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a) x2 - 5x + 6
= x2 - 2x - 3x + 6
=(x2 - 2x) - (3x + 6)
=x.(x - 2) - 3.(x - 2)
=(x-2).(x-3)
b) 3x2+9x-30
=3x2+15x-6x-30
=(3x2+15x) - (6x+30)
= 3x(x+5) - 6(x+5)
=(x+5).(3x-6)
c) x2-3x+2
=x2-2x-x+2
=(x2-2x) - (x-2)
=x(x-2)-(x-2)
=(x-2)(x-1)
a)x2 - 5x + 6
= x2 - 2x - 3x + 6
=x.(x - 2) - 3(x - 2)
=(x - 2).(x - 3)
b)3x2 +9x -30
=3x2 +15x - 6x -30
=3x.(x+5) - 6.(x + 5)
=(x+5).(3x - 6)
c)x2 - 3x +2
=x2 - 2x - x +2
=x.(x- 2) - 1.(x-2)
=(x-2).(x - 1)
d)x2 - 9x +18
=x2 - 6x -3x +18
=x.(x - 6) -3.(x - 6)
=(x - 6).(x - 3)
e)x2 - 6x +8
=x2 - 2x - 4x +8
=x.(x - 2)- 4.(x - 2)
=(x - 2).(x - 4)
f)x2 - 5x -14
=x2 + 2x - 7x - 14
=x.(x + 2) -7.(x + 2)
=(x + 2).(x - 7)
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
17x+15(x-1)=1-14(3x+1)
<=> 17x+15x-15=1-42x-14
<=>17x+15x+42x=1+15-14
<=>74x=2=> x = 2/74=1/34
vậy tập nghiệm S={1/34}
b. 2x(x+5) - (x-3)^2 = x^2 + 6
<=>\(2x^2+10x-x^2+6x-9=x^2+6\)
<=> \(2x^2-x^2-x^2+10x+6x=9+6\)
<=> 16x=15=> x=15/16
vậy tập nghiệm S={15/16}
Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)