36B

37C

38D

39B

40D

41A

42B

43B

44A

45B

46B

47A

48C

50B

51B

52B

53D

54C

55D

56C

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(3-\sqrt{x-1}\right)^2}=0\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|=0\)

Do \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1>0\) với mọi x thuộc TXĐ

\(\Rightarrow\) Phương trình đã cho vô nghiệm

11)11) 3x(x-5)²-(x+2)³+2(x-1)³-(2x+1)(4x²-2x+1)=3x(x²-10x+25)-(x³+6x²+12x+8)+2(x³-3x²+3x-1)-(8x³+1)=3x³-30x²+75x-x³-6x²-12x-8+2x³-6x²+6x-2-8x³-1=-4x³-42x²+63x-11

nhìn khó thế

a) x\(^3\)+3x\(^2\)-2x-6=0

⇔(x+3)(x+\(\sqrt{2}\))(x-\(\sqrt{2}\))

⇔x+3=0 hoặc x+\(\sqrt{2}\)=0 hoặc x-\(\sqrt{2}\)=0

⇔x=-3 hoặc x=-\(\sqrt{2}\) hoặc x=\(\sqrt{2}\)

​​​b)2x\(^3\)+7x\(^2\)+7x+2=0​

⇔(x+1)(2x+1)(x+2)=0

⇔x+1=0 hoặc2x+1=0 hoặc x+2=0

⇔x=-1 hoặcx=-\(\dfrac{1}{2}\) hoặc x=-2

c)x\(^3\)+7x\(^2\)-56x+48=0

⇔(x-1)(x-4)(x+12)=0

⇔x-1=0 hoặcx-4=0 hoặcx+12=0

⇔x =1hoặcx=4 hoặcx=-12