Tìm x biết:
a. 13329 : x = 3 b. x : 8 = 5678
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a: =>1/3x-2/5x=5
=>-1/15x=5
=>x=-75
b: =>4x=4
=>x=1
c: =>6*3^x-5*3^x=243
=>3^x=243
=>x=5
Lời giải:
a. $8-3x=(-7)^2:(-7)=(-7)$
$\Rightarrow 3x=8-(-7)=15$
$\Rightarrow x=15:3=5$
b.
$18\vdots x, 24\vdots x$ nên $x\in ƯC(18,24)$
$\Rightarrow ƯCLN(18,24)\vdots x$
Hay $6\vdots x$
$\Rightarrow x\in\left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
c: =>x+y-xy=-16
=>x+y-xy-1=-17
=>x(1-y)-(1-y)=-17
=>(1-y)(x-1)=-17
=>(x-1;y-1)=17
=>(x-1;y-1) thuộc {(1;17); (17;1); (-1;-17); (-17;-1)}
=>(x,y) thuộc {(2;18); (18;2); (0;-16); (-16;0)}
b: Tham khảo:
a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)
=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)
=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)
=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)
b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)
=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)
=>\(x\cdot\dfrac{-1}{6}=0,75\)
=>\(x=-0,75\cdot6=-4,5\)
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
a: \(13329:x=3\)
=>\(x=\dfrac{13329}{3}=4443\)
b: x:8=5678
=>\(x=5678\cdot8=45424\)