1, tìm x biết
\(x^2+5x=0\)
(x-1)(x+2)>0
(x-1)(x+2)<0
\(3\left(2x+3\right)\left(3x-5\right)<0\)
K
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Những câu hỏi liên quan
6 tháng 2 2016
a/ x2+5x=0
=> x2=5x=0
=> x=0
b/ 3(2x+3)(3x-5)<0
=> 2x+3 và 3x-5 phải khác dấu
x=0
câu này mk chỉ bít kết quả thui thông cảm nha
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
11 tháng 5 2020
\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
26 tháng 2 2020
a) Ta có: x(x-1)<0
\(\Leftrightarrow\)x; x-1 khác dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}x>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\Leftrightarrow0< x< 1\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x< 0\\x-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: 0<x<1
b) Ta có: (2-x)(3x-12)>0
\(\Leftrightarrow\)2-x; 3x-12 cùng dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}2-x>0\\3x-12>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\3x>12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>4\end{matrix}\right.\Leftrightarrow x>4\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}2-x< 0\\3x-12< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\3x< 12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< 4\end{matrix}\right.\Leftrightarrow x< 2\)
Vậy: 2<x<4
c) Ta có: \(\left(x+1\right)^2\cdot\left(5-2x\right)\le0\)
*Trường hợp 1:
\(\left(x+1\right)^2\cdot\left(5-2x\right)< 0\)
\(\Leftrightarrow\)(x+1)2; 5-2x khác dấu
-Trường hợp 1:
\(\left\{{}\begin{matrix}\left(x+1\right)^2< 0\\5-2x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1< 0\\2x< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< \frac{5}{2}\end{matrix}\right.\Leftrightarrow x< 1\)
-Trường hợp 2:
\(\left\{{}\begin{matrix}\left(x+1\right)^2>0\\5-2x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1>0\\2x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>\frac{5}{2}\end{matrix}\right.\Leftrightarrow x>\frac{5}{2}\)
Vậy: \(1< x< \frac{5}{2}\)
câu d tương tự nhé bạn
UT
19 tháng 3 2021
1, x(x-1)=2(x-1)
<=> x(x-1)-2(x-1)=0
<=> (x-2)(x-1)=0
<=>x=2 hoặc x=1
vậy ...
2, (x+2)(2x-3)=x^2 -4
<=>(x+2)(2x-3)=(x-2)(x+2)
<=> (x+2)(2x-3)-(x-2)(x+2)=0
<=> (x+2)(2x-3-x+2)=0
<=> x=-2 hoặc x=1
vây...
3,x^2 +3x +2=0
<=> x^2 +x+2x+2=0
<=>(x+2)(x+1)=0
<=> x=-2 hoặc x=-1
vậy ...
5, x^3+x^2-12x =0
<=> x(x^2+x-12)=0
<=>x(x^2-3x+4x-12)=0
<=>x(x+4)(x-3)=0
<=> x=0 hoặc x=-4 hoặc x=3
vậy ...
a) x2 + 5x = 0 <=> x(x + 5) = 0
=. X = 0 hoặc x = - 5
b) (x - 1).(x + 2) > 0
x - 1 > 0 hoặc x + 2 > 0
=> x > 1 hoặc x > - 2
c) (x - 1).(x + 2) < 0
x - 1 < 0 hoặc x + 2 < 0
=> x < 1 hoặc x < - 2
chọn x = 0
d) 3(2x + 3).(3x - 5) < 0
2x + 3 < 0 hoặc 3x - 5 < 0
=> x < -3/2 hoặc x < 5/3
chọn x <5/3