tìm x
\(6x^4+5x^3-38x^2+5x+6=0\)
phân tich da thuc thanh nhan tu
\(a^4+a^2+1\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)
\(=\left[6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right]\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)\)
\(=\left(x-2\right)\left(x+3\right)\left[2x\left(3x+1\right)-\left(3x+1\right)\right]\)
\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\left(3x+1\right)\)
a: Sửa đề: x^3-x^2+5x-5
=x^2(x-1)+5(x-1)
=(x-1)(x^2+5)
b: x^3+4x^2+x-6
=x^3-x^2+5x^2-5x+6x-6
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)
c: \(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
a)\(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
b)\(8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
c)\(8x^2-2x-1=8x^2+2x-4x-1=2x\left(4x+1\right)-\left(4x+1\right)=\left(2x-1\right)\left(4x+1\right)\)
\(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2+5x\right)\)
a) x =0 không là nghiệm
với x khác 0 ; chia cả 2 vế của pt cho x2
\(6x^2+5x-38+5.\frac{1}{x}+6.\frac{1}{x^2}=0\Leftrightarrow6\left(x^2+2+\frac{1}{x^2}\right)+5\left(x+\frac{1}{x}\right)-50=0\)
\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2+5\left(x+\frac{1}{x}\right)-50=0\) đặt \(t=\left(x+\frac{1}{x}\right)\)=> 6t2 +5t -50 =0 => t= -10/3 hoặc t =5/2
+x +1/x = -10/3 => 3x2 +10x+3 =0 => x =-3 ; x =-1/3
+x+1/x = 5/2 => 2x2 -5x +2 =0 => x=2; x =1/2
b) a4 +2a2 + 1 - a2 = ( a2 +1)2 -a2 = (a2 -a+1)(a2+a+1)
vao cau hoi tuong tu