(27x3-1) : (3x-1) Tính nhanh nha
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(27x3 – 1) : (3x – 1)
(Sử dụng HĐT để phân tích số bị chia thành tích)
= [(3x)3 – 1] : (3x – 1)
(Xuất hiện hằng đẳng thức (7))
= (3x – 1).[(3x)2 + 3x.1 + 12] : (3x – 1)
= (3x – 1).(9x2 + 3x + 1) : (3x – 1)
= 9x2 + 3x + 1
a) = x^2 - 9 - (x^2 + 3x - 10)
= -3x + 1
b) = 3x + 1 - 3x + 19
= 20
a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)
\(=x^2-9-x^2-3x+10\)
\(=-3x+1\)
b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)
\(=3x+1-3x+19\)
=20
Ta có
( 27 x 3 + 27 x 2 + 9 x + 1 ) : ( 3 x + 1 ) 2 = ( 3 x + 1 ) 3 : ( 3 x + 1 ) 2 = 3 x + 1
Đáp án cần chọn là: B
a.
\(\Leftrightarrow\left(3x-1\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow3x-1=-\dfrac{1}{2}\)
\(\Leftrightarrow3x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
b.
\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-1-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\\end{matrix}\right.\)
c.
\(\Leftrightarrow3x\left(5x-2\right)-2\left(5x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
\(a,=3xy^2\\ b,=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\\ c,=-2x^2+xy+5x^3y^2\\ d,=\left(3x-y\right)\left(9x^2+3xy+y^2\right):\left(3x-y\right)=9x^2+3xy+y^2\)
a: Ta có: \(27x^3-54x^2+36x=8\)
\(\Leftrightarrow27x^3-54x^2+36x-8=0\)
\(\Leftrightarrow\left(3x-2\right)^3=0\)
\(\Leftrightarrow3x-2=0\)
hay \(x=\dfrac{2}{3}\)
b: Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+5\right)=x^2+3x\)
\(\Leftrightarrow\left(x+3\right)\cdot\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x=-3\)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-3x+2y\right)\)
\(=0\cdot0\)
\(=0\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x+y\right)\left(x-y\right)\)
Làm thế này nha
\(\frac{27x^3-1}{3x-1}=\frac{\left(3x\right)^3-1}{3x-1}=\frac{\left(3x-1\right)\left(9x^2+3x+1\right)}{3x-1}=9x^2+3x+1\)
Ta có: \(\left(27x^3-1\right):\left(3x-1\right)=\left[\left(3x\right)^3-1^3\right]:\left(3x-1\right)=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)=9x^2+3x+1\)