Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
Bài 2
\(a,x^3+5x^2+3x-9\)
\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)
\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)
b,\(x^3-7x-6\)
\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)
\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
Bài 1:
a)\(28x^3+15x^2+75x+125=0\)
\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)
Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)
\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)
b)\(4x^2-x-5=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)
\(\Rightarrow x=-1;x=\frac{5}{4}\)
Bài 2:
a)\(x^3+5x^2+3x-9\)
\(=\left(x-1\right)\left(x+3\right)^2\)
b)\(x^3-7x-6\)
\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
a) 7xy^2+5x^2y
= xy(7y+5x)
b) x^2+2xy+y^2-11x-11y
= (x^2+2xy+y^2)-(11x+11y)
=(x+y)^2-11(x+y)
=(x+y)(x+y-11)
a) 7xy2 + 5x2y
= xy ( 7y + 5x )
b) x2 + 2xy + y2 - 11x - 11y
= ( x2 + 2xy + y2 ) - ( 11x + 11y )
= ( x + y )2 - 11 ( x + y )
= ( x + y )( x + y - 11 )
1. 2xy2 +x2y4+1 = (xy2+1)2
2. a)3x2+3x-10x-10=3x(x+1)-10(x+1)=(x+1)(3x-10)
b)2x2-5x-7=2x2+2x-7x-7=2x(x+1)-7(x+1)=(x+1)(2x-7)
Mong có thể giúp được bạn
\(x^4+9\)
\(=\left(x^2\right)^2+2.x^2.3+3^2-2.x^2.3\)
\(=\left(x^2\right)^2+6x^2+3^2-6x^2\)
\(=\left(x^2+3\right)^2-\left(\sqrt{6}x\right)^2\)
\(=\left(x^2+3-\sqrt{6}x\right)\left(x^2+3+\sqrt{6}x\right)\)
\(a.\left(x^2+5x+6\right)\left(x^2-15x+56\right)-144\\ =\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)-144\\ =\left[\left(x+2\right)\left(x-7\right)\right]\left[\left(x+3\right)\left(x-8\right)\right]-144\\ =\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\\ =\left(x^2-5x-19+5\right)\left(x^2-5x-19-5\right)-144\\ =\left(x^2-5x-19\right)^2-5^2-144\\ =\left(x^2-5x-19\right)^2-169\\ =\left(x^2-5x+19\right)^2-13^2\\ =\left(x^2-5x-19-13\right)\left(x^2-5x-19+13\right)\\ =\left(x^2-5x-32\right)\left(x^2-5x-6\right)\\ =\left(x^2-5x-32\right)\left(x+1\right)\left(x-6\right)\)
\(b.\left(x^2-11x+28\right)\left(x^2-7x+10\right)-72\\ =\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-2\right)-72\\ =\left[\left(x-4\right)\left(x-5\right)\right]\left[\left(x-7\right)\left(x-2\right)\right]-72\\ =\left(x^2-9x+20\right)\left(x^2-9x+14\right)-72\\ =\left(x^2-9x+17+3\right)\left(x^2-9x+17-3\right)-72\\ =\left(x^2-9x+17\right)^2-3^2-72\\ =\left(x^2-9x+17\right)^2-81\\ =\left(x^2-9x+17\right)^2-9^2\\ =\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)\\ =\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)