10000 câu trắc nghiệm tổng hợp môn Toán 2025 mới nhất (có đáp án) - Phần 20

Lời giải:

x⁵ – x⁴ + x³ – x²

= (x⁵ – x⁴) + (x³ – x²)

= x⁴(x – 1) + x²(x – 1)

= (x – 1)(x⁴ + x²)

= (x – 1)x²(x² + 1)

Lời giải:

\(\left\{ \begin{array}{l}x + my = 3\\mx + y = 2m + 1\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}x + my = 3\\{m^2}x + my = 2{m^2} + m\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}x + my = 3\\\left( {{m^2} - 1} \right)x = 2{m^2} + m - 3\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}x + my = 3\\x = \frac{{2{m^2} + m - 3}}{{{m^2} - 1}} = \frac{{\left( {2m + 3} \right)\left( {m - 1} \right)}}{{\left( {m - 1} \right)\left( {m + 1} \right)}} = \frac{{2m + 3}}{{m + 1}}\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}y = \frac{1}{{m + 1}}\\x = \frac{{2m + 3}}{{m + 1}}\end{array} \right.\)

Khi đó P = x² + 3y² =

\({\left( {\frac{{2m + 3}}{{m + 1}}} \right)^2} + 3.{\left( {\frac{1}{{m + 1}}} \right)^2} = 4 + \frac{4}{{m + 1}} + \frac{4}{{{{\left( {m + 1} \right)}^2}}} = {\left( {\frac{2}{{m + 1}} + 1} \right)^2} + 3 \ge 3\)

Vậy P_min = 3 khi m = -3

Lời giải:

\(\left\{ \begin{array}{l}x + y + \frac{1}{x} + \frac{1}{y} = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}\left( {x + \frac{1}{y}} \right) = xy + \frac{1}{{xy}}\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left( {x + \frac{1}{y}} \right) + \left( {y + \frac{1}{x}} \right) = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}\left( {x + \frac{1}{y}} \right) = \left( {x + \frac{1}{y}} \right)\left( {y + \frac{1}{x}} \right)\end{array} \right.\)

Đặt \(x + \frac{1}{y} = a;y + \frac{1}{x} = b\)

Khi đó ta có hệ: \(\left\{ \begin{array}{l}a + b = \frac{9}{2}\\\frac{1}{4} + \frac{3}{2}a = ab\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\\frac{1}{4} + \frac{3}{2}\left( {\frac{9}{2} - b} \right) = \left( {\frac{9}{2} - b} \right)b\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\7 - 6b + {b^2} = 0\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}a = \frac{9}{2} - b\\\left[ \begin{array}{l}b = 1\\b = - 7\end{array} \right.\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}a = \frac{7}{2}\\b = 1\end{array} \right.\\\left\{ \begin{array}{l}a = \frac{{23}}{2}\\b = - 7\end{array} \right.\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{7}{2}\\y + \frac{1}{x} = 1\end{array} \right.\\\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{{23}}{2}\\y + \frac{1}{x} = - 7\end{array} \right.\end{array} \right.\)

+ Với \[\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{7}{2}\\y + \frac{1}{x} = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{1}{{1 - \frac{1}{x}}} = \frac{7}{2}\\y = 1 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x\left( {1 - \frac{1}{x}} \right) + 1 = \frac{7}{2}\left( {1 - \frac{1}{x}} \right)\\y = 1 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{7}{2} - \frac{7}{{2x}}\\y = 1 - \frac{1}{x}\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}x + \frac{7}{{2x}} - \frac{7}{2} = 0\\y = 1 - \frac{1}{x}\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}2{x^2} + 7 - 2x = 0\left( {VN} \right)\\y = 1 - \frac{1}{x}\end{array} \right.\]

+ Với \(\left\{ \begin{array}{l}x + \frac{1}{y} = \frac{{23}}{2}\\y + \frac{1}{x} = - 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + \frac{1}{{ - 7 - \frac{1}{x}}} = \frac{{23}}{2}\\y = - 7 - \frac{1}{x}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}7x = \frac{{161x}}{{2x}} + \frac{{23}}{{2x}}\\y = - 7 - \frac{1}{x}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}49{x^2} - 161x - 23 = 0\\y = - 7 - \frac{1}{x}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = \frac{{23 + 3\sqrt {69} }}{{14}}\\y = \frac{{ - 161 - 21\sqrt {69} }}{{46}}\end{array} \right.\\\left\{ \begin{array}{l}x = \frac{{23 - 3\sqrt {69} }}{{14}}\\y = \frac{{ - 161 + 21\sqrt {69} }}{{46}}\end{array} \right.\end{array} \right.\)

Lời giải:

