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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
Gọi thương của phép chia là x ta có :
x . 4 + 2 = 9 . (x - 340) + 2
x . 4 + 2 = 9 . x - 340 . 9 + 2
x . 4 + 2 = 9 . x - 3060 + 2
x . 4 = 9 . x - 3060 ( cùng bớt cả 2 vế đi 2 đơn vị )
x . 9 - x .4 = 3060
x .(9 - 4) = 3060
x . 5 = 3060
x = 3060 : 5
x = 612
Chú ý : Dấu . là dấu nhân
Vậy thương của phép chia cho 4 là 612.
Số phải tìm là :
612 . 4 + 2 = 2450
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^2-4-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+2-3\right)=\left(x-2\right)\left(x-1\right)\)
x2 - 4 - 3(x - 2)
= (x - 2)(x + 2) - 3(x - 2)
= (x - 2)(x + 2 - 3)
= (x - 2)(x - 1)
Đặt \(A=x^4-2y^4-x^2y^2+x^2+y^2\)
\(\Rightarrow2A=2x^4-4y^4-2x^2y^2+2x^2+2y^2\)
\(\Rightarrow2A=\left(x^4+2x^2+1\right)-\left(y^4-2y^2+1\right)\)\(+\left(x^4-2x^2y^2+y^4\right)-4y^4\)
\(\Rightarrow2A=\left(x^2+1\right)^2-\left(y^2-1\right)^2+\left(x^2-y^2\right)^2-4y^4\)
\(\Rightarrow2A=\left[\left(x^2+1\right)^2-4y^4\right]+\left[\left(x^2-y^2\right)^2-\left(y^2-1\right)^2\right]\)
\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2\right)+\)\(\left(x^2-y^2+y^2-1\right)\left(x^2-y^2-y^2+1\right)\)
\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2\right)+\)\(\left(x^2-1\right)\left(x^2+1-2y^2\right)\)
\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2+x^2-1\right)\)
\(\Rightarrow2A=\left(x^2-2y^2+1\right)\left(2x^2+2y^2\right)\)
\(\Rightarrow2A=2\left(x^2-2y^2+1\right)\left(x^2+y^2\right)\)
\(\Rightarrow A=\left(x^2-y^2+1\right)\left(x^2+y^2\right)\)
Nhầm, tớ chốt lại: \(A=\left(x^2-2y^2+1\right)\left(x^2+y^2\right)\), đừng xem cái câu cuối ở tin 1, sai đấy.
=x2-22+(x-2)2=(x-2)(x+2)+(x-2)2=(x-2)(x+2+x-2)=(x-2)*2x
x2 - 4 + ( x - 2 )2 = ( x - 2 ) ( x + 2 ) + ( x + 2 ) ( x + 2 ) = ( x + 2 ) ( x - 2 + x + 2 ) = ( x + 2 ) 2x