Bài 3. Tìm x, biết 1. |x| < 10; 2. |x| > 11; 3. |x| ≥ 2x 4. |x| < −3x
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Bài 1:
a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)
\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{6}{7}\)
b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)
\(\Rightarrow x=\dfrac{1}{6}\)
\(10-x=-3\) \(-20-y=5\) \(12-z=-7\)
\(x\)\(=10-\left(-3\right)\) \(y\)\(=\left(-20\right)-5\) \(z\)\(=12-\left(-7\right)\)
\(x\) \(=13\) \(y\) \(=-25\) \(z\) \(=19\)
\(\text{Hok tốt!}\)
\(\text{@Kaito Kid}\)
(x+1)+(x+3)+...+(x+99)=0
Tổng các số hạng là: (99+1):2=50 (số hạng)
=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0
<=> 50.x+=0 <=> 50.x+2500=0 => x=-2500/50=-50
\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)
a: =>4/7<x<23/6
hay \(x\in\left\{1;2;3\right\}\)
b: =>19/10<x<27/5
hay \(x\in\left\{2;3;4;5\right\}\)
a) (125 - x) + 25 = 98
125 - x = 98 - 25
125 - x = 73
x = 125 - 73
x = 52
b) 100 : (x + 20) = 68 : 34
100 : (x + 20) = 2
x + 20 = 100 : 2
x + 20 = 50
x = 50 - 20
x = 30
c) (3x + 12) . 5 = 25 . 4 - 10
(3x + 12) . 5 = 100 - 10
(3x + 12) . 5 = 90
3x + 12 = 90 : 5
3x + 12 = 18
3x = 18 - 12
3x = 6
x = 6 : 2
x = 3
d) 210 : (2x - 3) - 20 = 10
210 : (2x - 3) = 10 + 20
210 : (2x - 3) = 30
2x - 3 = 210 : 30
2x - 3 = 70
2x = 70 + 3
2x = 73
x = 73/2
a, ( 125 - x ) + 25 = 98
⇒ 125 - x = 73
⇒ x = 52.
Vậy..
b, 100 : ( x + 20 ) = 68 : 34
⇒ 100 : ( x + 20 ) = 2
⇒ x + 20 = 50
⇒ x = 30
Vậy...
c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10
⇒ ( 3 . x + 12 ) . 5 = 90
⇒ 3 .x + 12 = 18
⇒ 3x = 6
⇒ x = 2.
Vậy...
d, 210 : ( 2 .x - 3 ) - 20 = 10
⇒ ( 2 .x - 3 ) - 20 = 21
⇒ 2x - 3 = 41
⇒ 2x = 44
⇒ x = 22.
Vậy..
bài 2: (x-3).(y+2) = -5
Vì x, y \(\in\)Z => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}
Ta có bảng:
x-3 | 5 | -5 | -1 | 1 |
y+2 | 1 | -1 | -5 | 5 |
x | 8 | -2 | 2 | 4 |
y | -1 | -3 | -7 | 3 |
bài 3: a(a+2)<0
TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)
TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
Vậy -2<a<0
Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)
TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2
TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại
Vậy 1<a<2
1) \(\left|x\right|< 10\)
\(\Leftrightarrow-10< x< 10\)
2) \(\left|x\right|>11\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -11\\x>11\end{matrix}\right.\)
3) \(\left|x\right|\ge2x\left(\forall x\ge0\right)\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x\le-2x\\x\ge2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x\le0\\x\le0\end{matrix}\right.\)
\(\Leftrightarrow x=0\) \(\left(thỏa.đk:x\ge0\right)\)
4) \(\left|x\right|\le-3x\left(\forall x\le0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\left(-3x\right)\\x\le-3x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\le0\\4x\le0\end{matrix}\right.\)
\(\Leftrightarrow x\le0\) \(\left(thỏa.đk\right)\)