a: \(27x^3-54x^2+36x-8=0\)

=>\(\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2-2^3=0\)

=>\(\left(3x-2\right)^3=0\)

=>3x-2=0

=>3x=2

=>\(x=\frac23\)

b: \(x^3+15x^2+75x+125=0\)

=>\(x^3+3\cdot x^2\cdot5+3\cdot x\cdot5^2+5^3=0\)

=>\(\left(x+5\right)^3=0\)

=>x+5=0

=>x=-5

c: \(x^3-18x^2+108x-216=0\)

=>\(x^3-3\cdot x^2\cdot6+3\cdot x\cdot6^2-6^3=0\)

=>\(\left(x-6\right)^3=0\)

=>x-6=0

=>x=6

d: \(\left(x-1\right)^3-x\left(x^2+3x\right)=2\)

=>\(x^3-3x^2+3x-1-x^3-3x^2=2\)

=>\(-6x^2+3x-3=0\)

=>\(2x^2-x+1=0\)

\(\Rightarrow x^2-\frac12x+\frac12=0\)

=>\(x^2-\frac12x+\frac{1}{16}+\frac{7}{16}=0\)

=>\(\left(x-\frac14\right)^2+\frac{7}{16}=0\) (vô lý)

=>x∈∅

e: \(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=3\)

=>\(x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3\)

=>15x-7=3

=>15x=10

=>\(x=\frac{10}{15}=\frac23\)

Giải:

\(A=x^3-15x^2+75x-125\)

\(\Leftrightarrow A=x^3-3.x^2.5+3.x.5^2-5^3\)

\(\Leftrightarrow A=\left(x-5\right)^3\)

Tại \(x=35\), giá trị của A là:

\(A=\left(35-5\right)^2=30^2=900\)

Vậy ...

ủa đang mũ 3 sao chuyển qua mũ 2 vậy?????/

\(A=x^3+15x^2+75x+125=\left(x+5\right)^3=-125\)

\(B=4x^2+12xy+9y^2=\left(2x+3y\right)^2=\left(3+6\right)^2=81\)

a) (x+4)(x² - 4x + 16 ) - x(x-5) = 264

x³ + 4³ - x(x² - 25) = 264

x³ + 64 - x³ + 25x= 264

64 + 25x = 264

25x = 264-64

25x= 200

x = \(\dfrac{200}{25}\) = 8

b) (x-2)³ - (x-2)(x² + 2x + 4 ) + 6(x-2)(x+2) = 60

x³ - 6x² + 12x - 8 - ( x³ - 2³ ) + 6(x² - 4 ) = 60

x³ - 6x² + 12x - 8 - x³ + 8 + 6x² -24 = 60

12x - 24 = 60

12x = 60 + 24

12x = 84

x = \(\dfrac{84}{12}\) = 7

Lời giải:

A) Tại $x=35$ thì \(x-35=0\)

\(A=x^3-15x^2+75x=x^3-35x^2+20x^2+75x\)

\(=x^3-35x^2+20x^2-700x+775x\)

\(=x^2(x-35)+20x(x-35)+775x\)

\(=775x=775.35=27125\)

B) \(x=-26\rightarrow x+26=0\)

\(B=x^3+18x^2+108x+16\)

\(=x^3+26x^2-8x^2-208x+316x+16\)

\(=x^2(x+26)-8x(x+26)+316x+16\)

\(=316x+16=316.-26+16=-8200\)

C)

\(C=(x^2-4y^2)(x^2-2xy+4y^2)(x^2+2xy+4y^2)\)

\(=(x-2y)(x+2y)(x^2-2xy+4y^2)(x^2+2xy+4y^2)\)

\(=[(x-2y)(x^2+2xy+4y^2)][(x+2y)(x^2-2xy+4y^2)]\)

\(=[x^3-(2y)^3][x^3+(2y)^3]\)

\(=(-8-1)(-8+1)=63\)

c)
\(x^3-3.x^2.6+3.x.6^2-6^3=0\)
\(\left(x-6\right)^3=0\)
x-6=0
x=6
d)
\(x^3-3.x^2.1+3.x.1^2-1-x^3-3x-2=0\)
\(x^3-3x^2+3x-1-x^3-3x^2-2=0\)
\(-6x^2-3=0\)
\(-3\left(2x^2+1\right)=0\)
\(2x^2+1=0\)
2x²=-1
x²=1/2
x=\(\dfrac{\sqrt{2}}{2}\)

a)
\(\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.2^2-2^3=0\)
\(\left(3x-2\right)^3=0\)
3x-2=0
3x=2
x=2/3
b)
\(x^3-3.x^2.5+3.x.5^2+5^3=0\)
\(\left(x-5\right)^3=0\)
x-5=0
x=5

\(a,A=\left(x+5\right)^3=\left(-10+5\right)^3=\left(-5\right)^3=-125\\ b,B=\left(2x+3y\right)^2=\left(2\cdot1+3\cdot2\right)^2=7^2=49\\ c,C=\left(3x-y\right)^3=\left(3\cdot1+2\right)^3=5^3=125\)

a. \(x^3+15x^2+75x+125\)\(=x^3+3.x^2.5+3.x.5^2+5^3=\left(x+5\right)^3\)

b. \(x^3-9x^2+27x-27=\)\(x^3-3.x^2.3+3x.3^2-27=\left(x-3\right)^3\)