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a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
\(5x+1\text{⋮}2x+3\)
\(\text{⇒}2\left(5x+1\right)\text{⋮}2x+3\)
\(\text{⇒}10x+2\text{⋮}2x+3\)
\(\text{⇒}10x+15-13\text{⋮}2x+3\)
\(\text{⇒}5\left(2x+3\right)-13\text{⋮}2x+3\)
\(\text{⇒}13\text{⋮}2x+3\)
\(\text{⇒}2x+3\text{∈}Ư\left(13\right)=\left\{13;-13;1;-1\right\}\)
\(\text{⇒}2x\text{∈ }\left\{10;-16;-2;-4\right\}\)
\(\text{⇒}x\text{∈}\left\{5;-8;-1;-2\right\}\)
Vậy \(x\text{∈}\left\{5;8;-1;-2\right\}\)
a, ( 1+x )^3 = (2x)^3
b, ( x-1 )^2=16
c, (x+1)^2=25
d, 4x^3+15=47
e,(2x-1)^5=x^5
Mn giải nhanh giúp mk vs
a,\(\left(1+x\right)^3=\left(2x\right)^3\)
=>\(1+x=2x\)
=>\(x-2x=-1\)
=>\(-x=-1\)
=>\(x=1\)
vậy \(x=1\)
b,\(\left(x-1\right)^2=16\)
=>\(\left(x-1\right)^2=4^2\)
=>\(x-1=4\)
=>\(x=4+1\)
=>\(x=5\)
Vậy\(x=5\)
c,\(\left(x+1\right)^2=25\)
=>\(\left(x+1\right)^2=5^2\)
=>\(x+1=5\)
=>\(x=5-1\)
=>\(x=4\)
Vậy \(x=4\)
d,\(4x^3+15=47\)
=>\(4x^3=47-15\)
=>\(4x^3=32\)
=>\(x^3=32:4\)
=>\(x^3=8\)
=>\(x^3=2^3\)
=>\(x=2\)
Vậy\(x=2\)
e,\(\left(2x-1\right)^5=x^5\)
=>\(2x-1=x\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy\(x=1\)
ĐÚNG K MÌNH NHA
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
a) \(2\left|x\right|-5=3\)
=> \(2\left|x\right|=3+5\)
=> \(2\left|x\right|=8\)
=> \(\left|x\right|=4\)
=> x = 4 hoặc x = -4
b) \(x-5=\left(-14\right)+2^3\)
=> \(x-5=\left(-14\right)+8\)
=> \(x-5=-6\)
=> \(x=-6+5=-1\)
c) \(10+2x=4^5:4^3\)
=> \(10+2x=4^{5-3}\)
=> \(10+2x=4^2\)
=> \(2x=4^2-10=16-10=6\)
=> \(2x=6\)
=> \(x=3\)
d) (x + 7) - 13 = 4
=> x + 7 = 17
=> x = 17 - 7 = 10
e) \(2x-10=2^4:2^2\)
=> \(2x-10=2^2\)
=> \(2x-10=4\)
=> \(2x=14\)
=> \(x=7\)
a) 2x - 3 = -12
=> 2x = -12 + 3 = -9
=> x = \(-\frac{9}{2}\)
b) \(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
=> \(\frac{1}{2}+2x=-\frac{5}{6}\cdot\frac{3}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\cdot\frac{1}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\)
=> \(2x=-\frac{5}{2}-\frac{1}{2}=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
c) \(1< \frac{x}{5}< 2\)
=> \(\frac{5}{5}< \frac{x}{5}< \frac{10}{5}\)
=> 5 < x < 10
=> x \(\in\){6,7,8,9}
Dù bạn có cho âm vào nx thì nó vẫn sai nhá
d) Đặt \(A=\frac{x+5}{x-2}=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)
=> \(x-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
+) x - 2 = 1 => x = 3(T/M)
x - 2 = -1 => x = -1 +2 = 1(t/m)
x - 2 = 7 => x = 9 (t/m)
x - 2 = -7 => x = -7 + 2 = -5(t/m)
e) làm nốt ...
a,\(2x-3=-12\)
\(< =>2x=-12+3=-9\)
\(< =>x=-\frac{9}{2}\)
b,\(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
\(< =>\frac{1}{2}+\frac{4x}{2}=-\frac{5}{6}.\frac{3}{2}\)\(< =>\frac{4x+1}{2}=-\frac{5}{4}\)
\(< =>\frac{8x+2}{4}=-\frac{5}{4}\)\(< =>8x+2=-5\)
\(< =>8x=-5-2=-7\)\(< =>x=-\frac{7}{8}\)
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
\(\dfrac{1}{2}x=\dfrac{-3}{10}-\dfrac{3}{4}\)
\(\dfrac{1}{2}x=\dfrac{-21}{20}\)
\(x=\dfrac{-21}{20}:\dfrac{1}{2}\)
\(x=\dfrac{-21}{10}\)
`1 / 2 x + 3 / 4 = [-3] / 10`
`1 / 2 x = [-3] / 10 - 3 / 4`
`1 / 2 x = [-21] / 20`
`x = [-21] / 20 : 1 / 2`
`x = [-21] / 10`