Giải các phương trình sau :
a) (x-3) (x+4) = 0
b) 2x (x-7) + 5 (x-7) = 0
c) 2x^3 – 2x = x^2 – 1
d) (2x – 5)^2 – (x-2)^2 = 0
e) 3x – 15 =7x (x-5)
f) 4x^2 + 4x + 1 = x^2
Giải các phương trình sau :
a) (x-3) (x+4) = 0
b) 2x (x-7) + 5 (x-7) = 0
c) 2x^3 – 2x = x^2 – 1
d) (2x – 5)^2 – (x-2)^2 = 0
e) 3x – 15 =7x (x-5)
f) 4x^2 + 4x + 1 = x^2
Gửi nè.Xin CTLHN CHO NHÓM Ạ
NO COPY.
Duyen Buii
Giải thích các bước giải:
`a,(x-3)(x+4)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
Vậy: `x∈{3;-4}`
`b,2x (x-7) + 5 (x-7) = 0`
`⇔(2x+5)(x-7)=0`
`⇔`\(\left[ \begin{array}{l}2x+5=0\\x-7=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\frac{5}{2}\\x=7\end{array} \right.\)
Vậy:`x∈{-5/2;7}`
`c) 2x^3 – 2x = x^2 – 1`
`⇔2x^3-2x-(x^2-1)=0`
`⇔2x(x^2-1)-(x^2-1)=0`
`⇔(2x-1)(x^2-1)=0`
`⇔`\(\left[ \begin{array}{l}2x-1=0\\x^2-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=±1\end{array} \right.\)
Vậy: `x∈{$\frac{1}{2}$;±1}`
`d,(2x – 5)^2 – (x-2)^2 = 0`
`⇔(2x-5)^2=(x-2)^2`
`⇔2x-5=x-2`
`⇔2x-x=-2+5`
`⇔x=3`
Vậy: `x=3`
`e,3x – 15 =7x (x-5)`
`⇔3(x-5)-7x(x-5)=0`
`⇔(3-7x)(x-5)=0`
`⇔`\(\left[ \begin{array}{l}3-7x=0\\x-5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{3}{7} \\x=5\end{array} \right.\)
Vậy: `x∈{\frac{3}{7};5}`
`f,4x^2 + 4x + 1 = x^2`
`⇔4x(x+1)-(x^2-1)=0`
`⇔4x(x+1)-(x-1)(x+1)=0`
`⇔(4x-x+1)(x+1)=0`
`⇔`\(\left[ \begin{array}{l}3x+1=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\frac{1}{3}\\x=-1\end{array} \right.\)
Vậy: `x∈{-\frac{1}{3};-1}`
Xin hay nhất cho nhóm nhen bn 🙂