b) (x-5)2 +9=0
=> (x-5)2 = -9 (vô lí)
=> Pt vô nghiệm
a)64x3+48x2+12x+1=27
<=> (4x)3 +3.(4x)2.1 +3.4x.1 +1=27
<=> (4x+1)3 =27
Mà 33 = 27
=> (4x+1)3=33
=>4x+1=3
=>4x=3-1
=>4x=2
=>x=2/4=1/2
a) 64x3 + 48x2 + 12x + 1 = 27
\(\Rightarrow\) (4x)3 + 3 . (4x)2 . 1 + 3 . 4x . 12 + 13 = 27
\(\Rightarrow\) (4x + 1)3 = 27
\(\Rightarrow\) 4x + 1 = 3
\(\Rightarrow\) 4x = 2
\(\Rightarrow\) x = 0,5
a,Ta có
\(64x^3+48x^2+12x+1=27\)
<=>\(\left(4x\right)^3+3\cdot\left(4x\right)^2\cdot1+3\cdot4x\cdot1^2+1^3=27\)
<=>\(\left(4x+1\right)^3=27\)
<=>4x+1=3
<=>x=-1/2
b) (x - 5)2 + 9 = 0
\(\Rightarrow\) (x - 5)2 = -9
\(\Rightarrow\) vô nghiệm
b.Ta có
\(\left(x-5\right)^2+9=0\)
<=>\(\left(x-5\right)^2=-9\)<0
Mà:\(\left(x-5\right)^2\ge0,\forall x\)
=>vô nghiệm
