64x ; 48x ; 88x và x lớn nhất.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ge0\\ PT\Leftrightarrow10\sqrt{x}+8\sqrt{x}-11\sqrt{x}=21\\ \Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\)
\(2\sqrt{25x}+\sqrt{64x}-\sqrt{121x}=21\left(x\ge0\right)\)
\(\Leftrightarrow10\sqrt{x}+8\sqrt{x}-11\sqrt{x}=21\)
\(\Leftrightarrow7\sqrt{x}=21\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) sin6(4x) + cos6(4x) + 2
= (sin2(4x) + cos2(4x))(sin4(4x) + cos4(4x) - sin2(4x)cos2(4x)) + 2
= 1.((sin2(4x) + cos2(4x))2 - 3sin2(4x)cos2(4x)) + 2
= - 3sin2(4x)cos2(4x) + 3
= -\(\dfrac{3}{4}\).(4sin2(4x)cos2(4x) + 3
= -\(\dfrac{3}{4}\).sin2(8x) + 3
vì -1 ≤ sin(8x) ≤ 1 nên 0 ≤ sin2(8x) ≤ 1.
=> min = 3 khi sin(8x) = 0
max = \(\dfrac{9}{4}\) khi sin(8x) = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
x^3-64x=0
<=>x^3-64x=(x-8)x(x+8)
=>(x-8)x(x+8)=0
Th1:x-8=0
=>x=8
Th2:x+8=0
=>x=-8
vậy pt có x=±8
x^3-64x=0
x.x.x-64x=0
=>x.x=64
Ta có:
x.x=64
x=\(\sqrt{64}\)
x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2\right)^2+2.8x^2.y^2+y^4-\left(4xy\right)^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy-y^2\right)\left(8x^2+4xy-y^2\right)\)
64x⁴ + 16x²y² + y⁴ - 16x²y²
= (8x²)² + 2.8x².y² + (y²)² - (4xy)²
= (8x² + y²)² - (4xy)²
= (8x² - 4xy + y²)(8x² + 4xy + y²)