HOC24
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Bài 5:
4: ta có: 5x(−x+1)−5(−x2−x)=05x(−x+1)−5(−x2−x)=0
⇔−5x(x−1)+5x(x+1)=0⇔−5x(x−1)+5x(x+1)=0
⇔5x(−x+1+x+1)=0⇔5x(−x+1+x+1)=0
hay x=0
H=x5−70x4−70x3−70x2−70x+29H=x5−70x4−70x3−70x2−70x+29
=x4(x−71)+x3(x−71)+x2(x−71)+x(x−71)+x+29=x4(x−71)+x3(x−71)+x2(x−71)+x(x−71)+x+29
=x4(71−71)+x3(71−71)+x2(71−71)+x(71−71)+71+29=x4(71−71)+x3(71−71)+x2(71−71)+x(71−71)+71+29
=0+71+29=100
nhiều thế
1.2x2−x−2x2−5=01.2x2−x−2x2−5=0
−x−5=0⇔x=−5−x−5=0⇔x=−5
2.−2x2−x+2x2−2=02.−2x2−x+2x2−2=0
−x−2=0⇔x=−2−x−2=0⇔x=−2
3.12x2+3x+24x−12x2=03.12x2+3x+24x−12x2=0
27x=0⇔x=027x=0⇔x=0
4.−5x2+5x+5x2+5x=04.−5x2+5x+5x2+5x=0
10x=0⇔x
−10x(2−x)−5x(x+2)=5x(x+3)⇒5x(x+3)+5x(x+2)+10x(2−x)=0⇒5x(x+3+x+2+4−2x)=0⇒5x.9=0⇒45x=0⇒x=0