TÌM X: a)2x ²(2x-3)-x ²(4x ²-6x+2)=0 b)(x-2) ²-(x+3) ²-4(x+1)=5 c)4x ²-4x+1=(5-x) ² d)4x ²-8x+4=2(1-x)(1+x) 27/08/2021 Bởi Eva TÌM X: a)2x ²(2x-3)-x ²(4x ²-6x+2)=0 b)(x-2) ²-(x+3) ²-4(x+1)=5 c)4x ²-4x+1=(5-x) ² d)4x ²-8x+4=2(1-x)(1+x)
Giải thích các bước giải: \[\begin{array}{l}a,\\2{x^2}\left( {2x – 3} \right) – {x^2}\left( {4{x^2} – 6x + 2} \right) = 0\\ \Leftrightarrow {x^2}\left( {4x – 6 – \left( {4{x^2} – 6x + 2} \right)} \right) = 0\\ \Leftrightarrow {x^2}\left( { – 4{x^2} + 10x – 8} \right) = 0\\ \Leftrightarrow {x^2}\left( { – \left( {2{x^2}} \right) + 2.2x.\frac{5}{2} – \frac{{25}}{4} – \frac{7}{4}} \right) = 0\\ \Leftrightarrow {x^2}\left( { – {{\left( {2x – \frac{5}{2}} \right)}^2} – \frac{7}{4}} \right) = 0\\ – {\left( {2x – \frac{5}{2}} \right)^2} – \frac{7}{4} < 0 \Rightarrow x = 0\\b,\\{\left( {x – 2} \right)^2} – {\left( {x + 3} \right)^2} – 4\left( {x + 1} \right) = 5\\ \Leftrightarrow \left( {x – 2 + x + 3} \right)\left( {x – 2 – x – 3} \right) – 4\left( {x + 1} \right) = 5\\ \Leftrightarrow – 5\left( {2x + 1} \right) – 4x – 4 = 5\\ \Leftrightarrow – 14x = 14\\ \Leftrightarrow x = – 1\\c,\\4{x^2} – 4x + 1 = {\left( {5 – x} \right)^2}\\ \Leftrightarrow {\left( {2x – 1} \right)^2} – {\left( {5 – x} \right)^2} = 0\\ \Leftrightarrow \left( {2x – 1 – 5 + x} \right)\left( {2x – 1 + 5 – x} \right) = 0\\ \Leftrightarrow \left( {3x – 6} \right)\left( {x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\\x = – 4\end{array} \right.\\d,\\4{x^2} – 8x + 4 = 2\left( {1 – x} \right)\left( {1 + x} \right)\\ \Leftrightarrow 4{x^2} – 8x + 4 = 2\left( {1 – {x^2}} \right)\\ \Leftrightarrow 6{x^2} – 8x + 2 = 0\\ \Leftrightarrow 2\left( {x – 1} \right)\left( {3x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = \frac{1}{3}\end{array} \right.\end{array}\] Bình luận
\[\begin{array}{l}a){x^2}(4x – 6 – 4{x^2} + 6x – 2) = 0\\ \to {x^2}( – 4{x^2} + 10x – 8) = 0\\ \to {x^2}({x^2} – 5x + 4) = 0\\ \to {x^2}(x – 1)(x – 4) = 0 \to x = 0;x = 1;x = 4\\b){x^2} – 4x + 4 – ({x^2} + 6x + 9) – 4x – 4 – 5 = 0\\ \to – 14x – 14 = 0 \to x = – 1\end{array}\] Bình luận
Giải thích các bước giải:
\[\begin{array}{l}
a,\\
2{x^2}\left( {2x – 3} \right) – {x^2}\left( {4{x^2} – 6x + 2} \right) = 0\\
\Leftrightarrow {x^2}\left( {4x – 6 – \left( {4{x^2} – 6x + 2} \right)} \right) = 0\\
\Leftrightarrow {x^2}\left( { – 4{x^2} + 10x – 8} \right) = 0\\
\Leftrightarrow {x^2}\left( { – \left( {2{x^2}} \right) + 2.2x.\frac{5}{2} – \frac{{25}}{4} – \frac{7}{4}} \right) = 0\\
\Leftrightarrow {x^2}\left( { – {{\left( {2x – \frac{5}{2}} \right)}^2} – \frac{7}{4}} \right) = 0\\
– {\left( {2x – \frac{5}{2}} \right)^2} – \frac{7}{4} < 0 \Rightarrow x = 0\\
b,\\
{\left( {x – 2} \right)^2} – {\left( {x + 3} \right)^2} – 4\left( {x + 1} \right) = 5\\
\Leftrightarrow \left( {x – 2 + x + 3} \right)\left( {x – 2 – x – 3} \right) – 4\left( {x + 1} \right) = 5\\
\Leftrightarrow – 5\left( {2x + 1} \right) – 4x – 4 = 5\\
\Leftrightarrow – 14x = 14\\
\Leftrightarrow x = – 1\\
c,\\
4{x^2} – 4x + 1 = {\left( {5 – x} \right)^2}\\
\Leftrightarrow {\left( {2x – 1} \right)^2} – {\left( {5 – x} \right)^2} = 0\\
\Leftrightarrow \left( {2x – 1 – 5 + x} \right)\left( {2x – 1 + 5 – x} \right) = 0\\
\Leftrightarrow \left( {3x – 6} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = – 4
\end{array} \right.\\
d,\\
4{x^2} – 8x + 4 = 2\left( {1 – x} \right)\left( {1 + x} \right)\\
\Leftrightarrow 4{x^2} – 8x + 4 = 2\left( {1 – {x^2}} \right)\\
\Leftrightarrow 6{x^2} – 8x + 2 = 0\\
\Leftrightarrow 2\left( {x – 1} \right)\left( {3x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{1}{3}
\end{array} \right.
\end{array}\]
\[\begin{array}{l}
a){x^2}(4x – 6 – 4{x^2} + 6x – 2) = 0\\
\to {x^2}( – 4{x^2} + 10x – 8) = 0\\
\to {x^2}({x^2} – 5x + 4) = 0\\
\to {x^2}(x – 1)(x – 4) = 0 \to x = 0;x = 1;x = 4\\
b){x^2} – 4x + 4 – ({x^2} + 6x + 9) – 4x – 4 – 5 = 0\\
\to – 14x – 14 = 0 \to x = – 1
\end{array}\]