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Mình ko bít là câu a) (\(\dfrac{x+1}{x+3}\)) + (\(\dfrac{x+12}{x+3}\)) hay (\(x\)\(+\dfrac{1}{x+3}\)) + (\(x+\dfrac{12}{x+3}\))
☺xin lỗi ạ , nhưng mình không biết viết phân số ạ
Nhưng đề đầu của bạn là đúng ý ạ
a) \(x.\left(x+7\right)=0\)
TH1:\(x=0\)
\(\Rightarrow0.\left(0+7\right)=0.7=0\)
\(TH2:x+7=0\)
\(\Rightarrow x=0-7\)
\(\Rightarrow x=-7\)
Vậy \(x\in\left\{-7;0\right\}\)
b) \(\left(x+12\right).\left(x-3\right)\)
\(TH1:x+12=0\)
\(\Rightarrow x=0-12\)
\(\Rightarrow x=-12\)
\(TH2:x-3=0\)
\(\Rightarrow x=0+3\)
\(\Rightarrow x=3\)
Vậy\(x\in\left\{-12;3\right\}\)
a)
− 12 . x = − 15 . − 4 − 12 − 12 . x = 60 − 12 − 12 . x = 48 x = 48 : − 12 x = − 4
b)
− 9 . x + 3 = − 2 . − 7 + 16 − 9 . x + 3 = 14 + 16 − 9 . x + 3 = 30 − 9 . x = 30 − 3 − 9 . x = 27 x = 27 : − 9 x = − 3
c)
− 12 . x − 34 = 2 − 12 . x = 2 + 34 − 12 . x = 36 x = 36 : − 12 x = − 3
a)
− 12 . x = − 15 . − 4 − 12 − 12 . x = 60 − 12 − 12 . x = 48 x = 48 : − 12 x = − 4
b)
− 9 . x + 3 = − 2 . − 7 + 16 − 9 . x + 3 = 14 + 16 − 9 . x + 3 = 30 − 9 . x = 30 − 3 − 9 . x = 27 x = 27 : − 9 x = − 3
c)
− 12 . x − 34 = 2 − 12 . x = 2 + 34 − 12 . x = 36 x = 36 : − 12 x = − 3
\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)
\(b,\left(10-x\right)\left(3-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
a: \(\Leftrightarrow\left(2x+1\right)^3=8\cdot25-75=125\)
=>2x+1=5
hay x=2
c: x=2; y=0
x = 0 nha bạn
Chúc bạn hok tốt
T.I.C.K cho mình nha