Question

Bài tập 1. Giải các phương trình sau :

a) 8t2 − 4t = 0.   e) 64t2 − 8t = 0.   i)  3t2 = 2t.   m) −t2 = 4t3.

b) 2t2 − 16 = 0.   f) 18t − 9t2 = 0.   j)  4t2 = 3t.   n) −7t2 = 14t3.

c) 5t2 + 7t = 0.   g) 2t2 = t.   k) t3 = t2.   o) t3 − 8t2 = 0.

d) −6t + 9t2 = 0.   h) −t = 3t2.   l) 2t3 = 3t2.   p) 27t2 − 54t3 = 0.

Answer 1

a: \(8t^2-4t=0\)

=>\(4t\left(2t-1\right)=0\)

=>t(2t-1)=0

=>\(\left[{}\begin{matrix}t=0\\2t-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}t=0\\t=\dfrac{1}{2}\end{matrix}\right.\)

b: \(2t^2-16=0\)

=>\(2t^2=16\)

=>\(t^2=8\)

=>\(t=\pm2\sqrt{2}\)

c: \(5t^2+7t=0\)

=>\(t\left(5t+7\right)=0\)

=>\(\left[{}\begin{matrix}t=0\\5t+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\dfrac{7}{5}\end{matrix}\right.\)

e: \(64t^2-8t=0\)

=>\(8\cdot\left(8t^2-t\right)=0\)

=>\(8t^2-t=0\)

=>\(t\left(8t-1\right)=0\)

=>\(\left[{}\begin{matrix}t=0\\8t-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\dfrac{1}{8}\end{matrix}\right.\)

f: \(18t-9t^2=0\)

=>\(t^2-2t=0\)

=>t(t-2)=0

=>\(\left[{}\begin{matrix}t=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\)

g: \(2t^2=t\)

=>\(2t^2-t=0\)

=>t(2t-1)=0

=>\(\left[{}\begin{matrix}t=0\\2t-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\dfrac{1}{2}\end{matrix}\right.\)

i: \(3t^2=2t\)

=>\(3t^2-2t=0\)

=>t(3t-2)=0

=>\(\left[{}\begin{matrix}t=0\\3t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\dfrac{2}{3}\end{matrix}\right.\)

j: \(4t^2=3t\)

=>\(4t^2-3t=0\)

=>t(4t-3)=0

=>\(\left[{}\begin{matrix}t=0\\4t-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\dfrac{3}{4}\end{matrix}\right.\)

k: \(t^3=t^2\)

=>\(t^2\left(t-1\right)=0\)

=>\(\left[{}\begin{matrix}t^2=0\\t-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=1\end{matrix}\right.\)

m: \(-t^2=4t^3\)

=>\(4t^3+t^2=0\)

=>\(t^2\left(4t+1\right)=0\)

=>\(\left[{}\begin{matrix}t^2=0\\4t+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}t=0\\t=-\dfrac{1}{4}\end{matrix}\right.\)

n: \(-7t^2=14t^3\)

=>\(14t^3+7t^2=0\)

=>\(2t^3+t^2=0\)

=>\(t^2\left(2t+1\right)=0\)

=>\(\left[{}\begin{matrix}t^2=0\\2t+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\dfrac{1}{2}\end{matrix}\right.\)

o: \(t^3-8t^2=0\)

=>\(t^2\left(t-8\right)=0\)

=>\(\left[{}\begin{matrix}t^2=0\\t-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t=8\end{matrix}\right.\)

Answer 2

`d)-6t+9t^2=0`

`<=>9t^2-6t=0`

`<=>3t(3t-2)=0`

`<=>3t=0` hoặc `3t-2=0`

`<=>t=0` hoặc `t=2/3`

`h)-t=3t^2`

`<=>3t^2+t=0`

`<=>t(3t+1)=0`

`<=>t=0` hoặc `3t+1=0`

`<=>t=0` hoặc `t=-1/3`

`l)2t^3=3t^2`

`<=>2t^3-3t^2=0`

`<=>t^2(2t-3)=0`

`<=>t^2=0` hoặc `2t-3=0`

`<=>t=0` hoặc `t=3/2`

`p)27t^2-54t^3=0`

`<=>54t^3-27t^2=0`

`<=>27t^2(3t-1)=0`

`<=>27t^2=0` hoặc `3t-1=0`

`<=>t=0` hoặc `t=1/3`