tìm x thuộc Q biết:
a, (x+1)(x-2)<0
b, (x-2)(x+\(\frac{2}{3}\))
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(a>\)\(\left(x+2\right)\) thuộc \(Ư\left(20\right)\)
\(\left(x+1\right)\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)
\(+>x+1=1\)
\(\Rightarrow x=0\)
\(+>x+1=2\)
\(\Rightarrow x=1\)
\(+>x+1=4\)
\(\Rightarrow x=3\)
\(+>x+1=5\)
\(\Rightarrow x=4\)
\(+>x+1=10\)
\(\Rightarrow x=9\)
\(+>x+1=20\)
\(\Rightarrow x=19\)
Vậy \(x\in\left\{0;1;3;4;9;19\right\}\)
\(b>\left(x-2\right)\) là ước của 6
\(\left(x-2\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
\(+>x-2=1\)
\(\Rightarrow x=3\)
\(+>x-2=2\)
\(\Rightarrow x=4\)
\(+>x-2=3\)
\(\Rightarrow x=5\)
\(+>x-2=6\)
\(\Rightarrow x=8\)
Vậy \(x\in\left\{3;4;5;8\right\}\)
\(c>\left(2x+3\right)\) là \(Ư\left(10\right)\)
\(\left(2x+3\right)\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)
\(+>2x+3=1\)
\(\Rightarrow x=-1\)
\(+>2x+3=2\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(+>2x+3=5\)
\(\Rightarrow x=1\)
\(+>2x+3=10\)
\(\Rightarrow x=\dfrac{7}{2}\)
Vậy \(x\in\left\{-1;-\dfrac{1}{2};1;\dfrac{7}{2}\right\}\)
-29-9(2x-1)\(^2\)= -110
(=) 9(2x-1)2 = (-29) +110
(=) 9(2x-1)2 = 81
(=) (2x-1)2 =81: 9
(=) (2x-1)2 =9
(=) (2x-1)2 = 32 =(-3)2
\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
vậy : ........
a,\(-29-9\left(2x-1\right)^2=-110\)
\(=>-29+110=9.\left(2x-1\right)^2\)
\(=>81=9.\left(2x-1\right)^2\)
\(=>\left(2x-1\right)^2=9\)
\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)
a)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
trog sach bai tap co do bn
bn là j