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a) \(3\left(2x-5\right)+125=134\)
\(\Leftrightarrow3\left(2x-5\right)=9\)
\(\Leftrightarrow2x-5=3\)
\(\Leftrightarrow2x=8\Leftrightarrow x=4\)
b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)
\(\Leftrightarrow6x+9=27\)
\(\Leftrightarrow6x=18\Leftrightarrow x=3\)
d) \(27\left(x-27\right)-27=0\)
\(\Leftrightarrow27\left(x-27\right)=27\)
\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
a) \(\left(2x+1\right)^3=27\)
\(\Leftrightarrow2x+1=3\)
\(\Leftrightarrow x=1\)
b) \(\left(2x-1\right)^3=125\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow x=3\)
c) \(\left(x+1\right)^4=\left(2x\right)^4\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow x=1\)
d) \(\left(2x-1\right)^5=x^5\)
\(\Leftrightarrow2x-1=x\)
\(\Leftrightarrow x=1\)
a. ( 2x + 1 )3 = 27
<=> ( 2x + 1 )3 = 33
<=> 2x + 1 = 3
<=> 2x = 2
<=> x = 1
b. ( 2x - 1 )3 = 125
<=> ( 2x - 1 )3 = 53
<=> 2x - 1 = 5
<=> 2x = 6
<=> x = 3
c. ( x + 1 )4 = 2x4
<=> x + 1 = 2x
<=> x = 1
d. ( 2x - 1 )5 = x5
<=> 2x - 1 = x
<=> x = 1
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
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 \(x=0\)
bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;
a) x + 12 = (-5) - x
=> chuyển vế : x + x = -12 - 5
=> 2x = -17
=> x = \(-\frac{17}{2}\)
b) 2 . (x - 1) + 3 . (x-2) = x - 4
=> 2x - 2 + 3x - 6 = x - 4
=> chuyển vế: 2x +3x - x = 2 + 6 - 4
=> 4x = 4 => x = 1
c) 4 . (2x +7) - 3 . (3x-2) = 24
=> 8x + 28 - 9x +6 = 24
=> - x = -28 - 6 + 24
=> x = 10
d) 3 . ( x-2) + 2x = 10 (Ý bạn c là x đúng không?)
=> 3x - 6 + 2x = 10
=> 5x = 6 + 10
=> 5x = 16 => x = \(\frac{16}{5}\)
Đúng thì k mik nha. Thanks!
\(a,x+12=\left(-5\right)-x\)
\(x+x=12-5\)
\(2x=7\Leftrightarrow x=\frac{7}{2}\)
\(b,2\left(x-1\right)+3\left(x-2\right)=x-4\)
\(2x-2+3x-6=x-4\)
\(2x+3x-x=-4+2+6\)
\(4x=4\Leftrightarrow x=1\)
\(c,4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(4x+28-9x+6=24\)
\(4x-9x=24-28-6\)
\(-5x=-10\Leftrightarrow x=2\)
\(d,3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x=10+6\)
\(5x=16\Leftrightarrow x=\frac{16}{5}\)
a,\(\left(x-1\right)^2=4=2^2=\left(-2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
b,\(\left(2x+1\right)^3=27=3^3\)
\(\Rightarrow2x+1=3\Rightarrow x=1\)
c,\(\left(2x-1\right)^5=x^5\Rightarrow2x-1=x\Rightarrow x=1\)
\(a,\left(x-1\right)^2=4\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)
\(b,\left(2x+1\right)^3=27\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow x=1\)
\(c,\left(2x-1\right)^5=x^5\)
\(\Rightarrow2x-1=x\)
\(\Rightarrow2x-x-1=0\)
\(\Rightarrow x=1\)
Học tốt!