12 câu Trắc nghiệm Toán 8 Bài 4: Bất phương trình một ẩn có đáp án (Vận dụng)

Đáp án B

Ta có: $B =\frac{\text{2x}-\text{4}}{\text{3}-\text{x}} < 0$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}\text{2x}-\text{4}>\text{0}\\ \text{3}-\text{x}<\text{0}\end{array}\right.\\ \left\{\begin{array}{l}\text{2x}-\text{4}<\text{0}\\ \text{3}-\text{x}>\text{0}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}\text{x}>\text{2}\\ \text{x}>\text{3}\end{array}\right.\\ \left\{\begin{array}{l}\text{x}<\text{2}\\ \text{x}<\text{3}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\text{x}<\text{2}\\ \text{x}>\text{3}\end{array}\right.$

Vậy với $\left[\begin{array}{l}\text{x}<\text{2}\\ \text{x}>\text{3}\end{array}\right.$ thì B âm

Đáp án A

Ta có: $B =\frac{\text{2x}-\text{4}}{\text{3}-\text{x}} \ge 0$

TH1: 

$$\begin{gathered} \left\{\begin{array}{l}\text{2x}-\text{4}\ge \text{0}\\ \text{3}-\text{x}>\text{0}\end{array}\right.\iff \left\{\begin{array}{l}\text{2x}\ge \text{4}\\ -\text{x}>-\text{3}\end{array}\right. \\ \iff \left\{\begin{array}{l}\text{x}\ge \text{2}\\ \text{x}<\text{3}\end{array}\right.\iff \text{2}\le \text{x}<\text{3} \end{gathered}$$

TH2:

$\left\{\begin{array}{l}\text{2x}-\text{4}\le \text{0}\\ \text{3}-\text{x}<\text{0}\end{array}\right.\iff \left\{\begin{array}{l}\text{2x}\le \text{4}\\ -\text{x}<-\text{3}\end{array}\right.\iff \left\{\begin{array}{l}\text{x}\le \text{2}\\ \text{x}>\text{3}\end{array}\right.$(không có x)

Vậy với 2 ≤ x < 3 thì B có giá trị không âm

Đáp án D

P > 1 $\iff \frac{\text{x}-\text{3}}{\text{x}+\text{1}}$  > 1 $\iff \frac{\text{x}-\text{3}}{\text{x}+\text{1}}$  - 1 > 0

$\iff \frac{\text{x}-\text{3-x-1}}{\text{x}+\text{1}}$  > 0 $\iff \frac{-\text{4}}{\text{x}+\text{1}}$  > 0

Vì -4 < 0 nên suy ra x + 1 < 0 $\iff$ x < -1

Đáp án C

P ≤ 1 $\iff \frac{\text{x}-\text{3}}{\text{x}+\text{1}}$  ≤ 1 $\iff \frac{\text{x}-\text{3}}{\text{x}+\text{1}}$  - 1 ≤ 0

$\iff \frac{\text{x}-\text{3-x-1}}{\text{x}+\text{1}}$  ≤ 0 $\iff \frac{-\text{4}}{\text{x}+\text{1}}$ ≤ 0

Vì -4 < 0 nên suy ra x + 1 > 0 $\iff$ x > -1

Đáp án A

* Ta có $\frac{\text{x}+\text{2}}{\text{5}}-\frac{\text{3x}-\text{7}}{\text{4}}>-\text{5}$

$\iff \frac{\text{4(x}+\text{2)}-\text{5(3x}-\text{7)}}{\text{20}}>\frac{-\text{100}}{\text{20}}$

$\iff$ 4x + 8 - 15x + 35 > -100

$\iff$ -11x > -143

$\iff$ x < 13 (1)

* Ta có $\frac{\text{3x}}{\text{5}}-\frac{\text{x}-\text{4}}{\text{3}}+\frac{\text{x}+\text{2}}{\text{6}}>\text{6}$

$\iff \frac{\text{6.3x}-\text{10(x}-\text{4)}+\text{5(x}+\text{2)}}{\text{30}}>\frac{\text{180}}{\text{30}}$

$\iff$ 18x - 10x + 40 + 5x + 10 > 180

$\iff$ 13x > 130

$\iff$ x > 10 (2)

Kết hợp (1) và (2) ta được: 10 < x < 13

Nên các số nguyên thỏa mãn là x = 11; x = 12

Đáp án A

* Ta có $\frac{\text{x}+\text{2}}{\text{5}}-\frac{\text{3x}-\text{7}}{\text{4}}>-\text{5}$

$\iff \frac{\text{4(x}+\text{2)}-\text{5(3x}-\text{7)}}{\text{20}}>\frac{-\text{100}}{\text{20}}$

$\iff$ 4x + 8 - 15x + 35 > -100

$\iff$ -11x > -143

$\iff$ x < 13 (1)

* Ta có $\frac{\text{3x}}{\text{5}}-\frac{\text{x}-\text{4}}{\text{3}}+\frac{\text{x}+\text{2}}{\text{6}}>\text{6}$

$\iff \frac{\text{6.3x}-\text{10(x}-\text{4)}+\text{5(x}+\text{2)}}{\text{30}}>\frac{\text{180}}{\text{30}}$

$\iff$ 18x - 10x + 40 + 5x + 10 > 180

$\iff$ 13x > 130

$\iff$ x > 10 (2)

Kết hợp (1) và (2) ta được: 10 < x < 13

Nên các số nguyên thỏa mãn là x = 11; x = 12.

