phan tich da thuc thanh nhan tu
a, x^5+x-1
b, (x^2+3x+2)(x62+7x+12)-24
Phan tich da thuc thanh nhan tu (x²+3x)^2+7x²+21x+10
Phan tich da thuc thanh nhan tu
a) 3x2 - 16x + 5
b) x3 - 7x + 6
c) 3x3 - 4x2 + 13x - 4
d) x2 + 7x + 12
Phan tich da thuc thanh nhan tu :x^2+7x-15
x^2 + 7x -15
= x^2 + 7x +12,25 -27,25
= (x+3,5)^2 - 27, 25
= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))
Phan tich da da thuc thanh nhan phan tu
(x^2+x+1)(x^2+x+2)-12
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
phan tich da thuc sau thanh nhan tu
x^3+3x^2+4x+2
6x^4-x^3-7x^2+x+1
giup minh nhe cam on nhieu
\(a.x^3+3x^2+4x+2\)
\(=x^3+x^2+2x^2+2x+2\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+2\right)\)
\(b.6x^4-x^3-7x^2+x+1\)
\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)
\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)
\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)
\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)\)
k giùm cái cho đỡ buồn!
phan tich da thuc sau thanh nhan tu bang cach ha bac-tách để xuất hiện ntc là (x-4)
x^4-4x^3-7x^2+22x+24
\(x^4-4x^3-7x^2+22x+24\)
\(=\left(x^4-4x^3\right)-\left(7x^2-28x\right)-\left(6x-24\right)\)
\(=x^4.\left(x-4\right)-7x.\left(x-4\right)-6.\left(x-4\right)\)
\(=\left(x-4\right).\left(x^4-7x-6\right)\)
Tham khảo nhé~
Tim x biet
a) 28x^3+15x^2+75x+125=0
b)4x^2-x-5=0
Phan tich da thuc thanh nhan tu
a) x^3+5x^2+3x-9
b)x^3-7x-6
c)3x^3-7x^2+17x-5
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
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,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
\(4x^2-x-5=0\)
<=> \(4x^2+4x-5x-5=0\)
<=> \(4x\left(x+1\right)-5\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(4x-5\right)=0\)
tự lm tiếp
Tim x biet
a) 28x^3+15x^2+75x+125=0
b)4x^2-x-5=0
Phan tich da thuc thanh nhan tu
a) x^3+5x^2+3x-9
b)x^3-7x-6
c)3x^3-7x^2+17x-5
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)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(4x^2-x-5=0\)
<=> \(4x^2+4x-5x-5=0\)
<=> \(4x\left(x+1\right)-5\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(4x-5\right)=0\)
tự giải nốt
phan tich da thuc thanh nhan tu 1-3x-x^3+3x^2
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)