Trắc nghiệm Toán Bài 1: Phân thức đại số có lời giải chi tiết (Phần 1)

 \(\frac{\text{3x(x}+{\text{y)}}^{\text{3}}}{{\text{9x}}^{\text{2}}{(x+y)}^{2}}\)

 \(\frac{\text{3x(x}+{\text{y)}}^{\text{3}}}{{\text{9x}}^{\text{2}}{(x+y)}^{3}}\)

 \(\frac{\text{3x(x}+{\text{y)}}^{\text{2}}}{\text{9x}{(x+y)}^{2}}\)

 \(\frac{\text{3x(x}+{\text{y)}}^{\text{3}}}{\text{9x}{(x+y)}^{2}}\)

 \(\frac{\text{x-3}}{{\text{x}}^{\text{2}}-\text{9}}=\frac{\text{1}}{\text{x}+\text{3}}\)

 \(\frac{\text{3x-3}}{\text{3x}}=\frac{\text{x}-\text{1}}{\text{x}}\)

 \(\frac{{\text{x}}^{\text{2}}-\text{6x}+\text{9}}{{\text{9-x}}^{\text{2}}}=\frac{\text{3}-\text{x}}{\text{x}+\text{3}}\)

 \(\frac{{\text{x(x}}^{\text{2}}-\text{4)}}{\text{2-x}}=\text{x(x}+\text{2)}\)

x ≤ 2

x ≠ 1

x = 2

x ≠ 2

x = 1

x = -1

x = -1; x = 1

x = 0

 \(\frac{\text{3a(x}+{\text{y)}}^{\text{2}}}{\text{9a}(x+y)}\)(x ≠ -y)

 \(\frac{\text{-x}-\text{y}}{\text{3a}}\)

 \(\frac{\text{-x}+\text{y}}{\text{3a}}\)

 \(\frac{\text{3a(x}+{\text{y)}}^{\text{2}}}{{\text{9a}}^{\text{2}}(x+y)}\)(x ≠ -y)

1

2

3

-1

 \(\frac{\text{5x}+\text{5}}{\text{5x}}=\frac{\text{x}+\text{1}}{\text{x}}\)

 \(\frac{{\text{x}}^{\text{2}}-\text{9}}{\text{x}+\text{3}}=\text{ }x\text{ }-\text{ }3\)

 \(\frac{\text{x}+\text{3}}{{\text{x}}^{\text{2}}-\text{9}}=\frac{\text{1}}{\text{x}-\text{3}}\)

 \(\frac{\text{5x}+\text{5}}{\text{5x}}=\text{5}\)

0

2

3

1

\(x = \frac{1}{16}\)

\(x = \frac{-1}{16}\)

\(x = \frac{1}{4}\)

Không có x thỏa mãn

 \(\frac{\text{x-1}}{\text{x}+\text{3}}\)

 \(\frac{\text{x+1}}{\text{x-3}}\)

 \(\frac{\text{x-1}}{\text{x-3}}\)

 \(\frac{\text{x+1}}{\text{x}+\text{3}}\)

 \(\frac{{\text{-x}}^{\text{2}}+3x}{\text{3x-9}}\)

 \(\frac{{\text{x}}^{\text{2}}}{\text{3}}\)

 \(\frac{{\text{-x}}^{\text{3}}-{x}^{2}}{\text{3x}+\text{3}}\)

 \(\frac{-2{\text{x}}^{\text{3}}}{\text{6x}}\)

x ≠ -4.

x ≠ 3.

x ≠ 4.

x ≠ 2.

A. B = C. D

A. C = B. D

A. D = B. C

A. C < B. D

\(M = 6{x}^{2} + 9x\)

M = -3x

M = 3x

M = 2x + 3