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a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)
=>2x=-3/2
hay x=-3/4
b: 2x+3=5
=>2x=2
hay x=1
c: =>3(x-2)=4(5+x)
=>4x+20=3x-6
=>x=-26
a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)
=>2x=-45/22
hay x=-45/44
b: =>x/7=-1/28:1/4=-1/7
=>x=-1
a)(7/2+2x).11/3=16/3
7/2+2x=16/3:11/3
7/2+2x=16/3.3/11
7/2+2x=16/11
2x=16/11-7/2
2x= -45/22
x= -45/22:2
x= -45/44
Vậy x= -45/44
b)x/7 +1/4= -1/28
x/7= -1/28-1/4
x/7= -2/7
=>x= -2
\(a,\left(x-\dfrac{5}{8}\right).\dfrac{5}{8}=-\dfrac{15}{36}\)
\(\left(x-\dfrac{5}{8}\right)=-\dfrac{15}{36}\div\dfrac{5}{8}\)
\(x-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+\dfrac{5}{8}\)
\(x=-\dfrac{1}{24}\)
\(b,\left(x-\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}\)
\(a,\left(x-\dfrac{5}{8}\right)\cdot\dfrac{8}{18}=-\dfrac{15}{16}\\ x-\dfrac{5}{8}=-\dfrac{15}{36}:\dfrac{8}{18}\\ x-\dfrac{5}{8}=-\dfrac{15}{16}\\ x=-\dfrac{15}{16}+\dfrac{5}{8}\\ x=-\dfrac{15}{16}+\dfrac{10}{16}\\ x=-\dfrac{5}{16}\\ b,x-\dfrac{1}{3}=\dfrac{5}{6}\\ x=\dfrac{5}{6}+\dfrac{1}{3}\\ x=\dfrac{5}{6}+\dfrac{2}{6}\\ x=\dfrac{7}{6}\)
a.
\(10⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(10\right)\)
\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)
b.
\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;1;3;9\right\}\)
c.
\(\left(3x+8\right)⋮\left(x-1\right)\)
\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)
\(\Rightarrow3\left(x-1\right)+11⋮x-1\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x=\left\{-10;0;2;12\right\}\)
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
a, | x - 3 | - ( -3 ) = 4
= | x - 3 | + 3 = 4
|x - 3| = 4 - 3
|x - 3| = 1
=> x - 3 = 1 hoặc x - 3 = ( - 1 )
x = 1 + 3 = 4 x = ( -1 ) + 3 = 2
vậy x = 4 hoặc x = 2
b, x - (1 - x) = 5 + (-1+ x )
= x - 1 + x = 5 + -1 + x
= x -x +x = 5 -1 + 1
= x = 5
a)\(\left|x-3\right|-\left(-3\right)=4\)
\(\left|x-3\right|+3=4\)
\(\left|x-3\right|=4-3\)
\(\left|x-3\right|=1\)
\(x-3=1\) hoặc \(x-3=-1\)
Tự giải TIẾP NHA