Question

Tìm x:
a)3.(x-2)+2.(x-3)=5
b)(2x-8)²-16=0
c)(2x-1)²-(4x+1).(x-3)=3

Answer 1

a)3(x-2)+2(x-3)=5

=>3x-6+2x-6=5

=>5x=17

=>x=17/5

Answer 2

b)(2x-8)^2=16

TH1:2x-8=4=>x=6

TH2:2x-8=-4=>x=2

Answer 3

\(a,\Leftrightarrow3x-6+2x-6=5\\ \Leftrightarrow5x-12=5\\ \Leftrightarrow x=\dfrac{17}{5}\\ b,\left(2x-8\right)^2-16=0\\ \Leftrightarrow\left(2x-12\right)\left(2x-4\right)=0\\ \Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-4x+1-4x^2+11x+3=3\\ \Leftrightarrow7x=-1\\ \Leftrightarrow x=-\dfrac{1}{7}\)

Answer 4

c)4x^2−4x+1−4x^2+11x+3=3

=>7x=−1

=>x=−1/7

Answer 5

a: ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)