Tìm x biết x–8=(–3)–8
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a) \(\dfrac{3}{5}:x=3\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{3}{5}\times\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
b) \(\dfrac{x}{5}:\dfrac{3}{8}=\dfrac{8}{5}\)
\(\dfrac{x}{5}=\dfrac{8}{5}\times\dfrac{3}{8}\)
\(\dfrac{x}{5}=\dfrac{3}{5}\)
\(=>x=3\)
x – 8 = (–3) – 8
x = (–3) – 8 + 8 (chuyển –8 từ vế trái sang vế phải)
x = –3 + 8 – 8
x = –3.
Vậy x = –3.
\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
x.8+56+x.4-28=x.3+9+x+4*(1-3)
=>12x+(56-28)=(3x+x)+9+4*(-2)
=>12x+28=4x+9-8=4x+1
=>12x-4x=1-28
8x=-27
x=-27/8
\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)
\(\frac{8}{3}:x=\frac{16}{9}.\frac{3}{8}\)
\(\frac{8}{3}:x=\frac{2}{3}\)
\(x=\frac{8}{3}:\frac{2}{3}\)
\(x=\frac{8}{3}.\frac{3}{2}\)
\(x=4\)
Vậy x = 4
x-8=-3-8
x-8=-3+(-8)
x-8=-11
x=-11+8
x=-3
vậy..............
x - 8 = -3 - 8
x - 8 = -11
=>x =-11+8 = -3