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a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)
\(\Rightarrow x^3+27-x^3+4x=27\)
\(\Rightarrow27+4x=27\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\)
b) \(2x^2+7x+3=0\)
\(\Rightarrow2x^2+x+6x+3=0\)
\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x\cdot3+3^2\right)\)
\(=x^3-3^3=x^3-27\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x\cdot2+2^2\right)\)
\(=x^3-2^3=x^3-8\)
c) Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)
\(=x^3+4^3=x^3+64\)
d) Ta có: \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x\cdot3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
e) Ta có: \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\)
\(=\left(x^2-\frac{1}{3}\right)\left[\left(x^2\right)^2+x^2\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3\)
\(=x^6-\frac{1}{27}\)
f) Ta có: \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\frac{1}{27}x^3+8y^3\)
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 6
x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 6
x2 + 5x + 6 - x2 - 3x + 10 = 6
2x +16 = 6
\(\Rightarrow\) 2x = -10
\(\Rightarrow\) x = -5
b) (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
6x2 + 27x + 4x + 18 - (6x2 + x + 12x + 2) = x + 1 - x + 6
6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
18x + 16 = 7
\(\Rightarrow\) 18x = -9
\(\Rightarrow\) x = -0.5
c) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
3(6x2 - 2x - 3x + 1) - (18x2 - 2x - 27x + 3) = 0
3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
18x2 - 15x + 3 - 18x2 + 29x - 3 = 0
14x = 0
\(\Rightarrow\) x = 0
tính nha