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a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a) (3x-15)7 = 0
3x-15 = 0
3x = 0+15
3x = 15
x = 15:3
x = 5
b) 42x-6 = 1
2x-6 = 0
2x = 0+6
2x = 6
x = 6:2
x = 3
c) Tớ ko bít
d) (x - 6)3 = (x - 6)2
Th1:
x - 6 = 1
x = 1 + 6
x = 7
Th2:
x - 6 = 0
x = 6
Vậy x = 7
x = 6
--thodagbun--
a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5
b, 42x+6=1 <=> 16x=-5 <=>x=-5/16
c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)
TH1: 10x-20 = 0 <=> x=2
TH2: 3-x=1 <=> x=2
Vậy x=2
d, (x-6)^3 = (x-6)^2
<=> (x-6)^2.[(x-6)-1]=0
<=> (x-6)^2=0 hoặc (x-6)-1=0
<=> x=6 hoặc x=7
a,(2x+7)+135=0 b, 1/2x-2/5=1/5
2x+7=0-135 1/2x=1/5+2/5
2x+7=-135 1/2x=3/5
2x=-135-7 x=3/5:1/2
2x=-142 x=6/5
x=-142:2 Vậy x=6/5
x=-71
Vậy x=-71
c, 10-|x+1|=5 d, 1/2x+150%x=2014
|x+1|=10-5 2x=2014
|x+1|=5 x=2014:2
*TH1:x+1=5 *TH2:x+1=-5 x=1007
x=5-1 x=-5-1 Vậy x=1007
x=4 x=-6
Vậy x=4 hoặc x=-6
\(5^{x+1}+5^{x-1}=130\)
\(5^x\cdot5^1+5^x\div5^1=130\)
\(5^x\cdot5^1+5^x\cdot\dfrac{1}{5}=130\)
\(5^x\cdot\left(5+\dfrac{1}{5}\right)=130\)
\(5^x\cdot\dfrac{26}{5}=130\)
\(5^x=130\div\dfrac{26}{5}\)
\(5^x=130\cdot\dfrac{5}{26}\)
\(5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Mọi người còn câu trả lời nào khác không cứ trả lời đi mik tick cho
a: x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b: 2x(x+3)=0
=>x(x+3)=0
=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
c: \(\left(6-x\right)\left(x+10\right)=0\)
=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)
d: \(\left(5x+20\right)\left(x^2+1\right)=0\)
=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)
=>5x=-20
=>x=-4