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Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
Câu 2:
a: \(\left(x-1\right)\left(x+1\right)-x\left(x+3\right)+7=0\)
=>\(x^2-1-x^2-3x+7=0\)
=>-3x+6=0
=>-3x=-6
=>\(x=\dfrac{-6}{-3}=2\)
b: \(2x^3-22x^2+36x=0\)
=>\(2x\left(x^2-11x+18\right)=0\)
=>\(x\left(x^2-11x+18\right)=0\)
=>\(x\left(x^2-2x-9x+18\right)=0\)
=>\(x\left[x\left(x-2\right)-9\left(x-2\right)\right]=0\)
=>\(x\left(x-2\right)\left(x-9\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=9\end{matrix}\right.\)
Câu 4:
1: Diện tích cỏ cần thay là:
\(105\cdot68=7140\left(m^2\right)\)
Số tiền BQL sân cần trả là:
\(7140\cdot120000=856800000\left(đồng\right)\)
2:
a: Xét tứ giác ABDC có
M là trung điểm chung của AD và BC
=>ABDC là hình bình hành
Hình bình hành ABDC có \(\widehat{BAC}=90^0\)
nên ABDC là hình chữ nhật
b: Xét ΔADE có
H,M lần lượt là trung điểm của AE,AD
=>HM là đường trung bình của ΔADE
=>HM//DE
=>BC//DE
=>\(\widehat{EDB}=\widehat{DBM}\)(hai góc so le trong)(1)
Ta có: ABDC là hình chữ nhật
=>AD=BC
mà \(MD=\dfrac{AD}{2};MB=\dfrac{BC}{2}\)
nên MD=MB
=>ΔMBD cân tại M
=>\(\widehat{MDB}=\widehat{MBD}\left(2\right)\)
Từ (1) và (2) suy ra \(\widehat{MDB}=\widehat{EDB}\)
=>\(\widehat{ADB}=\widehat{EDB}\)
=>DB là phân giác của góc ADE
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
\(1,=3xy\left(x^2+2xy+y^2\right)=3xy\left(x+y\right)^2\\ 2,=7xy\left(2x-3y+4xy\right)\\ 3,=\left(x-1\right)\left(x^2-4\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\\ 4,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ 5,=\left(b-c\right)\left(8a-6b\right)=2\left(4a-3b\right)\left(b-c\right)\\ 6,=\left(x-1\right)\left(x^2-16\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\\ 7,=x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(x+5\right)\\ 8,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\\ 9,=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\\ 10,=\left(x-1\right)^2-4y^2=\left(x-2y-1\right)\left(x+2y-1\right)\)