Phân tích thành nhân tử:
a) x4+3x3-7x2-27x-18
b) x3-8x2+x+42
c) x4+5x3-7x2-41x-30
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)
\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)
Giải phương trình??? sử dụng Hooc-ne cho nhanh nhá :v
1) \(x^4-8x^2+4x+3=0\)
( dùng máy tính ta đoán được 1 nghiệm chính xác là -3 )
\(\Leftrightarrow\left(x+3\right)\left(x^3-3x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3-3x^2+x+1=0\left(2\right)\end{matrix}\right.\)
Tiếp tục dùng máy tính ta tìm được 1 nghiệm chính xác của pt ( 2 ) là 1
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)
rồi mấy câu còn lại tương tự
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
= x2(5x-7)+2(5x-7)
= (5x-7)(x2+2)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+3x+9+3x)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
=(5x3+10x)-(7x2+14)
=5x(x2+2)-7(x2+2)
=(x2+2)(5x-7)
\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
a) Ta có: \(x^4+3x^3-7x^2-27x-18\)
\(=x^4-3x^3+6x^3-18x^2+11x^2-33x+6x-18\)
\(=x^3\left(x-3\right)+6x^2\left(x-3\right)+11x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+6x^2+11x+6\right)\)
\(=\left(x-3\right)\left(x^3+x^2+5x^2+5x+6x+6\right)\)
\(=\left(x-3\right)\left[x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b) Ta có: \(x^3-8x^2+x+42\)
\(=x^3-7x^2-x^2+7x-6x+42\)
\(=x^2\left(x-7\right)-x\left(x-7\right)-6\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2-x-6\right)\)
\(=\left(x-7\right)\left(x-3\right)\left(x+2\right)\)
c) Ta có: \(x^4+5x^3-7x^2-41x-30\)
\(=x^4+5x^3-7x^2-35x-6x-30\)
\(=x^3\left(x+5\right)-7x\left(x+5\right)-6\left(x+5\right)\)
\(=\left(x+5\right)\left(x^3-7x-6\right)\)
\(=\left(x+5\right)\left(x^3-x-6x-6\right)\)
\(=\left(x+5\right)\left[x\left(x^2-1\right)-6\left(x+1\right)\right]\)
\(=\left(x+5\right)\left[x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\right]\)
\(=\left(x+5\right)\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+5\right)\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
a ) \(==>x^3.\left(x+3\right)-\left(7x^2+27x+18\right)\)
ko xét phần x^3.( x+3 ) nữa mà mik phân tích trong ngoặc nha zo thi ko lm như vậy mà ghi lại phần đó nha
\(7x^2+21x+6x+18\)
\(7x\left(x+3\right)+6\left(x+3\right)\)
\(\left(x+3\right)\left(7x+6\right)\)
==> \(x^3.\left(x+3\right)-\left(x+3\right)\left(7x+6\right)\)
==>\(\left(x+3\right)\left(x^3-7x-6\right)\)