Vậy có 2 giá trị nguyên của x thỏa mãn bài toán

Đáp án C

Đáp án B

Đáp án A

Ta có $(x^2 - 4)(x - 3) \ge 0$ $\iff$ (x - 2)(x + 2)(x - 3) ≥ 0

Ta có x - 2 = 0 $\iff$ x = 2; x - 3 = 0

$\iff$ x = 3; x + 2 = 0 $\iff$ x = -2

Bảng xét dấu:

Từ bảng xét dấu ta có $(x^2 - 4)(x - 3) \ge 0$ $\iff$ -2 ≤ x ≤ 2 hoặc x ≥ 3

Đáp án C

Đáp án B

Ta có

$$\begin{gathered} \frac{\text{1987}-\text{x}}{\text{15}}+\frac{\text{1988}-\text{x}}{\text{16}}+\frac{\text{27}+\text{x}}{\text{1999}}+\frac{\text{28}+\text{x}}{\text{2000}}>\text{4} \\ \iff \frac{\text{1987}-\text{x}}{\text{15}}+\frac{\text{1988}-\text{x}}{\text{16}}+\frac{\text{27}+\text{x}}{\text{1999}}+\frac{\text{28}+\text{x}}{\text{2000}}-\text{4}>\text{0} \\ \iff \left(\frac{\text{1987}-\text{x}}{\text{15}}-\text{1}\right)+\left(\frac{\text{1988}-\text{x}}{\text{16}}-\text{1}\right)+\left(\frac{\text{27}+\text{x}}{\text{1999}}-\text{1}\right)+\left(\frac{\text{28}+\text{x}}{\text{2000}}-\text{1}\right)>\text{0} \\ \iff \frac{\text{1972}-\text{x}}{\text{15}}+\frac{\text{1972}-\text{x}}{\text{16}}+\frac{\text{x}-\text{1972}}{\text{1999}}+\frac{\text{x}-\text{1972}}{\text{2000}}>\text{0} \\ \iff (1972 - x)\left(\frac{\text{1}}{\text{15}}+\frac{\text{1}}{\text{16}}-\frac{\text{1}}{\text{1999}}-\frac{\text{1}}{\text{2000}}\right)>0 \end{gathered}$$

Mà $\frac{\text{1}}{\text{15}}+\frac{\text{1}}{\text{16}}-\frac{\text{1}}{\text{1999}}-\frac{\text{1}}{\text{2000}}> 0$ nên 1972 - x > 0

$\iff$ x < 1972

Vậy x < 1972

Đáp án D

Ta có

$$\begin{gathered} \frac{\text{2017}-\text{x}}{\text{15}}+\frac{\text{2018}-\text{x}}{\text{16}}+\frac{\text{17}+\text{x}}{\text{2019}}+\frac{\text{18}+\text{x}}{\text{2020}}\le \text{4} \\ \iff \frac{\text{2017}-\text{x}}{\text{15}}+\frac{\text{2018}-\text{x}}{\text{16}}+\frac{\text{17}+\text{x}}{\text{2019}}+\frac{\text{18}+\text{x}}{\text{2020}}-\text{4}\le \text{0} \\ \iff \frac{\text{2017}-\text{x}}{\text{15}}-1+\frac{\text{2018}-\text{x}}{\text{16}}-1+\frac{\text{17}+\text{x}}{\text{2019}}-1+\frac{\text{18}+\text{x}}{\text{2020}}-1\le \text{0} \\ \iff \frac{\text{2002}-\text{x}}{\text{15}}+\frac{\text{2002}-\text{x}}{\text{16}}+\frac{\text{x-2002}}{\text{2019}}+\frac{\text{x-2002}}{\text{2020}}\le \text{0} \\ \iff -\frac{\text{x-2002}}{\text{15}}-\frac{\text{x-2002}}{\text{16}}+\frac{\text{x-2002}}{\text{2019}}+\frac{\text{x-2002}}{\text{2020}}\le \text{0} \\ \iff (x - 2002)\left(-\frac{\text{1}}{\text{15}}-\frac{\text{1}}{\text{16}}+\frac{\text{1}}{\text{2019}}+\frac{\text{1}}{\text{2020}}\right)\le 0 \end{gathered}$$

Mà $\frac{\text{1}}{\text{2019}}+\frac{\text{1}}{\text{2020}}-\frac{\text{1}}{\text{15}}-\frac{\text{1}}{\text{16}} < 0$  nên

x - 2002 ≥ 0 $\iff$ x ≥ 2002

Vậy giá trị nguyên nhỏ nhất của x là 2002