b) Thay x=-1; y=1 và z=-2 vào B, ta được:

\(B=\dfrac{3\cdot\left(-1\right)\cdot1\cdot\left(-2\right)-2\cdot\left(-2\right)^2}{\left(-1\right)^2+1}=\dfrac{6-8}{1+1}=\dfrac{-2}{2}=-1\)

Lời giải:
$P=(x+1)^3-(x+1)^3-[(x-1)^2+(x+1)^2]$

$=-[(x-1)^2+(x+1)^2]=-[(x^2-2x+1)+(x^2+2x+1)]=-2(x^2+1)$ phụ thuộc vào giá trị của biến nhé. Bạn xem lại đề.

$Q=(2x)^3-y^3+(2x)^3+y^3-16x^3$

$=8x^3-y^3+8x^3+y^3-16x^3=(8x^3+8x^3-16x^3)+(-y^3+y^3)=0+0=0$ không phụ thuộc vào giá trị của biến (đpcm)

$P=(x+1)^3-(x-1)^3-3[(x-1)^2+(x+1)^2]$

$=(x^3+3x^2+3x+1)-(x^3-3x^2+3x-1)-3[(x^2-2x+1)+(x^2+2x+1)]$

$=6x^2+2-3(2x^2+1)=3(2x^2+1)-3(2x^2+1)=0$ là giá trị không phụ thuộc vào giá trị của biến.

Thay x = -1,76; y = 3/25

⇒ P = 1/2

Lời giải:

a) 

$A=4x^2+4x+11=(4x^2+4x+1)+10=(2x+1)^2+10\geq 10$

Vậy $A_{\min}=10$. Giá trị này đạt tại $(2x+1)^2=0$

$\Leftrightarrow x=-\frac{1}{2}$

b) 

$C=x^2-2x+y^2-4y+7=(x^2-2x+1)+(y^2-4y+4)+2$

$=(x-1)^2+(y-2)^2+2\geq 2$

Vậy $C_{\min}=2$. Giá trị này đạt tại $(x-1)^2=(y-2)^2=0$

$\Leftrightarrow x=1; y=2$

a: \(A=\left(2x-1\right)\left(4x^2+2x+1\right)-7\left(x^3+1\right)\)

\(=\left(2x\right)^3-1^3-7x^3-7\)

\(=8x^3-1-7x^3-7=x^3-8\)

b: Thay x=-1/2 vào A, ta được:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{1}{8}-8=-\dfrac{65}{8}\)

c: \(A=x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

Để A là số nguyên tố thì x-2=1

=>x=3

ghi đề hẳn hoi coi

Đáp án B

Ta có x 2 + 1 x 2 - 1 ≥ 2 x 2 . 1 x 2 - 1 = 1 ⇒ 4 x 2 + 1 x 2 - 1 ≥ 4 14 - y - 2 y + 1 ≤ 16 ⇒ log 2 14 - y - 2 y + 1 ≤ 4  

Theo giả thiết  4 x 2 + 1 x 2 - 1 = log 2 14 - y - 2 y + 1 ⇒ x 2 = 1 x 2 y = 0 ⇔ x 2 = 1 y = 0

Vậy giá trị biểu thức P = x 2 + y 2 - x y + 1 = 2 .

`# \text {04th5}`

`a.`

`P = (5x^2 - 2xy + y^2) - (x^2 + y^2) - (4x^2 - 5xy + 1)`

`= 5x^2 - 2xy + y^2 - x^2 - y^2 - 4x^2 + 5xy - 1`

`= (5x^2 - x^2 - 4x^2) + (-2xy + 5xy) + (y^2 - y^2) - 1`

`= 3xy - 1`

`b.`

\((x^2-5x+4)(2x+3)-(2x^2-x-10)(x-3)\)

`= x^2(2x + 3) - 5x(2x + 3) + 4(2x + 3) - [ 2x^2(x - 3) - x(x - 3) - 10(x - 3)]`

`= 2x^3 + 3x^2 - 10x^2 - 15x + 8x + 12 - (2x^3 - 6x^2 - x^2 + 3x - 19x + 30)`

`= 2x^3 -7x^2 - 7x + 12 - (2x^3 - 7x^2 - 7x + 30)`

`= 2x^3 - 7x^2 - 7x + 12 - 2x^3 + 7x^2 + 7x -30`

`= -30`

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

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c) tự làm, đkxđ: x≠1;x≠−1

ê k bn với mk ik

😘 😘 😘 😘