Question

Giải các phương trình sau:

a)x^3 - 3x^2 + 4=0

b)x^4 + x^3 - 4x^2 + 5x -3=0

c)4^x - 10.2^x + 16=0 

 

Answer 1

a, pt <=> (x^3+x^2)-(4x^2-4) = 0

<=> (x+1).(x^2-4x+4) = 0

<=> (x+1).(x-2)^2 = 0

<=> x+1=0 hoặc x-2=0

<=> x=-1 hoặc x=2

b, pt <=> (x^4-x^3)+(2x^3-2x^2)-(2x^2-2x)+(3x-3) = 0

<=> (x-1).(x^3+2x^2-2x+3) = 0

<=> (x-1).[(x^3+3x^2)-(x^2+3x)+(3x+3)] = 0

<=> (x-1).(x+3).(x^2-3x+3) = 0

<=> x-1=0 hoặc x+3=0 ( vì x^2-3x+3 > 0 )

<=> x=1 hoặc x=-3

c, pt <=> (4^x-10.2^x+25)-9 =0

<=> (2^x-5)^2-9 = 0

<=> (2^x-5-3).(2^x-5+3) = 0

<=> (2^x-8).(2^x-2) = 0

<=> 2^x-8=0 hoặc 2^x-2=0

<=> x=3 hoặc x=1

Tk mk nha

Answer 2

a)   \(x^3-3x^2+4=0\)

\(\Leftrightarrow\)\(x^3+x^2-4x^2+4=0\)

\(\Leftrightarrow\)\(x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

Vậy....

Answer 3

a, pt <=> (x^3+x^2)-(4x^2-4) = 0

<=> (x+1).(x^2-4x+4) = 0

<=> (x+1).(x-2)^2 = 0

<=> x+1=0 hoặc x-2=0

<=> x=-1 hoặc x=2

b, pt <=> (x^4-x^3)+(2x^3-2x^2)-(2x^2-2x)+(3x-3) = 0

<=> (x-1).(x^3+2x^2-2x+3) = 0

<=> (x-1).[(x^3+3x^2)-(x^2+3x)+(3x+3)] = 0

<=> (x-1).(x+3).(x^2-3x+3) = 0

<=> x-1=0 hoặc x+3=0 ( vì x^2-3x+3 > 0 )

<=> x=1 hoặc x=-3

c, pt <=> (4^x-10.2^x+25)-9 =0

<=> (2^x-5)^2-9 = 0

<=> (2^x-5-3).(2^x-5+3) = 0

<=> (2^x-8).(2^x-2) = 0

<=> 2^x-8=0 hoặc 2^x-2=0

<=> x=3 hoặc x=1

Tk mk nha