Tìm x biết x3 =64
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x3 = 64
Ta có: 64 = 43. Do đó x3 = 43 nên x = 4.
Vậy x = 4 là giá trị cần tìm.
Ta có
x 3 – 12 x 2 + 48 x – 64 = 0 ⇔ x 3 – 3 . x 2 . 4 + 3 . x . 4 2 – 4 3 = 0 ⇔ ( x – 4 ) 3 = 0
ó x – 4 = 0 ó x = 4
Vậy x = 4
Đáp án cần chọn là: B
\(ĐK:x\ge0\\ PT\Leftrightarrow\left(x-\dfrac{3}{4}\right)\left(x^2+\dfrac{3}{4}x+\dfrac{9}{16}\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(n\right)\\\sqrt{x}=3\left(n\right)\\x^2+2\cdot\dfrac{3}{8}x+\dfrac{9}{64}+\dfrac{27}{64}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\\\left(x+\dfrac{3}{8}\right)^2+\dfrac{27}{64}=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\end{matrix}\right.\)
\(x^3+\dfrac{3}{4}x+\dfrac{3}{2}x^2+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
2:
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
bài 2 :
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
<=> \(\frac{1.2.3....31}{4.6.8....64}=2^n\Rightarrow\frac{1.2.3....30.31}{2\left(2.3.4.5...31\right).32}=2^n\Leftrightarrow\frac{1}{2.32}=2^n\Leftrightarrow\frac{1}{2^6}=2^n\)
=> 2^6.2^n = 1
=> 2^ (n + 6 ) = 2^0
=> n+ 6 = 0
=> n = - 6
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{31}{64}=\frac{1.2.3....31}{4.6.8....64}=\frac{1.2.3....31}{2.3.2.4....2.32}=\frac{1.2.3....31}{2^{30}.\left(3.4....32\right)}=\frac{2}{2^{30}.32}=\frac{1}{2^{34}}=2^{-34}=2^n=>n=-34\)
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4