(x-1/3)mũ 3=-8/27
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a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
a.\(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)
b.\(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)
c.\(\left(2x-1\right)^3+8=\left(2x-1\right)^3+2^3=\left(2x+1\right)\left[\left(2x-1\right)^2-2\left(2x-1\right)+4\right]\)
d.\(x^6+6^3=\left(x^2+6\right)\left(x^4-6x+36\right)\)
e.\(1-27x^3=1-\left(3x\right)^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)
j.\(\left(x-3\right)^3-27=\left(x-3\right)^3-3^3=\left(x-6\right)\left[\left(x-3\right)^2+3\left(x-3\right)+9\right]\)
g.\(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
t.\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
u.\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{x}{3}+\frac{1}{9}\right)\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
Tìm x hả bạn ?
a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)
\(\left(3x+\frac{1}{4}\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+\frac{1}{4}=-3\)
\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)
\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)
Vậy x = \(-\frac{13}{12}\)
cho em hoi câu này xin các anh chị:
10mux x+4y = 2013
Ta có 2x= 4y-1 hay 2x=22(y-1) => x=2(y-1)=2y-2 (1)
27y=3x+8 hay 33y = 3x+8 => 3y=x+8 (2)
thay (1) vào (2) ta được 3y= 2y-2+8
=> 3y-2y=6 => y=6
ta có x=2y-2=2. 6 -2=10
x=10 ; y=6
\(\left(x-\dfrac{1}{3}\right)^3=-\dfrac{8}{27}\)
=>\(\left(x-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{3}\right)^3\)
=>\(x-\dfrac{1}{3}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)
\(\frac{3^2.3^8}{27^3}=3^x\)
=> \(\frac{3^{2+8}}{3^9}=3^x\)
=> \(\frac{3^{10}}{3^9}=3^x\)
=> \(3^x=3\)
=> \(3^x=3^1\)
=> x = 1
`#3107.101107`
`(x - 1/3)^3 = -8/27`
`=> (x - 1/3)^3 = (-2/3)^3`
`=> x - 1/3 = -2/3`
`=> x = -2/3 + 1/3`
`=> x = -1/3`
Vậy, `x = -1/3.`
\(\left(\dfrac{x-1}{3}\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(\dfrac{x-1}{3}\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{x-1}{3}=\dfrac{-2}{3}\)
\(\Rightarrow x-1=-2\)
\(\Rightarrow x=-2+1\)
\(\Rightarrow x=-1\)
Vậy $x=-1$.