Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) làm lúc này
b) 3x3=9x
=> 3x2=9
=> x2=3
=> x=\(\sqrt{3}\)
c)
c, (x - 3 )2=(3-x)3
=> (x-3)2= - (x-3)3
=> - (x-3)3: (x-3)2=1
=> -x-3=1
=> -x=2
=> x=-2
lik-e minh nhé
a) (x2 - 121) . (2x + 3) = 0
=>x2-121=0 hoặc 2x+3=0
+)Nếu x2-121=0
=>x2=0+121=121
=>x2=(-11)2 hoặc x2=112
=>x=-11 hoặc x=11
+)Nếu 2x+3=0
=>2x=0-3=-3
=>x=(-3):2=\(\frac{-3}{2}\)
Vậy x=-11 hoặc x=11 hoặc x=\(\frac{-3}{2}\)
b) 2x2 - 8x = 0
=>2x(x-4)=0
=>x=0 hoặc x-4=0
Nếu x-4=0
=>x=0+4=4
Vậy x=0 hoặc x=4
c) (3x + 1)5 = (3x + 1)4
=>(3x+1)5:(3x+1)4=(3x+1)4:(3x+1)4
=>3x+1=1
=>3x=1-1=0
=>x=0:3=0
Vậy x=0
a)(x2 - 121) . (2x + 3) = 0
=>x2-121=0 hoặc 2x+3=0
- Với x2-121=0
<=>x2=121 <=>x=±11
- Với 2x+3=0
<=>2x=-3 <=>x=-3/2
b) 2x2 - 8x = 0
=>2x(x-4)=0
=>2x=0 hoặc x-4=0
=>x=0 hoặc x=4
a, \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x = 2 hoặc x = -3
b, \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2x-5=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy...
1.
a) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x=2 hoặc x=-3
b) \(8x^3-50x=0\)
\(x\left(8x^2-50\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}=\left(\pm\dfrac{5}{2}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{-5}{2}\) hoặc \(x=\dfrac{5}{2}\)
tik mik nhé !!!
\(\text{2^5.3 - 3.(x+1)=42}\)
\(3\left(x+1\right)=54\)
\(x+1=18\)
\(x=17\)
\(\text{(4x + 5) : 3 - 11^2:11=2^2}\)
\((4x+5):3-11=4\)
\((4x+5):3=15\)
\((4x+5)=5\)
\(4x=0\)
\(x=0\)
\(\text{3x - 15 . 4 = 6.(19 - x)}\)
\(\text{3x - 60 = 6.(19 - x)}\)
\(\text{3x - 60 :(19-x)= 6}\)
\(\Rightarrow x=6\)
\(\text{(x - 7).(2x - 16)=0}\)
\(\Rightarrow\hept{\begin{cases}x-7=0\\2x-16=0\end{cases}\Rightarrow\hept{\begin{cases}x=7\\8\end{cases}}}\)
a) \(\left(x-1\right)^3=125\)
\(\Leftrightarrow\left(x-1\right)^3=5^3\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
b) \(2^{x+2}-2^x=96\)
\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)
\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)
\(\Leftrightarrow3\cdot2^x=96\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
c) \(\left(2x+1\right)^3=343\)
\(\Leftrightarrow\left(2x+1\right)^3=7^3\)
\(\Leftrightarrow2x+1=7\)
\(\Leftrightarrow2x=7-1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)
\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)
\(\Leftrightarrow720:\left(46-2x\right)=40\)
\(\Leftrightarrow\dfrac{720}{46-2x}=40\)
\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)
\(\Leftrightarrow\dfrac{360}{23-x}=40\)
\(\Leftrightarrow360=40\left(23-x\right)\)
\(\Leftrightarrow9=23-x\)
\(\Leftrightarrow x=23-9\)
\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
e) \(2^x\cdot7=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
f) \(\left(3x+5\right)^2=289\)
\(\Leftrightarrow3x+5=\pm17\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{22}{3};x_2=4\)
a)\(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Leftrightarrow x-1=5\Leftrightarrow x=6\)b)\(2^{x+2}-2^x=96\Leftrightarrow2^x.2^2-2^x=96\Leftrightarrow2^x\left(2^2-1\right)=96\Leftrightarrow2^x.3=96\Leftrightarrow2^x=32\Leftrightarrow x=5\)c)\(\left(2x-1\right)^3=343\Leftrightarrow\left(2x-1\right)^3=7^3\Leftrightarrow2x-1=7\Rightarrow2x=8\Rightarrow x=4\)d)\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=40\Leftrightarrow\left[41-\left(2x-5\right)\right]=720:40=18\)
\(\Leftrightarrow41-2x+5=18\Leftrightarrow36-2x=18\Leftrightarrow2x=18\Leftrightarrow x=9\)
e)\(2^x.7=224\Leftrightarrow2^x=224:7=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
f) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)
x2+2x2+3x2+.......+50x2=10200
=> x2(1+2+3+4+....+50)=10200
=> x2.1275=10200
=> x2=10200:1275
=> x2=8
??????????