Trang chủ » Hỏi đáp » Môn Toán Tìm x, biết: a, (4x+1)²=8x+2 b, x²+12x+8=0 c, x²-6x-1=0 d, (5x-1)²=(x+1)² e, x³+7x=4x+28 08/01/2025 Tìm x, biết: a, (4x+1)²=8x+2 b, x²+12x+8=0 c, x²-6x-1=0 d, (5x-1)²=(x+1)² e, x³+7x=4x+28 f, x+x+1
Giải đáp:$\begin{array}{l}a)x = – \dfrac{1}{4};x = \dfrac{1}{4}\\b)x = \pm 2\sqrt 7 – 6\\c)x = 3 \pm \sqrt {10} \\d)x = \dfrac{1}{2};x = 0\\e)x = – 7;x = 4\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a){\left( {4x + 1} \right)^2} = 8x + 2\\ \Leftrightarrow {\left( {4x + 1} \right)^2} = 2\left( {4x + 1} \right)\\ \Leftrightarrow {\left( {4x + 1} \right)^2} – 2\left( {4x + 1} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {4x + 1 – 2} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {4x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}4x + 1 = 0\\4x – 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{1}{4}\\x = \dfrac{1}{4}\end{array} \right.\\Vậy\,x = – \dfrac{1}{4};x = \dfrac{1}{4}\\b){x^2} + 12x + 8 = 0\\ \Leftrightarrow {x^2} + 2.x.6 + 36 – 28 = 0\\ \Leftrightarrow {\left( {x + 6} \right)^2} = 28\\ \Leftrightarrow \left[ \begin{array}{l}x + 6 = 2\sqrt 7 \\x + 6 = – 2\sqrt 7 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\sqrt 7 – 6\\x = – 2\sqrt 7 – 6\end{array} \right.\\Vậy\,x = \pm 2\sqrt 7 – 6\\c){x^2} – 6x – 1 = 0\\ \Leftrightarrow {x^2} – 6x + 9 – 10 = 0\\ \Leftrightarrow {\left( {x – 3} \right)^2} = 10\\ \Leftrightarrow \left[ \begin{array}{l}x – 3 = \sqrt {10} \\x – 3 = – \sqrt {10} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 3 + \sqrt {10} \\x = 3 – \sqrt {10} \end{array} \right.\\Vậy\,x = 3 \pm \sqrt {10} \\d){\left( {5x – 1} \right)^2} = {\left( {x + 1} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l}5x – 1 = x + 1\\5x – 1 = – x – 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}4x = 2\\6x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{1}{2}\\x = 0\end{array} \right.\\Vậy\,x = \dfrac{1}{2};x = 0\\e){x^2} + 7x = 4x + 28\\ \Leftrightarrow x\left( {x + 7} \right) = 4\left( {x + 7} \right)\\ \Leftrightarrow x\left( {x + 7} \right) – 4\left( {x + 7} \right) = 0\\ \Leftrightarrow \left( {x + 7} \right)\left( {x – 4} \right) = 0\\ \Leftrightarrow x = – 7;x = 4\\Vậy\,x = – 7;x = 4\\f)x + x + 1??\end{array}$ Trả lời
a)x = – \dfrac{1}{4};x = \dfrac{1}{4}\\
b)x = \pm 2\sqrt 7 – 6\\
c)x = 3 \pm \sqrt {10} \\
d)x = \dfrac{1}{2};x = 0\\
e)x = – 7;x = 4
\end{array}$
a){\left( {4x + 1} \right)^2} = 8x + 2\\
\Leftrightarrow {\left( {4x + 1} \right)^2} = 2\left( {4x + 1} \right)\\
\Leftrightarrow {\left( {4x + 1} \right)^2} – 2\left( {4x + 1} \right) = 0\\
\Leftrightarrow \left( {4x + 1} \right)\left( {4x + 1 – 2} \right) = 0\\
\Leftrightarrow \left( {4x + 1} \right)\left( {4x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x + 1 = 0\\
4x – 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{1}{4}\\
x = \dfrac{1}{4}
\end{array} \right.\\
Vậy\,x = – \dfrac{1}{4};x = \dfrac{1}{4}\\
b){x^2} + 12x + 8 = 0\\
\Leftrightarrow {x^2} + 2.x.6 + 36 – 28 = 0\\
\Leftrightarrow {\left( {x + 6} \right)^2} = 28\\
\Leftrightarrow \left[ \begin{array}{l}
x + 6 = 2\sqrt 7 \\
x + 6 = – 2\sqrt 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\sqrt 7 – 6\\
x = – 2\sqrt 7 – 6
\end{array} \right.\\
Vậy\,x = \pm 2\sqrt 7 – 6\\
c){x^2} – 6x – 1 = 0\\
\Leftrightarrow {x^2} – 6x + 9 – 10 = 0\\
\Leftrightarrow {\left( {x – 3} \right)^2} = 10\\
\Leftrightarrow \left[ \begin{array}{l}
x – 3 = \sqrt {10} \\
x – 3 = – \sqrt {10}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3 + \sqrt {10} \\
x = 3 – \sqrt {10}
\end{array} \right.\\
Vậy\,x = 3 \pm \sqrt {10} \\
d){\left( {5x – 1} \right)^2} = {\left( {x + 1} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
5x – 1 = x + 1\\
5x – 1 = – x – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = 2\\
6x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 0
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2};x = 0\\
e){x^2} + 7x = 4x + 28\\
\Leftrightarrow x\left( {x + 7} \right) = 4\left( {x + 7} \right)\\
\Leftrightarrow x\left( {x + 7} \right) – 4\left( {x + 7} \right) = 0\\
\Leftrightarrow \left( {x + 7} \right)\left( {x – 4} \right) = 0\\
\Leftrightarrow x = – 7;x = 4\\
Vậy\,x = – 7;x = 4\\
f)x + x + 1??
\end{array}$