(\(x^2-2x+4\)).(\(x^4\) - 2\(x^2\) + 8) = 21

[(\(x^2-2x+1\)) + 3].[(\(x^4\) - 2\(x^2\) + 1) + 7] = 21

[(\(x-1)^2+3]\).[(\(x^2-1)^2+7]\) = 21

Vì (\(x-1)^2\) ≥ 0 ∀ \(x\); (\(x^2-1\))\(^2\) ≥ 0 \(\) ∀ \(x\) nên:

(\(x-1)^2+3\) ≥ 3; (\(x^2-1)^2+7\) ≥ 7 ∀ \(x\)

[(\(x^2-2x+1\)) + 3].[(\(x^4\) - 2\(x^2\) + 1) + 7] ≥ 3 x 7 = 21

Dấu bằng xảy ra khi \(\begin{cases}x-1=0\\ x^2-1=0\end{cases}\)

⇒ \(x\) = 1

5,6 x 2 + 2,8 x 8 +11,2 x 2 - 50,4

= 5,6 x 2 + (2,8 x 2) x 4+ (11,2:2) x (2 x 2) - 50,4

= 5,6 x 2 + 5,6 x 4+ 5,6 x 4 - 5,6 x 9

= 5,6 x (2+4+4-9)

= 5,6 x 1= 5,6

quá chuẩn luôn !!!!!!!!

NHỚ L.I.K.E cho mk nha

 a) (x+2)(x^2-2x+4)-x(x^2+2)=15 
<=> x^3 + 8 - x^3 - 2x = 15 
<=> -2x = 7 
<=> x = -7/2 

b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28 
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28 
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0 
<=> 3x² + 26x = 0 
<=> x(3x + 26) = 0 
Vậy x = 0 và x = -26/3 

c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0 
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0 
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0 
<=> -(x-1)(x+1)3x² = 0 
Vậy nghiệm là x = 1 ; -1 ; 0

Ta có : 2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-82x+2x+1+2x+2+...+2x+2015=22019−8

\Leftrightarrow2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8⇔2x(1+2+22+...+22015)=22019−8 (1)

Đặt : A=1+2+2^2+...+2^{2015}A=1+2+22+...+22015

\Rightarrow2A=2+2^2+2^3+...+2^{2016}⇒2A=2+22+23+...+22016

\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)⇒2A−A=(2+22+23+...+22016<...

$2^x+2^{x+1}+...+2^{x+2015}=2^{2019}-8$

$\to 2^x.1+2^x.2+....+2^x.2^{2015}=2^{2019}-8$

$\to 2^x.(1+2+...+2^{2015})=2^{2019}-8$

Đặt:

$A=1+2+...+2^{2015}$

$2A=2.(1+2+...+2^{}$

1+2*2+1*2+8=15

ok

1+1*2+1*2+8

=1+2+2+8

=1+4+8

=5+8

=13

Ta có; \(\left(\frac{a}{2}-b\right)^2\ge0;\forall x\)

\(\Rightarrow\frac{a^2}{4}+b^2\ge2.\frac{a}{2}.b=ab\)

đpcm

\(\frac{a^2}{4}+b^2\ge ab\)

<=>  \(\frac{a^2}{4}+b^2-ab\ge0\)

<=>  \(\frac{a^2+4b^2-4ab}{4}\ge0\)

<=>  \(\frac{\left(a-2b\right)^2}{4}\ge0\) luôn đúng

Dấu "=" xảy ra <=>  \(a=2b\)

\(D=1-4+4^2-4^3+4^4-...+4^{2016}-4^{2017}\) (sửa \(2^{2017}\) thành \(4^{2017}\))

\(\Rightarrow D=\left(1-4\right)+4^2\left(1-4\right)+4^4\left(1-4\right)-...+4^{2016}\left(1-4\right)\)

\(\Rightarrow D=\left(-3\right)+4^2.\left(-3\right)+4^4.\left(-3\right)-...+4^{2016}.\left(-3\right)\)

\(\Rightarrow D=\left(-3\right)\left(1+4^2+4^4+...+4^{2016}\right)\)

\(\Rightarrow4D=\left(-3\right)\left(4+4^3+4^5+...+4^{2017}\right)\)

\(\Rightarrow4D+D=\left(-3\right)\left(1+4+4^2+4^3+4^4+...+4^{2016}+4^{2017}\right)\)

\(\Rightarrow5D=\left(-3\right)\dfrac{4^{2017+1}-1}{4-1}\)

\(\Rightarrow D=\left(-3\right)\dfrac{4^{2018}-1}{3.5}\)

\(\Rightarrow D=\left(-1\right)\dfrac{4^{2018}-1}{5}=\dfrac{1-4^{2018}}{5}\)

B1: 

số gạo 3 tuần đaafu họ quyên góp là :

8400.1/2=4200

so gạo quyên góp lần 2 là : 

8400.2/3= 5600

so gạo quyên góp lần 3 là : 

8400.1/4= 2100

Tổng số gạo họ đax quyên góp là:

4200+5600+2100= 11900

so gạo vượt chỉ tiêu là:

11900-8400= 3500

\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)

\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)