Giải các phương trình sau:
a) (x-1)^2 – (2x+5)^2 =0
b) x^2 – (1-x)^2 – (2x-1)^2 =0
c) x^3 +8 = -2x (x+2)
d) 4x^2 + 8x – 5 =0
Giải các phương trình sau:
a) (x-1)^2 – (2x+5)^2 =0
b) x^2 – (1-x)^2 – (2x-1)^2 =0
c) x^3 +8 = -2x (x+2)
d) 4x^2 + 8x – 5 =0
Đáp án:
Giải thích các bước giải:
`a) (x-1)^2 – (2x+5)^2 =0`
`↔(x-1-2x-5)(x-1+2x+5)=0`
`↔(-x-6)(3x+4)=0`
`↔` \(\left[ \begin{array}{l}-x-6=0\\3x+4=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=-6\\x=-\dfrac{4}{3}\end{array} \right.\)
Vậy `S={-6;-4/3}`
`b)x^2 – (1-x)^2 – (2x-1)^2 =0`
`↔-4x^2+6x-2=0`
`↔2(-2x+1)(x-1)=0`
`↔` \(\left[ \begin{array}{l}-2x+1=0\\x-1=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={1/2;1}`
`c)x^3+8=-2x(x+2)`
`↔(x+2)(x^2-2x+4)+2x(x+2)=0`
`↔(x+2)(x^2+4)=0`
`x^2≥0→x^2+4≥4`
`↔x+2=0`
`↔x=-2`
Vậy `S={-2}`
`d)4x^2+8x-5=0`
`↔(2x-1)(2x+5)=0`
`↔` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{array} \right.\)
Vậy `S={1/2;-5/2}`
`a) (x-1)^2 – (2x+5)^2 =0`
`<=> (x-1-2x-5) (x-1+2x+5) =0`
`<=> (-x-6) (3x+4) =0`
`<=>`\(\left[ \begin{array}{l}x=-6\\x=\cfrac{-4}3\end{array} \right.\)
`b) x^2 – (1-x)^2 – (2x-1)^2 =0`
` <=>(x-1+x) (x+1-x) – (2x-1)^2 =0`
` <=>(2x-1) .1 – (2x-1)^2 =0`
` <=>(2x-1)( 1 – 2x+1) =0`
` <=>(2x-1)( 2 – 2x) =0`
`<=>`\(\left[ \begin{array}{l}x=\cfrac12\\x=1\end{array} \right.\)
`c) x^3 +8 = -2x (x+2) `
`<=> (x +2)(x^2-2x+4) +2x (x+2) =0`
`<=> (x +2)(x^2-2x+4 +2x) =0`
`<=>(x+2)(x^2+4)=0`
Vì `x^2+4>0`
`<=>x+2=0`
`<=>x=-2`
`d) 4x^2 + 8x – 5 =0`
`<=>( 4x^2 + 8x +4)-9 =0`
`<=>( 2x +2)^2-9 =0`
`<=>( 2x +2-3)(2x+2+3) =0`
`<=>( 2x -1)(2x+5) =0`
`<=>`\(\left[ \begin{array}{l}x=\cfrac12\\x=\cfrac{-5}{2}\end{array} \right.\)