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1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)
\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)
\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
4. \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
5. \(a^3x-ab+b-x\)
\(=a^3x-x-ab+b\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)
6. \(x^3-64=x^3-4^3\)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
7. \(0,125\left(a+1\right)^3-1\)
\(=\left[0,5\left(a+1\right)\right]^3-1^3\)
\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)
\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)
\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)
8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
11. \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
1.
\(x^2-22x+12\) : biểu thức không phân tích được thành nhân tử nữa.
2.
\(9x^2+6x+1=(3x)^2+2.3x.1+1^2=(3x+1)^2\)
3.
\(x^2-10x+2\): không p. tích được thành nhân tử.
4.
\(x^3+1=x^3+1^3=(x+1)(x^2-x+1)\)
5.
\(8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)[(2x)^2+(2x)(3y)+(3y)^2]\)
\(=(2x-3y)(4x^2+6xy+9y^2)\)
6.
\((x+3y)^2-(3y+1)^2=[(x+3y)-(3y+1)][(x+3y)+(3y+1)]\)
\(=(x-1)(x+6y+1)\)
7.
\(4y^2-36x^2=(2y)^2-(6x)^2=(2y-6x)(2y+6x)=4(y-3x)(y+3x)\)
8.
\(27-(x+4)^3=3^3-(x+4)^3=[3-(x+4)][3^2+3(x+4)+(x+4)^2]\)
\(=-(x+1)(37+x^2+11x)\)
9.
\(25x^2-10xy+y^2=(5x)^2-2.5x.y+y^2=(5x-y)^2\)
10.
\(9x^6-12x^7+4x^8=x^6(9-12x+4x^2)=x^6[3^2-2.3.2x+(2x)^2]\)
\(=x^6(3-2x)^2\)
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....