30 câu Trắc nghiệm Lập phương của một tổng. Lập phương của một hiệu (có đáp án 2024) – Toán 8 Kết nối tri thức

Đáp án đúng là: A

${\left(\text{A + B}\right)}^{\text{3}}{\text{= A}}^{\text{3}}{\text{+ 3A}}^{\text{2}}{\text{B + 3AB}}^{\text{2}}{\text{+ B}}^{\text{3}}$

Đáp án đúng là: A

${\text{x}}^{\text{3}}{\text{+ 3x}}^{\text{2}}\text{+ 3}x\text{+ 1 =}{\left(\text{x + 1}\right)}^{\text{3}}$

Đáp án đúng là: A

${\left(\text{x}-\text{2}\right)}^{\text{3}}{\text{= x}}^{\text{3}}-{\text{3.x}}^{\text{2}}{\text{.2 + 3.x.2}}^{\text{2}}-{\text{2}}^{\text{3}}{\text{= x}}^{\text{3}}-{\text{6x}}^{\text{2}}\text{+ 12x}-\text{8}$

Đáp án đúng là: B

$={\left(-\frac{1}{2}\text{x}\right)}^3+3.{\left(-\frac{1}{2}\text{x}\right)}^2.1+3.\left(-\frac{1}{2}\text{x}\right){.1}^2+1^3$

$={\left(-\frac{\text{x}}{2}+1\right)}^3$

$\Rightarrow \text{A}={\left(-\frac{1}{2}\text{x}\right)}^3=-\frac{{\text{x}}^3}{8};\text{B}=-\frac{1}{2}\text{x}=-\frac{\text{x}}{2}$

Vậy P là một số chẵn.

Đáp án đúng là: B

${\text{8 − 36x + 54x}}^2-27{\text{x}}^3=2^3-3{.2}^2.\left(\text{3x}\right)+3.2.{\left(\text{3x}\right)}^2-{\left(\text{3x}\right)}^3$

$={\left(2-3\text{x}\right)}^3$

Đáp án đúng là: C

Ta có: $\left({\text{x}}^2-2\text{x}+1\right)\left(\text{x}–1\right)={\left(\text{x}-1\right)}^2\left(\text{x}-1\right)$

$\text{=}{\left(\text{x}-1\right)}^3={\text{x}}^3-3{\text{x}}^2+3\text{x}-1.$

Đáp án đúng là: C

$\text{H =}\left(\text{x+}5\right)\left({\text{x}}^2-5\text{x}+25\right)-{\left(\text{2x}+1\right)}^3+7{\left(\text{x}-1\right)}^3$ $-3\text{x}\left(-11\text{x}+5\right)$

${\text{= x}}^3-5{\text{x}}^2+25\text{x+}5{\text{x}}^2-25\text{x+}125$$-\left(8{\text{x}}^3+12{\text{x}}^2+\text{6x}+1\right)$ $\text{+7}\left({\text{x}}^3-3{\text{x}}^2+3\text{x}-1\right)+{\text{33x}}^2-1\text{5x}$

${\text{= x}}^3+125-8{\text{x}}^3-12{\text{x}}^2-6\text{x}-1+7{\text{x}}^3-21{\text{x}}^2$ $+2\text{1x}-7+3{\text{3x}}^2-1\text{5x}$

$\text{=}\left({\text{x}}^3-8{\text{x}}^3+7{\text{x}}^3\right)+\left(-1{\text{2x}}^2-2{\text{1x}}^2+3{\text{3x}}^2\right)$ $+\left(5^3-1-7\right)$

= 117

Vậy H là một số lẻ.

Đáp án đúng là: C

$\text{M=}{\left(\text{x}+2\text{y}\right)}^3-6{\left(\text{x}+2\text{y}\right)}^2+12\left(\text{x}+2\text{y}\right)-8$

$\text{=}{\left(\text{x}+2\text{y}\right)}^3-3.{\left(x+2\text{y}\right)}^2.2+3.\left(\text{x}+2\text{y}\right).2^2-2^3$

$={\left(x+2y-2\right)}^3$

Thay x = 20, y = 1 vào biểu thức M ta có $\text{M}={\left(20+2.1-2\right)}^3={20}^3=8000$

Đáp án đúng là: C

$P={\left(4x+1\right)}^3-\left(4x+3\right)\left(16x^2+3\right)$

$={\left(4x\right)}^3+3.{\left(4x\right)}^2.1+3.4x{.1}^2+1^3-\left(64x^3+12x+48x^2+9\right)$

