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(4x-12)(x3+64)=0
=> [x3+64=0=>x=4x-12=0=>4x=12=>x=3 olm bị lỗi nên em đừng có viết cách ra 1 quãng như kia nhé !
vậy x thuộc {3;4}
(3x-12)(x2-4)=0
=>[x2-4=0=>x2=4=>x=2 hoặc x=-23x-12=0=>3x=12=>x=4
vậy x thuộc {4;2;-2}
(x+3)3:3-1=-10
(x+3)3:3=-9
(x+3)3=-9.3
=>(x+3)3=-27
=>x+3=-3
=>x=-6
(3x-1)3-2=-66
(3x-1)3=-64
(3x-1)3=-43
=>3x-1=-4
=>3x=-3
=>x=-1
\(\left(4x-12\right)\left(x^3+64\right)=0\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=0+12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12\div4\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow x^3+64=0\)
\(\Leftrightarrow x^3=0=64\)
\(\Leftrightarrow x^3=\left(-64\right)\)
\(\Leftrightarrow x^3=\left(-4\right)^3\)
\(\Leftrightarrow x=\left(-4\right)\)
\(\Rightarrow x\in\left\{-4;3\right\}\)
\(\Leftrightarrow\left(3x-12\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=0+12\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=12\div3\)
\(x=4\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=0+4\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x\in\left\{2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;2;4\right\}\)
Các câu khác tương tự nhé !
(428-4x) .27=1728
<=> 428-4x = 1728 : 27
<=> 428-4x = 64
<=> 4x = 428 - 64
<=> 4x = 364
<=> x= 364 :4
<=> x= 91
Vậy x= 91
x.x2.x3=64
<=> x1+2+3 = 64
<=> x6 = 64
<=> x6= (+ - 2 )6
=> x = cộng trừ 2
Vậy x= cộng trừ 2
\(318-5\left(x-64\right)=103\)
\(\Rightarrow5\left(x-64\right)=318-103\)
\(\Rightarrow5\left(x-64\right)=215\)
\(\Rightarrow x-64=43\)
\(\Rightarrow x=43+64\)
\(\Rightarrow x=107\)
_____________
\(4^x\cdot5+216=296\)
\(\Rightarrow4^x\cdot5=296-216\)
\(\Rightarrow4^x\cdot5-80\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
___________
\(376-6^x:3=364\)
\(\Rightarrow6^x:3=376-364\)
\(\Rightarrow6^x:3=12\)
\(\Rightarrow6^x=36\)
\(\Rightarrow6^x=6^2\)
\(\Rightarrow x=2\)
___________
\(\left(4x-1\right)^2=121\)
\(\Rightarrow\left(4x-1\right)^2=11^2\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=11\\4x-1=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=12\\4x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
(64-4x).3=24
64-4x=24:3
64-4x=8
4x=64-8
4x=56
x=56:4
x=14
vậy x=14
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
a) 2x-13=15
2x = 15+13
2x = 18
x = 18:2
x =9
c. /3x-7/=2
+) Nếu 3x-7=2 thì x = 3
+)Nếu 3x-7=-2 thì x = 5/3 không thuộc Z
Vậy x = 3
a,\(\left(3x-4\right).2^3=64\)
(=) \(\left(3x-4\right)=64:8\)
(=) \(3x-4=8\)
(=) \(3x=8+4\)
(=) \(3x=12\)
(=) \(x=12:3\)
(=) \(x=4\)
b, Ta có \(2^3.3^2-4x=4^3\)
(=) \(8.9-4x=64\)
(=) \(72-4x=64\)
(=) \(4x=72-64\)
(=) \(4x=8\)
(=) \(x=8:4\)
(=) \(x=2\)
a) ( 3x - 4 ) . 23 = 64
=> ( 3x - 4 ) = 64 : 23
=> ( 3x - 4 ) = 26 : 23
=> ( 3x - 4 ) = 23
=> 3x - 4 = 8
=> 3x = 12
=> x = 4
b) 23 . 32 - 4x = 43
=> 8 . 9 - 4x = 64
=> 72 - 4x = 64
=> 4x = 8
=> x = 2
a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)
\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)
\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)
\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)
Vậy \(x=7\)
c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)
\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)
\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
a: 3x=81
nên x=27
b: \(5\cdot4^x=80\)
\(\Leftrightarrow4^x=16\)
hay x=2
c: \(2^x=4^5:4^3\)
\(\Leftrightarrow2^x=2^4\)
hay x=4
a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50
b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1
c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10
d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13
ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0