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Áp dụng công thức : (A + B)3 = A3 + 3A2B + 3AB2 + B3
(A - B)3 = A3 - 3A2B + 3AB2 -B3
a) (3x + 1)3 = (3x)3 + 3.(3x)2.1 + 3.3x.1 + 13 = 27x3 + 27x2 + 9x + 1
b) \(\left(\frac{x}{3}-1\right)^3=\left(\frac{x}{3}\right)^3-3\cdot\left(\frac{x}{3}\right)^2\cdot1+3\cdot\left(\frac{x}{3}\right)\cdot1^2-1^3\)
\(=\frac{x^3}{27}-3\cdot\frac{x^2}{9}\cdot1+3\cdot\frac{x}{3}\cdot1-1\)
= \(\frac{x^3}{27}-\frac{x^2}{3}+x-1\)
c) \(\left(2x-\frac{1}{x}\right)^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot\frac{1}{x}+3\cdot2x\cdot\left(\frac{1}{x}\right)^2-\left(\frac{1}{x}\right)^3\)
\(=8x^3-3\cdot4x^2\cdot\frac{1}{x}+6x\cdot\frac{1}{x^2}-\frac{1}{x^3}\)
\(=8x^3-12x+\frac{6}{x}-\frac{1}{x^3}\)
d) \(\left(-y^2+3x\right)^3=\left(3x-y^2\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot y^2+3\cdot3x\cdot y^4-y^6\)
= 27x3 - 27x2y2 + 9xy4 - y6
= -y6 + 9xy4 - 27x2y2 + 27x3
Tương tự câu cuối :>
a, (2x-5)3 = (2x)3 - 3.(2x)2.5 + 3.2x.52 - 53
= 8x3 - 60x2 + 150x - 125
Câu b, câu c làm tương tự như câu a
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
câu 1:
\(a,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
=> \(25x^2+10x+1-\left(25x^2-9\right)=30\)
=> \(25x^2+10x+1-25x^2+9=30\)
=> \(10x+10=30\)
=> \(10x=20\)
=> \(x=2\)
Vậy..........
\(b,\left(2x+3\right)^2-\left(2x-3\right)^2+4\left(x^2-6x\right)=64\)
=> \(6.4x+4x^2-24x=64\)
=> \(24x+4x^2-24x=64\)
=> \(4x^2=64\)
=> \(x^2=64:4=16\)
=> \(\left|x\right|=\sqrt{16}\)
=> \(x=\pm4\)
Vậy \(x\in\left\{4;-4\right\}\)
mấy cả +23 thành -23 nữa nhé
\(\left(3x-2y\right)^3=\left(3x\right)^3-3.\left(3x\right)^2.2y+3.3x.\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^3-54x^2y+36xy^2-8y^3\)