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a) (x + 5)(2x - 4) = 0
x + 5 = 0 hoặc 2x - 4 = 0
*) x + 5 = 0
x = -5
*) 2x - 4 = 0
2x = 4
x = 4 : 2
x = 2
Vậy x = -5; x = 2
b) (x - 3)(5x - 10) = 0
x - 3 = 0 hoặc 5x - 10 = 0
*) x - 3 = 0
x = 3
*) 5x - 10 = 0
5x = 10
x = 10 : 5
x = 2
Vậy x = 2; x = 3
A , x+5 hoặc 2x-4 =0
tương tự bài b
CHÚC BẠN HỌC TỐT
`|x+3|+10-5x=0`
`<=>|x+3|=5x-10(x>=2)`
`+)x+3=5x-10`
`<=>4x=13`
`<=>x=13/4(tm)`
`+)x-3=10-5x`
`<=>6x=13`
`<=>x=13/6(tm)`
Vậy `S={13/4,13/6}`
\(\left|x+3\right|+10-5x=0\)
\(\Leftrightarrow\left|x+3\right|=5x-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-10\ge0\\\left[{}\begin{matrix}x+3=5x-10\\x+3=10-5x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{13}{4}\left(N\right)\\x=\dfrac{7}{6}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)
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) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
a) \(\left(3-x\right)\left(5x+10\right)=0\)
\(\left(3-x\right).5.\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
vậy...
b) \(\left|5x+2\right|-4x=7\)
\(\left|5x+2\right|=7+4x\)
\(\Rightarrow\orbr{\begin{cases}5x+2=7+4x\\5x+2=-7-4x\end{cases}}\Rightarrow\orbr{\begin{cases}5x-4x=7-2\\5x+4x=-7-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\9x=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
vậy....
k nha
a, \(\left(3-x\right)\left(5x+10\right)=0\)
\(\Rightarrow3-x=0\) hoặc \(5x+10=0\)
\(\Rightarrow x=3\) hoặc \(x=-2\)
\(a,80-\left(10x-5\right)=45\\ \Rightarrow80-10x+5=45\\ \Rightarrow-10x=-40\\ \Rightarrow x=4\)
\(b,\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(c,\left|5+5x\right|=2^2.5\\ \Rightarrow\left|5+5x\right|=20\\ \Rightarrow\left[{}\begin{matrix}5+5x=20\\5+5x=-20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x=15\\5x=-25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(d,-10-\left(-5\right)+\left(3-x\right)=-8\\ \Rightarrow-10+5+3-x=-8\\ \Rightarrow-x=-6\\ \Rightarrow x=6\)
a) 80-(10x-5)=45
=> 80 - 10 x + 5 =45
=>-10x = -40
=>x=4
b)(x+1)×(x-2)=0
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x=2+0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
c) |5+5x|=2^2×5
=>|5+5x|=20
\(\Rightarrow\left[{}\begin{matrix}5+5x=-20\\5+5x=20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
d) -10-(-5)+(3-x)=-8
=>-10+5 +3-x=-8
=>2+x=8
=>x=6
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
a) (5x - 12 ) : 4 = 318 : 315
( 5x - 12 ) : 4 = 33
( 5x - 12 ) : 4 = 27
5x - 12 = 108
5x = 120
x = 24
b) 5x3 - 102 = 35
5x3 - 100 = 35
5x3 = 135
x3 = 27
x3 = 33
=> x = 3
c) 65 - 4x+2 = 20180
65 - 4x+2 = 1
4x+2 = 64
4x+2 = 43
x + 2 = 3
=> x = 1
d) 2x+1 - 2x = 32
2x . ( 2 - 1 ) = 32
2x . 1 = 32
2x = 32
2x = 25
=> x = 5
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
( x - 3) - ( 10 - 5x ) = 0
x - 3 = 10 - 5x
-10 - 3 = -5x - x
-13 = -6x
=> x = 13/6
(x-3)-(10-5x)=0
x-3-10+5x=0
6x-13=0
6x=13
x=13/6