ai mần tr mình cho tick

a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x^2-x+1>0\forall x\)

\(1-2+3-4+5-6+...+2019-2020+2021\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021\)

\(=-1-1-1-..-1+2021\)

\(=-1\cdot1010+2021\)

\(=-1010+2021\)

\(=-1011\)

$1-2+3-4+5-6+...+2019-2020+2021$

$=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021$

$=-1-1-1-..-1+2021$

$=-1\cdot 1010+2021$

$=-1010+2021$

$=-1011$

\(1000=2^3\cdot5^3\)

\(7200=2^5\cdot3^2\cdot5^2\)

\(810=2\cdot3^4\cdot5\)

\(8000=2^6\cdot5^3\)

1.

$3x^4-48=3(x^4-16)=3[(x^2)^2-4^2]=3(x^2-4)(x^2+4)$

$=3(x-2)(x+2)(x^2+4)$

2. 

$x^3-4x^2+8x-8=(x^3-8)-(4x^2-8x)$
$=(x-2)(x^2+2x+4)-4x(x-2)=(x-2)(x^2+2x+4-4x)=(x-2)(x^2-2x+4)$

3.

$x^4-27x=x(x^3-27)=x(x^3-3^3)=x(x-3)(x^2+3x+9)$

4.

$x^4-x^3+x^2-1=(x^4-x^3)+(x^2-1)=x^3(x-1)+(x-1)(x+1)$

$=(x-1)(x^3+x+1)$

5.

$x^5+x^3-x^2-1=(x^5+x^3)-(x^2+1)=x^3(x^2+1)-(x^2+1)$
$=(x^2+1)(x^3-1)=(x^2+1)(x-1)(x^2+x+1)$

6.

$x^3+2x^2+2x+1=(x^3+x^2)+(x^2+2x+1)=x^2(x+1)+(x+1)^2$
$=(x+1)(x^2+x+1)$

7.

$x^3+x^2-4x-4=(x^3+x^2)-(4x+4)=x^2(x+1)-4(x+1)$

$=(x+1)(x^2-4)=(x+1)(x-2)(x+2)$

8.

$x^4-2x^3+2x-1=(x^4-2x^3+x^2)-(x^2-2x+1)$

$=(x^2-x)^2-(x-1)^2=x^2(x-1)^2-(x-1)^2=(x-1)^2(x^2-1)=(x-1)^2(x-1)(x+1)$

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Tính

Lời giải:
Số hsg là: $32.\frac{1}{4}=8$ (hs) 

Số hsk là: $(32-8).\frac{3}{8}=9$ (hs) 

Số hstb là: $32-8-9=15$ (hs)

8¹⁹⁷⁵ = (8⁴)⁴⁹³.8³ = \(\overline{..6}\)⁴⁹³. \(\overline{...2}\) = \(\overline{..2}\)

8^1975=8^1972*8^3

8^1975=(8^4)^493*512

8^1975= ...6^493*512

8^1975=....6*512

=>8^1975=...2