Giúp với nha, cảm ơn trước ạ. Mong được các chuyên gia giúp đỡ.!!???? November 4, 2020 by Hưng Gia Giúp với nha, cảm ơn trước ạ. Mong được các chuyên gia giúp đỡ.!!????
Đáp án: n. \(\left[ \begin{array}{l}x = 2\\x = – 1\\x = – 2\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}g.2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0\\ \to \left( {x + 5} \right)\left( {2 – x} \right) = 0\\ \to \left[ \begin{array}{l}x + 5 = 0\\2 – x = 0\end{array} \right. \to \left[ \begin{array}{l}x = – 5\\x = 2\end{array} \right.\\h.3x\left( {{x^2} – 16} \right) = 0\\ \to 3x\left( {x – 4} \right)\left( {x + 4} \right) = 0\\ \to \left[ \begin{array}{l}x = 0\\x – 4 = 0\\x + 4 = 0\end{array} \right. \to \left[ \begin{array}{l}x = 0\\x = 4\\x = – 4\end{array} \right.\\i.{\left( {2x – 3} \right)^2} = {\left( {x – 1} \right)^2}\\ \to \left| {2x – 3} \right| = \left| {x – 1} \right|\\ \to \left[ \begin{array}{l}2x – 3 = x – 1\\2x – 3 = – x + 1\end{array} \right.\\ \to \left[ \begin{array}{l}x = 2\\3x = 4\end{array} \right.\\ \to \left[ \begin{array}{l}x = 2\\x = \dfrac{4}{3}\end{array} \right.\\k.{x^2} + 6x + 9 + {x^2} – 4x + 4 = 2{x^2}\\ \to 2x = – 13\\ \to x = – \dfrac{{13}}{2}\\m.3{x^2} – 12x – 5x – 3{x^2} = – 34\\ \to – 17x = – 34\\ \to x = 2\\n.{x^3} – 2{x^2} + 3{x^2} – 6x + 2x – 4 = 0\\ \to {x^2}\left( {x – 2} \right) + 3x\left( {x – 2} \right) + 2\left( {x – 2} \right) = 0\\ \to \left( {x – 2} \right)\left( {{x^2} + 3x + 2} \right) = 0\\ \to \left( {x – 2} \right)\left( {{x^2} + x + 2x + 2} \right) = 0\\ \to \left( {x – 2} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 0\\ \to \left[ \begin{array}{l}x = 2\\x = – 1\\x = – 2\end{array} \right.\end{array}\) Reply
h) 3x³ – 48x = 0 ⇔ 3(x³ – 16x) = 0 ⇔ x³ – 16x = 0 ⇔ x(x²-16) = 0 ⇔ x(x-4)(x+4) = 0 ⇔ \(\left[ \begin{array}{l}x=0\\ x+4=0\\x-4=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0\\ x=-4\\x=4\end{array} \right.\) i) (2x-3)² = x² – 2x + 1 ⇔ (2x-3)² = (x-1)² ⇔ 2x-3 = x-1 ⇔ x = 2 k) (x+3)² + (x-2)² = 2x² ⇔ x² + 6x + 9 + x² – 4x + 4 – 2x² = 0 ⇔ 6x + 9 -4x + 4 = 0 ⇔ 2x + 13 = 0 ⇔ 2x = -13 ⇔ x = $\frac{-13}{2}$ m) 3x(x-4) – x(5+3x) = -34 ⇔ 3x² – 12² – 5x – 3x² = -34 ⇔ -5x = -34 + 144 = 126 ⇔ x = – 25,2 n) x³ + x² – 4x = 4 ⇔ x³ + x² – 4x – 4 = 0 ⇔ x²(x+1) -4(x+1) = 0 ⇔ (x+1)(x²-4) =0 ⇔ (x-1)(x-2)(x+2) =0 ⇔ \(\left[ \begin{array}{l}x-1=0\\ x-2=0\\x+2=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=1\\ x=2\\x=-2\end{array} \right.\) Reply
Đáp án:
n. \(\left[ \begin{array}{l}
x = 2\\
x = – 1\\
x = – 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
g.2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0\\
\to \left( {x + 5} \right)\left( {2 – x} \right) = 0\\
\to \left[ \begin{array}{l}
x + 5 = 0\\
2 – x = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = – 5\\
x = 2
\end{array} \right.\\
h.3x\left( {{x^2} – 16} \right) = 0\\
\to 3x\left( {x – 4} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x – 4 = 0\\
x + 4 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
x = 4\\
x = – 4
\end{array} \right.\\
i.{\left( {2x – 3} \right)^2} = {\left( {x – 1} \right)^2}\\
\to \left| {2x – 3} \right| = \left| {x – 1} \right|\\
\to \left[ \begin{array}{l}
2x – 3 = x – 1\\
2x – 3 = – x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
3x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{4}{3}
\end{array} \right.\\
k.{x^2} + 6x + 9 + {x^2} – 4x + 4 = 2{x^2}\\
\to 2x = – 13\\
\to x = – \dfrac{{13}}{2}\\
m.3{x^2} – 12x – 5x – 3{x^2} = – 34\\
\to – 17x = – 34\\
\to x = 2\\
n.{x^3} – 2{x^2} + 3{x^2} – 6x + 2x – 4 = 0\\
\to {x^2}\left( {x – 2} \right) + 3x\left( {x – 2} \right) + 2\left( {x – 2} \right) = 0\\
\to \left( {x – 2} \right)\left( {{x^2} + 3x + 2} \right) = 0\\
\to \left( {x – 2} \right)\left( {{x^2} + x + 2x + 2} \right) = 0\\
\to \left( {x – 2} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 1\\
x = – 2
\end{array} \right.
\end{array}\)
h) 3x³ – 48x = 0
⇔ 3(x³ – 16x) = 0
⇔ x³ – 16x = 0
⇔ x(x²-16) = 0
⇔ x(x-4)(x+4) = 0
⇔ \(\left[ \begin{array}{l}x=0\\ x+4=0\\x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\ x=-4\\x=4\end{array} \right.\)
i) (2x-3)² = x² – 2x + 1
⇔ (2x-3)² = (x-1)²
⇔ 2x-3 = x-1
⇔ x = 2
k) (x+3)² + (x-2)² = 2x²
⇔ x² + 6x + 9 + x² – 4x + 4 – 2x² = 0
⇔ 6x + 9 -4x + 4 = 0
⇔ 2x + 13 = 0
⇔ 2x = -13
⇔ x = $\frac{-13}{2}$
m) 3x(x-4) – x(5+3x) = -34
⇔ 3x² – 12² – 5x – 3x² = -34
⇔ -5x = -34 + 144 = 126
⇔ x = – 25,2
n) x³ + x² – 4x = 4
⇔ x³ + x² – 4x – 4 = 0
⇔ x²(x+1) -4(x+1) = 0
⇔ (x+1)(x²-4) =0
⇔ (x-1)(x-2)(x+2) =0
⇔ \(\left[ \begin{array}{l}x-1=0\\ x-2=0\\x+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\ x=2\\x=-2\end{array} \right.\)