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1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
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Ngày mai em bị ăn đòn mất!!!hu hu
a) Bạn xem lại vế phải của PT là $x^2-1$ hay $x^3-1$?
b) ĐK: $x\neq \pm 4$
PT \(\Leftrightarrow 5+\frac{48}{x-8}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}=\frac{2(x+4)-9}{x+4}+\frac{3(x-4)+11}{x-4}\)
\(\Leftrightarrow 5+\frac{48}{x-8}=2-\frac{9}{x+4}+3+\frac{11}{x-4}\)
\(\Leftrightarrow \frac{48}{x-8}=\frac{11}{x-4}-\frac{9}{x+4}=\frac{11(x+4)-9(x-4)}{(x-4)(x+4)}=\frac{2x+80}{x^2-16}\)
\(\Leftrightarrow \frac{24}{x-8}=\frac{x+40}{x^2-16}\Rightarrow 24(x^2-16)=(x-8)(x+40)\)
\(\Leftrightarrow 24x^2-384=x^2+32x-320\)
\(\Leftrightarrow 23x^2-32x-64=0\Rightarrow x=\frac{16\pm 24\sqrt{3}}{23}\) (cảm giác đề cứ sai sai)
c)
ĐK: $x\neq \pm \frac{2}{3}$
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow \frac{(3x+2)^2-6(3x-2)}{(3x-2)(3x+2)}=\frac{9x^2}{(3x-2)(3x+2)}\)
\(\Rightarrow (3x+2)^2-6(3x-2)=9x^2\)
\(\Leftrightarrow 9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow -6x+16=0\Rightarrow x=\frac{8}{3}\)
\(a,\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)
\(b,\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
\(c,64-48x+12x^2-x^3=\left(4-x\right)^3\)
\(d,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
\(e,8x^3-60x^2y+150xy^2-125y^3=\left(2x-5y\right)^3\)
bài 1 điền vào chỗ trống
a) x2 + 4x + 4
= (x + 2)2
b) x2 - 8x + 16
= (x - 4)2
c) x3 +12x2 + 48x + 64
= (x + 4)3
d) x3 - 6x + 12x - 8
= (x - 2)3
e) x2 + 2x + 1
= (x + 1)2
f) x2 - 1
= (x - 1)(x + 1)
g) x2 - 4x + 4
= (x - 2)2
h) x2 - 4
= (x - 2)(x + 2)
i) x2 + 6x + 9
= (x + 3)2
j) 4x2 - 9
= (2x - 3)(2x + 3)
k) 16x2 - 8x + 1
= (4x - 1)2
l) 9x2 + 6x + 1
= (3x + 1)2
m) 36x2 + 36x + 9
= (6x + 3)2
n) x3 + 27
= (x + 3)(x2 - 3x + 9)
o) 17x3 + 27 (Đề sai)
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
a) (x + 3)2 - (x - 2)2 = 2x
=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x
=> 5(2x + 1) = 2x
=> 10x + 5 = 2x
=> 10x - 2x = -5
=> 8x = -5
=> x = -5/8
b) 7x(x - 2) = x - 2
=> 7x(x - 2) - (x - 2) = 0
=> (7x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)
c) 8x3 - 12x2 + 6x - 1 = 0
=> (2x - 1)3 = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
a) \(x^2-6x+9=\left(x-3\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(\left(x-3\right)^2-16=\left(x-3-4\right)\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)
d) \(64+16x+x^2=\left(x+8\right)^2\)
e) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
f) mk chỉnh đề
\(8-36x+54x^2-27x^3=\left(2-3x\right)^3\)
g) \(8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)