$=64x^3+48x^2+12x+1-64x^3-12x-48x^2-9$

= – 8

$Q={\left(x-2\right)}^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)+5x$

$=x^3-3.x^2.2+3x{.2}^2-2^3-x\left(x^2-1\right)+6x^2-18x+5x$

$=x^3-6x^2+12x-8-x^3+x+6x^2-18x+5x$

= - 8

$\Rightarrow$P = Q

Đáp án đúng là: C

$P=8x^3-12x^{2}y+6x{y}^2-y^3+12x^2-12xy+3y^2+6x-3y+11$

$={\left(2x-y\right)}^3+3{\left(2x-y\right)}^2+3\left(2x-y\right)+1+10$

$={\left(2x-y+1\right)}^3+10$

Đáp án đúng là: C

Ta có

$\text{Q}={\left(2\text{x}-1\right)}^3-8\text{x}\left(\text{x}+1\right)\left(\text{x}-1\right)+2\text{x}\left(6\text{x}-5\right)$ $=\text{ax}-\text{b}\left(\text{a, b}\in \mathbb{Z}\right)$

$=8x^3-12x^2+6x-1-8x\left(x^2-1\right)+12x^2-10$

$=8x^3-12x^2+6x-1-8x^3+8x+12x^2-10x$

$=4x-1$

$\Rightarrow a=4;b=1$

Đáp án đúng là: C

Ta có

$\text{P =}{\left(\text{4x + 1}\right)}^{\text{3}}-\left(\text{4x + 3}\right){\text{(16x}}^{\text{2}}\text{+ 3)}$

$={\left(4x\right)}^3+3.{\left(4x\right)}^2.1+3.4x{.1}^2+1^3-(64x^3+12x+48x^2+9)$

$=64x^3+48x^2+12x+1-64x^3-12x-48x^2-9$

= - 8

$\text{Q =}{\left(\text{x}-\text{2}\right)}^{\text{3}}-\text{x}\left(\text{x + 1}\right)\left(\text{x}-\text{1}\right)\text{+ 6x}\left(\text{x}-\text{3}\right)\text{+ 5x}$

$=x^3-3.x^2.2+3x{.2}^2-2^3-x(x^2-1)+6x^2-18x+5x$

$=x^3-6x^2+12x-8-x^3+x+6x^2-18x+5x$

= - 8

$\Rightarrow$P = Q

Đáp án đúng là: C

Ta có

$A=8x^3-12x^{2}y+6x{y}^2-y^3+12x^2-12xy+3y^2+6x-3y+11$

$={\left(2x\right)}^3-3.{\left(2x\right)}^2.y+3.2x.y+y^3+3(4x^2-4xy+y^2)+3(2x-y)+11$

$={(2x-y)}^3+3{(2x-y)}^2+3(2x-y)+1+10$

$={(2x-y+1)}^3+10$

Thay 2x - y = 9 vào biểu thức A ta có $\text{A}={\left(9+1\right)}^3+10=1010$

Đáp án đúng là: C

Ta có

${\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3=a^3-b^3-3ab\left(a-b\right)$

$\Rightarrow a^3-b^3={\left(a-b\right)}^3+3ab\left(a-b\right)$

$\iff Q={\left(a-b\right)}^3+3ab\left(a-b\right)$

Thay a + b = 5 và ab = − 3 vào Q ta có

$Q={\left(a-b\right)}^3+3ab\left(a-b\right)$

$=4^3+3.(-3).4$

= 64 - 36 = 28

Đáp án đúng là: A

${\left(a+b\right)}^3=a^3+3a^{2}b+3a{b}^2+b^3=a^3+b^3+3ab\left(a+b\right)$

$\Rightarrow a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)$

Ta có

$B=a^3+b^3+c^3-3abc$

$={\left(a+b\right)}^3-3ab\left(a+b\right)+c^3-3abc$

$={\left(a+b\right)}^3+x^3-3ab\left(a+b+c\right)$

Tương tự, ta có ${\left(a+b+c\right)}^3-3\left(a+b\right)c\left(c+b+c\right)$

$\iff B={\left(a+b+c\right)}^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)$

Mà a + b + c = 0 nên $\text{B = 0}-\text{3(a + b)c.0}-\text{3ab.0 = 0}$.