮d\Leftrightarrow d=1\Rightarrow\frac{x-8}{2x-17}toigian\forall x\)
\(ĐK:x\ne-1;Goi:d=\left(x-4,x+1\right)\Rightarrow\left\{{}\begin{matrix}x-4⋮d\\x+1⋮d\end{matrix}\right.\Rightarrow x+1-\left(x-4\right)⋮d\Leftrightarrow5⋮d\Rightarrow d\in\left\{1;5\right\}\) gia sư: \(x+1⋮5\Rightarrow x\ne5k+1\left(k\in N\right)\)