Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, (2x-3)3 = -64
=> (2x-3)3 = -43
=> 2x-3=-4
=> 2x = -1
=> x = -1 : 2
=> x = -1/2
b, (2x-3)2 =25
=> (2x-3)2 =5^2
=> 2x-3 = 5
=> 2x = 8
=> x = 4
c, (3x-4)2 =36
=> (3x-4)2 =62
=> 3x-4 = 6
=> 3x = 10
=> x = 3.(3)
d, 2x+1 = 64
=> 2x+1 = 26
=> x+1 = 6
=> x = 5
+) \(\left(x-3\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=4^2\\\left(x-3\right)^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
Vậy x = 7 hoặc x = -1
+) \(\left(1-3x\right)^3=-64\)
\(\Rightarrow\left(1-3x\right)^3=\left(-4\right)^3\)
\(\Rightarrow1-3x=-4\)
\(\Rightarrow3x=1+4\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=5:3\)
\(\Rightarrow x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
+) \(x^{13}=27.x^{10}\)
\(\Rightarrow x^{13}:x^{10}=27\)
\(\Rightarrow x^3=27\)
\(\Rightarrow x^3=3^3\)
\(\Rightarrow x=3\)
Vậy x = 3
+) \(\left(4x-1\right)^2=\left(1-4x\right)^4\)
\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)
\(\Rightarrow\left(4x-1\right)^2-\left(4x-1\right)^4=0\)
\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\\left(4x-1\right)^2=1\end{cases}}\)
TH 1 : \(\left(4x-1\right)^2=0\Rightarrow4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\)
TH 2 : \(\left(4x-1\right)^2=1\Rightarrow\orbr{\begin{cases}4x-1=1\\4x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4x=2\\4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{4};\frac{1}{2};0\right\}\)
_Chúc bạn học tốt_
a, (x-3)^2 = 16
=> (x-3)^2=4^2
=> x-3=4
=> x= 4+3
=> x = 7 .Vậy x =7
b,(1-3x)^3 = 64
=> ( 1-3x)^3 = 4^3
=> 1-3x = 4
=> 3x = 1-4
=> 3x = -3
=> x = -1 . Vậy x = -1
c, x^13 = 27.x^10
=> x^13 : x^10 = 27
=> x^3 = 3^3
=> x = 3 . Vậy x = 3
a. x2 - 1/4 = 0
x2 = 1/4
x2 = (1/2)2
=>x=1/2
b. x2 + 16 = 0
=>x2= -16 (vô lí)
=>ko tồn tại x tm~
c. x3 + 27 = 0
x3= -27
x3= (-3)3
=>x= -3
d. 2x3 - 16 = 0
x3 - 8 = 0
x3=8=23
=>x=2
e.[( - 0,5)3] = 1/64 =>????
h. (2n)2 = 64
22n=26
=>2n=6 => n=3
a) x = 1/2 hoặc x = -1/2
b) Ko có giá trị của x thỏa mãn
c) x = -3
d) x = 2 hoặc x = -2
e) Ko thấy x thì sao giải đc
h) n = 3
a) \(x^3-x^2-4=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)
b) \(x^4-64=\left(x^2-8\right)\left(x^2+8\right)\)
c) \(81x^4+4y^4=\left(9x^2+2y^2\right)^2-36x^2y^2=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)
d) \(x^7-x^2-1=\left(x^2-x+1\right)\left(x^5+x^4-x^2-x-1\right)\)
a) \(2^{x+3}+5.2^{x+2}=224\)
\(2.2^{x+2}+5.2^{x+2}=224\)
\(7.2^{x+2}=224\)
\(2^{x+2}=32=2^5\)
\(x+2=5\Leftrightarrow x=3\)
b) \(4^{x+1}-4^x=48\)
\(4.4^x-4^x=48\)
\(3.4^x=48\)
\(4^x=16=4^2\)
vậy x=2
c) \(2^{2\left(x+1\right)}+3.4^{x+1}=64\)
\(4^{x+1}+3.4^{x+1}=64\)
\(4.4^{x+1}=64\)
\(4^{x+2}=64=4^3\)
x+2=3
x=1
d) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\left(\dfrac{2}{3}\right)^2\right)^3\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
(23 : 4).2x + 1 = 64
23 : 22 . 2x + 1 = 64
2.2x + 1 = 26
2x + 2 = 26
x + 2 = 6
=> x = 4
Vậy x = 4
(23 : 4). 2x+1 = 64
=> (8 : 4) . 2x+1 = 64
=> 2.2x+1 = 64
=> 2x+1 = 32
=> 2x+1 = 25
=> x + 1 = 5 => x = 4
Vậy x = 4
\(\left(2^3\div4\right).2^{\left(x+1\right)}=64\Leftrightarrow\left(8\div4\right).2^x.2^1=2^6\)
\(\Leftrightarrow2^1.2^x.2^1=2^2.2^x=2^6\Leftrightarrow2^{\left(2+x\right)}=2^6\)
\(\Leftrightarrow2+x=6\Leftrightarrow x=6-2=4\)
Vậy \(x=4\)