(X-2)mũ 8=(x-2)mũ 3
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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
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) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
dai the giai bao gio xong chac mua quyt nam sau moi giai xong
Có phải viết như này ko bn?:)
\(x^3.x^2=2^8:2^{^{ }3}\)
a, \(x^2-4-3\left(x-2\right)=\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
b, \(x^2-xy+5y-25=\left(x-5\right)\left(x+5\right)-y\left(x-5\right)=\left(x+5-y\right)\left(x-5\right)\)
c, \(x^3+x^2-2x-8=\left(x-2\right)\left(x^2+2x+4\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+3x+4\right)\)
d, \(x^3-4x^2-8x+8=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)=\left(x^2-6x+4\right)\left(x+2\right)\)
Trả lời:
1, x2 - 4 - 3 ( x - 2 )
= ( x2 - 4 ) - 3 ( x - 2 )
= ( x - 2 ) ( x + 2 ) - 3 ( x - 2 )
= ( x - 2 ) ( x + 2 - 3 )
= ( x - 2 ) ( x - 1 )
2, x2 - xy + 5y - 25
= ( x2 - 25 ) - ( xy - 5y )
= ( x - 5 ) ( x + 5 ) - y ( x - 5 )
= ( x - 5 ) ( x + 5 - y )
3, x3 + x2 - 2x - 8
= ( x3 - 8 ) + ( x2 - 2x )
= ( x - 2 ) ( x2 + 2x + 4 ) + x ( x - 2 )
= ( x - 2 ) ( x2 + 2x + 4 + x )
= ( x - 2 ) ( x2 + 3x + 4 )
4, x3 - 4x2 - 8x + 8
= ( x3 + 8 ) - ( 4x2 + 8x )
= ( x + 2 ) ( x2 - 2x + 4 ) - 4x ( x + 2 )
= ( x + 2 ) ( x2 - 2x + 4 - 4x )
= ( x + 2 ) ( x2 - 6x + 4 )
(x-2)^8 = (x-2)^3
=> (x-2)^8 - (x-2)^3 = 0
=> (x-2)^3.[(x-2)^5 - 1] = 0
=> (x-2)^3 = 0 hoặc (x-2)^5 = 1
=> x - 2 = 0 hoặc x - 2 = 1
=> x = 2 hoặc x = 3
\(\left(x-2\right)^8=\left(x-2\right)^3\)
\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)^3\left[\left(x-2\right)^5-1\right]=0\)
\(\Rightarrow\left(x-2\right)^3=0\) hoặc \(\left(x-2\right)^5-1=0\)
\(x-2=0\) \(\left(x-2\right)^5=1\)
\(x=2\) \(x-2=1\)
\(x=3\)
Vậy ...