Question

Bài 1: Tìm x biết :

a) (x+2)^2 - (3x-7)^2=0

b) (4x+1) -(5x-3)^2=0

c) 25(x-3)^2 - 49(2x+1)^2=0

d) 9(3x-2)^2=121(1-4x)^2

e) (x-5/4)^2=(5x+1/2)^2

Answer 1

a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x-7\\x+2=-3x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-2-7\\x+3x=-2+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Mấy câu kia tương tự.

Answer 2

a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)

\(\Leftrightarrow\left(x+2-3x+7\right)\left(x+2+3x-7\right)=0\)

\(\Leftrightarrow\left(-2x+9\right)\left(4x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+9=0\\4x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{-2}=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Vậy \(x=\dfrac{9}{2}\) hoặc \(x=\dfrac{5}{4}\)

b) lộn đề à

c) \(25\left(x-3\right)^2-49\left(2x+1\right)^2=0\)

\(\Leftrightarrow5^2\left(x-3\right)^2-7^2\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[5\left(x-3\right)\right]^2-\left[7\left(2x+1\right)\right]^2=0\)

\(\Leftrightarrow\left(5x-15\right)^2-\left(14x+7\right)^2=0\)

\(\Leftrightarrow\left(5x-15-14x-7\right)\left(5x-15+14x+7\right)=0\)

\(\Leftrightarrow\left(-9x-22\right)\left(19x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-9x-22=0\\19x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-9x=22\\19x=8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{22}{-9}=\dfrac{-22}{9}\\x=\dfrac{8}{19}\end{matrix}\right.\)

Vậy \(x=\dfrac{-22}{9}\) hoặc \(x=\dfrac{8}{19}\)

d) \(9\left(3x-2\right)^2=121\left(1-4x\right)^2\)

\(\Leftrightarrow9\left(3x-2\right)^2-121\left(1-4x\right)^2=0\)

\(\Leftrightarrow3^2\left(3x-2\right)^2-11^2\left(1-4x\right)^2=0\)

\(\Leftrightarrow\left[3\left(3x-2\right)\right]^2-\left[11\left(1-4x\right)\right]^2=0\)

\(\Leftrightarrow\left(9x-6\right)^2-\left(11-44x\right)^2=0\)

\(\Leftrightarrow\left(9x-6-11+44x\right)\left(9x-6+11-44x\right)=0\)

\(\Leftrightarrow\left(53x-17\right)\left(-35x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}53x-17=0\\-35x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}53x=17\\-35x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{53}\\x=\dfrac{-5}{-35}=\dfrac{1}{7}\end{matrix}\right.\)

Vậy \(x=\dfrac{17}{53}\) hoặc \(x=\dfrac{1}{7}\)