Tháng Năm 19, 2024

Tìm \(x\) biết: \(a){{\left( x+5 \right)}^{2}}-2\left( x+5 \right)\left( x-2 \right)+{{\left( x-2 \right)}^{2}}=49\) \(b)\ 4{{x}^{2}}+12x+9=0\) \(c)\ 9{{x}^{2}}-16=0\) \(d)\ {{x}^{3}}-3{{x}^{2}}+3x+63=0\)

Tìm \(x\) biết:

\(a){{\left( x+5 \right)}^{2}}-2\left( x+5 \right)\left( x-2 \right)+{{\left( x-2 \right)}^{2}}=49\)

\(b)\ 4{{x}^{2}}+12x+9=0\)

\(c)\ 9{{x}^{2}}-16=0\)

\(d)\ {{x}^{3}}-3{{x}^{2}}+3x+63=0\)

Lời giải chi tiết:

Hướng dẫn chi tiết:

\(\begin{array}{l}a)\;{\left( {x + 5} \right)^2} – 2\left( {x + 5} \right)\left( {x – 2} \right) + {\left( {x – 2} \right)^2} = 49\\ \Leftrightarrow {\left( {\left( {x + 5} \right) – \left( {x – 2} \right)} \right)^2} = 49\\ \Leftrightarrow {\left( {x + 5 – x + 2} \right)^2} = 49\\ \Leftrightarrow {7^2} = 49\end{array}\)

Vậy với mọi \(x\) đều thỏa mãn.

\(\begin{array}{l}b)\;4{x^2} + 12x + 9 = 0\\ \Leftrightarrow {\left( {2x} \right)^2} + 2.2x.3 + {3^2} = 0\\ \Leftrightarrow {\left( {2x + 3} \right)^2} = 0\\ \Leftrightarrow 2x + 3 = 0\\ \Leftrightarrow x = – \frac{3}{2}\end{array}\)

Vậy \(x=-\frac{3}{2}\)

\(\begin{array}{l}c)\;9{x^2} – 16 = 0\\ \Leftrightarrow {\left( {3x} \right)^2} – {4^2} = 0\\ \Leftrightarrow \left( {3x – 4} \right)\left( {3x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}3x – 4 = 0\\3x + 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{4}{3}\\x = – \frac{4}{3}\end{array} \right.\end{array}\)

Vậy \(x=\frac{4}{3}\) hoặc \(x=-\frac{4}{3}\).

\(\begin{array}{l}d)\;{x^3} – 3{x^2} + 3x + 63 = 0\\ \Leftrightarrow {x^3} + 3.{x^2}.\left( { – 1} \right) + 3.x.{\left( { – 1} \right)^2} + {\left( { – 1} \right)^3} + 64 = 0\\ \Leftrightarrow {\left( {x – 1} \right)^3} + 64 = 0\\ \Leftrightarrow {\left( {x – 1} \right)^3} = – 64\\ \Leftrightarrow {\left( {x – 1} \right)^3} = {\left( { – 4} \right)^3}\\ \Leftrightarrow x – 1 = – 4\\ \Leftrightarrow x = – 3\end{array}\)

Vậy \(x=-3\)