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Đặt 2x-2013=a; 2013-3y=b
Theo đề, ta có: \(a^3+b^3-\left(a+b\right)^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3\)
\(=-3ab\left(a+b\right)\)
\(=-3\left(2x-2013\right)\left(2013-3y\right)\left(2x-3y\right)\)
\(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)
=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)2
=(2x-3y)(x-y-2y+3x)-(4x-3y)2
=(2x-3y)(4x-3y)-(4x-3y)2
=(4x-3y)(2x-3y-4x+3y)
=(4x-3y))(-2x)
3*(\(4x^2-4xy+y^2\))-10(2x-y)+8
3*(2x-y)^2-10(2x-y)+8
3*(2x-y)^2-6(2x-y)-4(2x-y)+8
3(2x-y)(2x-y-2)-4(2x-y-2)
(2x-y-2)(6x-3y-40
\(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)
\(=\left(12x^2-6xy-6xy+3y^2\right)-10\left(2x-y\right)+8\)
\(=\left[6x\left(2x-y\right)-3y\left(2x-y\right)\right]-10\left(2x-y\right)+8\)
\(=\left(2x-y\right)\left(6x-3y\right)-10\left(2x-y\right)+8\)
\(=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)
Đặt \(2x-y=a\), khi đó biểu thức có dạng:
\(3a^2-10a+8=3a^2-6a-4a+8\)
\(=3a\left(a-2\right)-4\left(a-2\right)=\left(a-2\right)\left(3a-4\right)\)
\(=\left(2x-y-2\right)\left(6x-3y-4\right).\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
a/Dùng hằng đẳng thức A2-B2=(A+B)(A-B) phân tích được ngay
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
=\(\left(3x-2y+3\right)\left(4-x-4y\right)\)
b/Chắc chỉ phân tích hằng đẳng thức (A-B)2=A2-2ab+B2
\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)
\(=40y^2-395y+748\)
Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa
c/Khai triển biểu thức ban đầu ta được
\(x\left(x-y\right)+y\left(y-x\right)=x^2-xy+y^2-xy=x^2-2xy+y^2=\left(x-y\right)^2\)
\(\left(2x-2013\right)^3+\left(2013-3y\right)^3+\left(3y-2x\right)^3.\)
\(=\left(2x-2013+2013-3y\right)\left[\left(2x-2013\right)^2-\left(2x-2013\right).\left(2013-3y\right)+\left(2013-3y\right)^2\right]+\left(3y-2x\right)^3\)
\(=\left(2x-3y\right).\left(4x^2-8042x+2013^2-4026x+6xy+2013^2-6039y+2013^2-12078y+9y^2\right)+\left(3y-2x\right)^3.\)
\(=\left(2x-3y\right).\left(4x^2+9y^2-12078x-18117y+6xy+3.2013^2\right)-\left(2x-3y\right)^3.\)
\(=\left(2x-3y\right).\left[4x^2+9y^2-12078x-18117y+6xy+3.2013^2-\left(2x-3y\right)^2\right]\)
\(=\left(2x-3y\right).\left(4x^2+9y^2-12078x-18117y+6xy+3.2013^2-4x^2+12xy-9y^2\right).\)
\(=\left(2x-3y\right).\left(-12078x-18117y+18xy+3.2013^2\right).\)
\(=\left(2x-3y\right).9.\left(-1342x-2013y+2xy+1350723\right).\)
\(=9.\left(2x-3y\right).\left[\left(-1342x+2xy\right)+\left(1350723-2013y\right)\right]\)
\(=9.\left(2x-3y\right).\left[-2x\left(671-y\right)+2013\left(671-y\right)\right]\)
\(=9.\left(2x-3y\right).\left(-2x+2013\right)\left(671-y\right)\)