(x+5).(x-4)=0
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Khách
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20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
4L
2
16 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=5\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\\ c,\Leftrightarrow\left(x+1\right)\left(3x-6\right)=0\\ \Leftrightarrow3\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x-3\right)\left(5x-10\right)=0\\ \Leftrightarrow5\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
16 tháng 10 2021
a) \(\left(x+8\right)\left(x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x+8=0\\x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-8\\x=5\end{matrix}\right.\)
b) \(x\left(x-4\right)+5\left(x-4\right)=0\) \(\Rightarrow\left(x-4\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
c) \(3x\left(x+1\right)-6\left(x+1\right)=0\) \(\Rightarrow\left(3x-6\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-6=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
d) \(5x\left(x-3\right)+10\left(3-x\right)=0\) \(\Rightarrow5x\left(x-3\right)-10\left(x-3\right)=0\)
\(\Rightarrow\left(5x-10\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-10=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
30 tháng 7 2021
a) Ta có: \(x^4-16x^2=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b) Ta có: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4\left(x^4+36\right)=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
c) Ta có: \(\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
d) Ta có: \(5\left(x-2\right)-x^2+4=0\)
\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
DL
3
H
1 tháng 8 2023
\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ \left(x-5\right)\left(25-5x?\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\25-5x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\5x=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=5\end{matrix}\right.\\ \left(x-4\right)\left(2x-8\right)\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\2x-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=4\end{matrix}\right.\)
1 tháng 8 2023
a,(x-2)(4x-20)=0
=>x-2=0 hoặc 4x-20=0
=>x=2 hoặc x=5
b,(x-5)(25-5)=0
=>x-5=0 ( vì 25-5 ≠0)
=>x=5
c,(x-4)(2x-8)=0
=>x-4=0 hoặc 2x-8=0
=>x=4
BL
2
6 tháng 9 2019
Bài làm
a) 2( x + 1 ) - 4x = 6
=> 2x + 2 - 4x = 6
=> ( 2x - 4x ) + 2 = 6
=> -2x + 2 = 6
=> -2x = 4
=> x = -2
Vậy x = -2
b) 3( 2 - x ) + 4( 5 - x ) = 4
=> 6 - 3x + 20 - 4x = 4
=> ( 6 +20 ) + ( -3x - 4x ) = 4
=> 26 - 7x = 4
=> 7x = 22
=> x = 22/7
Vậy x = 22/7
c) Cũng phân tích như hai câu trên rồi rút gọn ra, sử dụng tính chất phân phối đó, do là phân số nên mik k muốn làm.
d) ( x + 1 )( x - 3 ) = 0
=> \(\hept{\begin{cases}x+1=0\Rightarrow x=-1\\x-3=0\Rightarrow x=3\end{cases}}\)
Vậy x = -1; x = 3
# Học tốt #
XO
6 tháng 9 2019
Tìm x biết :
a) \(2\left(x+1\right)-4x=6\)
\(\Rightarrow2x+2-4x=6\)
\(\Rightarrow2x-4x=6-2\)
\(\Rightarrow-2x=4\)
\(\Rightarrow x=-2\)
b) \(3\left(2-x\right)+4\left(5-x\right)=4\)
\(\Rightarrow6-3x+20-4x=4\)
\(\Rightarrow-3x-4x=4-6-20\)
\(\Rightarrow-7x=22\)
\(\Rightarrow x=-\frac{22}{7}\)
c) \(\frac{7}{3}.\left(x-\frac{4}{3}\right)+\frac{2}{5}.\left(4-\frac{1}{3}x\right)=0\)
\(\Rightarrow\frac{7}{3}x-\frac{28}{9}+\frac{8}{5}-\frac{2}{15}x=0\)
\(\Rightarrow\left(\frac{7}{3}x-\frac{2}{15}x\right)-\left(\frac{28}{9}-\frac{8}{5}\right)=0\)
\(\Rightarrow\frac{33}{15}x-\frac{68}{45}=0\)
\(\Rightarrow\frac{33}{15}.x=\frac{68}{45}\)
\(\Rightarrow x=\frac{68}{45}:\frac{33}{15}\)
\(\Rightarrow x=\frac{68}{99}\)
d) \(\left(x+1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
NN
1
ND
Nguyễn Đức Trí
VIP
22 tháng 7 2023
a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{1}{2}\)
c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)
TH
2
LM
Lê Minh Vũ
CTVHS
VIP
19 tháng 9 2023
\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)
Th1:
\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)
\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)
( 1 )
Th2:
\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)
\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)
( 2 )
Từ ( 1 ) và ( 2 ), ta có:
\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)
LM
Lê Minh Vũ
CTVHS
VIP
19 tháng 9 2023
\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
Th1:
\(2-x>0\Leftrightarrow x>2\)
\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)
( Loại )
Th2:
\(2-x< 0\Leftrightarrow x< 2\)
\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)
=> \(\dfrac{4}{5}< x< 2\)
18 tháng 9 2021
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
18 tháng 9 2021
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
NQ
3
22 tháng 12 2021
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
`(x+5).(x-4)=0`
`@TH1:`
`x+5=0`
`x=0-5`
`x=-5`
`@TH2:`
`x-4=0`
`x=0+4`
`x=4`
Vậy `x={-5;4}`
no