Tìm x
a) 2x(x + 3) + 5(x + 3) = 0
b) x(x + 4) = 2(-x – 4)
c) x^3 = x
d) x^2 – 10x + 25 = 0
e) x^2 – 10x + 24 = 0
f) 2(x + 1)/ 3 – 1 = 3/2 – 1-2x / 4
g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5
h) x^2 – x – 2 = 0
Tìm x
a) 2x(x + 3) + 5(x + 3) = 0
b) x(x + 4) = 2(-x – 4)
c) x^3 = x
d) x^2 – 10x + 25 = 0
e) x^2 – 10x + 24 = 0
f) 2(x + 1)/ 3 – 1 = 3/2 – 1-2x / 4
g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5
h) x^2 – x – 2 = 0
Đáp án:
Giải thích các bước giải:
a) 2x(x + 3) + 5(x + 3) = 0
⇔(x+3)(2x+5)=0
⇔\(\left[ \begin{array}{l}x+3=0\\2x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=-5/2\end{array} \right.\)
Vậy x=-3; x=-5/2
b,x(x + 4) = 2(-x – 4)
⇔x(x+4)-2(-x-4)=0
⇔x(x+4)+2(x+4)=0
⇔(x+4)(x+2)=0
⇔\(\left[ \begin{array}{l}x+4=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-4\\x=-2\end{array} \right.\)
Vậy x=-4; x=-2
c,x³=x
⇔x³-x=0
⇔x(x²-1)=0
⇔x(x-1)(x+1)=0
⇔\(\left[ \begin{array}{l}x=0\\x-1=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy x=0;x=±1
d,x²-10x+25=0
⇔(x-5)²=0
⇔x-5=0
⇔x=5
Vậy x=5
e,x²-10x+24=0
⇔(x²-10x+25)-1=0
⇔(x-5)²-1²=0
⇔(x-5-1)(x-5+1)=0
⇔(x-6)(x-4)=0
⇔\(\left[ \begin{array}{l}x-6=0\\x-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=6\\x=4\end{array} \right.\)
Vậy x=6;x=4
f,$\frac{2(x+1)}{3}$ -1=$\frac{3}{2}$ -$\frac{1-2x}{4}$
⇔$\frac{8(x+1)}{12}$ -12/12=$\frac{18}{12}$ -$\frac{3(1-2x)}{12}$
⇒8x+8-12=18-3+6x
⇔2x=19
⇔x=19/2
Vậy x=19/2
g,$\frac{3(x+1)}{10}$ = $\frac{2(x-1)}{3}$ + $\frac{4}{5}$
<=>$\frac{9(x+1)}{30}$= $\frac{20(x-1)}{30}$ +$\frac{24}{30}$
=>9x+9=20x-20+24
=>-11x=-5
<=>x=5/11
Vậy x=5/11
h,x²-x-2=0
⇔(x-2)(x+1)=0
⇔\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy x=2;x=-1