Theo hình vẽ ta có: a ⊥ c; b ⊥ c nên a // b

Suy ra: x + y = 180°

Mà 2x = 3y nên \(\frac{x}{3} = \frac{y}{2} = \frac{{x + y}}{{3 + 2}} = \frac{{180^\circ }}{5} = 36^\circ \)

⇒ x = 36° . 3 = 108°

y = 36° . 2 = 72°

Lời giải:

\(\left\{ \begin{array}{l}x + y + xy + 1 = 0\\{x^2} + {y^2} - x - y = 22\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}y\left( {x + 1} \right) + x + 1 = 0\\{y^2} - y + {x^2} - x = 22\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left( {y + 1} \right)\left( {x + 1} \right) = 0\\{y^2} - y + {x^2} - x = 22\end{array} \right.\)

⇔ \(\left\{ \begin{array}{l}\left[ \begin{array}{l}x = - 1\\y = - 1\end{array} \right.\\{y^2} - y + {x^2} - x = 22\end{array} \right.\)

+ Nếu x = -1 thì: y² – y + 1 + 2 – 22 = 0

⇔ y² – y – 20 = 0

⇔ \(\left[ \begin{array}{l}y = 5\\y = - 4\end{array} \right.\)

+ Nếu y = -1 thì x² – x – 20 = 0

⇔ \(\left[ \begin{array}{l}x = 5\\x = - 4\end{array} \right.\)

Vậy tập nghiệm của phương trình là {−1;5};{−1;−4};{5;−1};{−4;−1}

Lời giải:

Ta có: x + y + z = 1 nên \(\left\{ \begin{array}{l}x = 1 - y - z\\y = 1 - z - x\\z = 1 - x - y\end{array} \right.\) và \(\left\{ \begin{array}{l}x + y = 1 - z\\y + z = 1 - x\\z + x = 1 - y\end{array} \right.\)

\(P = \frac{{{{\left( {x + y} \right)}^2}}}{{xy + z}}.\frac{{{{\left( {y + z} \right)}^2}}}{{yz + x}}.\frac{{{{\left( {x + z} \right)}^2}}}{{zx + y}}\)

\(P = \frac{{{{\left( {x + y} \right)}^2}}}{{xy + 1 - x - y}}.\frac{{{{\left( {y + z} \right)}^2}}}{{yz + 1 - y - z}}.\frac{{{{\left( {x + z} \right)}^2}}}{{zx + 1 - z - x}}\)

\(P = \frac{{{{\left( {x + y} \right)}^2}}}{{\left( {y - 1} \right)\left( {x - 1} \right)}}.\frac{{{{\left( {y + z} \right)}^2}}}{{\left( {y - 1} \right)\left( {z - 1} \right)}}.\frac{{{{\left( {x + z} \right)}^2}}}{{\left( {z - 1} \right)\left( {x - 1} \right)}}\)

\(P = \frac{{{{\left( {x + y} \right)}^2}{{\left( {y + z} \right)}^2}{{\left( {z + x} \right)}^2}}}{{{{\left[ {\left( {x - 1} \right)\left( {y - 1} \right)\left( {z - 1} \right)} \right]}^2}}}\)

\[P = \frac{{{{\left( {z - 1} \right)}^2}{{\left( {x - 1} \right)}^2}{{\left( {y - 1} \right)}^2}}}{{{{\left[ {\left( {x - 1} \right)\left( {y - 1} \right)\left( {z - 1} \right)} \right]}^2}}}\]

P = 1

Vậy giá trị biểu thức P không phụ thuộc vào giá trị của biến.

Lời giải:

Xét x⁴ + y⁴ + z⁴ =

\[\frac{{{x^4} + {y^4}}}{2} + \frac{{{y^4} + {z^4}}}{2} + \frac{{{z^4} + {x^4}}}{2} \ge {x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} = \frac{{{x^2}{y^2} + {y^2}{z^2}}}{2} + \frac{{{y^2}{z^2} + {x^2}{z^2}}}{2} + \frac{{{x^2}{y^2} + {x^2}{z^2}}}{2}\]

\( \ge x{y^2}z + xy{z^2} + {x^2}yz = xyz\left( {x + y + z} \right) = xyz\)

Dấu “=” xảy ra khi x = y = z

Mà x + y + z = 1 nên \(x = y = z = \frac{1}{3}\)

Lời giải:

Xét x + y = 3(x – y)

Ta có: x + y = 3(x – y)

⇔ x + y – 3x + 3y = 0

⇔ 4y – 2x = 0

⇔ 2(2y – x) = 0

⇔ 2y = x

Thay vào ta được:

2y + y = 3(2y – y) = \(\frac{{4y}}{y} = 4\)

Suy ra: \(x = \frac{8}{3};y = \frac{4}{3}\)