(x-3).(y-2)=6
Giúp mik nhanh nhé
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\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{2y-4}{6}=\dfrac{x-1+2y-4-z+2}{5+6-2}=\dfrac{6-5}{9}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}9x-9=5\\9y-18=3\\9z-18=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{9}\\y=\dfrac{7}{3}\\z=\dfrac{20}{9}\end{matrix}\right.\)
0,5x-\(\dfrac{2}{3}.\left(x-2\right)=\dfrac{1}{6}\)
\(\dfrac{1}{2}x-\dfrac{2}{3}x-\dfrac{4}{3}=\dfrac{1}{6}\)
\(x.\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{1}{6}+\dfrac{4}{3}\)
\(x.\dfrac{7}{6}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:\dfrac{7}{6}\)
\(x=\dfrac{9}{7}\)
`0,5x-2/3 .(x-2)=1/6`
`0,5x - 2/3 x + 4/3= 1/6`
` -1/6x = -7/6`
`x=7`
Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^6-y^6\)
\(\Leftrightarrow x^{6n}-y^{6n}=-x^6-y^6\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=-1\\n=1\end{matrix}\right.\Leftrightarrow n\in\varnothing\)
\(a,\dfrac{3}{4}:\dfrac{6}{x}=\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{4}:\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{5}\\ x=6:\dfrac{3}{5}\\ x=10\\ b,\left(x+\dfrac{7}{4}\right)\times\dfrac{2}{3}=6\\ x+\dfrac{7}{4}=6:\dfrac{2}{3}\\x+\dfrac{7}{4}=9\\ x=9-\dfrac{7}{4}\\ x=\dfrac{29}{4}\)
\(\left|16-x\right|-2x+8=-6\)
\(\left|16-x\right|=2x-14\left(x\ge7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}16-x=2x-14\\16-x=14-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=14\left(N\right)\\x=-2\left(L\right)\end{matrix}\right.\)
\(S=\left\{14\right\}\)
x+y+z=6 = 1+2+3 <=> (x-1) +(y-2) +(z-3) = 0
mũ 3 lên ra pt cần CM
Đặt: x - 1 = a; y - 2 = b; c - 3 = z
=> a + b + c = 0
=> a + b = - c
=> (a + b)3 = - c3
a3 + b3 + c3
= a3 + b3 - (a + b)3
= a3 + b3 - a3 - 3ab(a + b) - b3
= - 3ab(a + b) = - 3ab(-c) = 3abc
Thay trở lại đc:
\(\left(x-1\right)^3+\left(y-2\right)^3+\left(z-3\right)^3=3\left(x-1\right)\left(y-2\right)\left(z-3\right)\)