Tim x biêt a,3x+2 (5-x)=0 b,x (2x-1)(x+5)-(2x^+1)x+4,5)=3,5 c,(x-2)^-(x-3)(x+3)=6 d,4 (x-3)^-(2x-1)(2x+1)=10

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
3x + 2\left( {5 – x} \right) = 0\\
 \Leftrightarrow 3x + 10 – 2x = 0\\
 \Leftrightarrow x + 10 = 0\\
 \Leftrightarrow x =  – 10\\
b,\\
x\left( {2x – 1} \right)\left( {x + 5} \right) – \left( {2{x^2} + 1} \right)x + 4,5 = 3,5\\
 \Leftrightarrow x\left( {2{x^2} + 9x – 5} \right) – \left( {2{x^2} + 1} \right)x – 1 = 0\\
 \Leftrightarrow x\left[ {2{x^2} + 9x – 5 – 2{x^2} – 1} \right] – 1 = 0\\
 \Leftrightarrow x\left( {9x – 6} \right) – 1 = 0\\
 \Leftrightarrow 9{x^2} – 6x – 1 = 0\\
 \Leftrightarrow x = \frac{{1 \pm \sqrt 2 }}{3}\\
c,\\
{\left( {x – 2} \right)^2} – \left( {x – 3} \right)\left( {x + 3} \right) = 6\\
 \Leftrightarrow \left( {{x^2} – 4x + 4} \right) – \left( {{x^2} – 9} \right) = 6\\
 \Leftrightarrow  – 4x + 13 = 6\\
 \Leftrightarrow 4x = 7\\
 \Leftrightarrow x = \frac{7}{4}\\
d,\\
4{\left( {x – 3} \right)^2} – \left( {2x – 1} \right)\left( {2x + 1} \right) = 10\\
 \Leftrightarrow 4\left( {{x^2} – 6x + 9} \right) – \left( {4{x^2} – 1} \right) = 10\\
 \Leftrightarrow  – 24x + 27 = 0\\
 \Leftrightarrow x = \frac{{27}}{{24}} = \frac{9}{8}
\end{array}\)