`2 1/2 xx 7/5 + (-9/10) xx 2/3`

`= 5/2 xx 7/10 + (-9/10) xx 2/3`

`= 35/20 + (-18/30)`

`= 7/4 + (-3/5)`

`= 23/20`

__

`x + 45% =2 1/2 - 5/3`

`=> x+ 45/100 = 5/2 - 5/3`

`=>x+ 9/20= 15/6-10/6`

`=> x+9/20 = 5/6`

`=>x= 5/6 - 9/20`

`=>x=23/60`

__

`80% +x=-5/2 +3/4`

`=> 80/100 + x= -10/4 +3/4`

`=> 4/5 + x= -7/4`

`=>x= -7/4-4/5`

`=>x=-51/20`

__

`4/25 -x=-5/2 +(-3/10)`

`=> 4/25 -x= -25/10 +(-3/10)`

`=> 4/25 -x= -28/10`

`=>x= 4/25 -(14/5)`

`=>x= 4/25 + 14/5`

`=>x=74/25`

• 2(x+5)(x-5)-(x+2)(2x-3)+x(x^2-8)=(x+1)(x^2-x+1)

<=> 2(x^2-25) - 2x^2+3x-4x+6 + x^3-8x = x^3+1

=>2x^2-50 - 2x^2 -9x+6+x^3-x^3-1 = 0

<=>-9x - 45 =0

<=>-9x=45

<=>x=-5

Còn phần b và c bạn cứ khai triển ra,mình phải đi học nên không có thời gian giải cho bạn

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-80=\left(x^2-5x+4\right)\left(x^2-5x+6\right)-80\)

Đặt \(x^2-5x+4=t\), ta có:

\(t\left(t+2\right)-80=t^2-2t+1-81=\left(t-1\right)^2-9^2=\left(t-1-9\right)\left(t-1+9\right)=\left(t-10\right)\left(t+8\right)\)

\(=\left(x^2-5x+4-10\right)\left(x^2-5x+4+8\right)=\left(x^2-5x-6\right)\left(x^2-5x+12\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt x² + x + 1 = t, ta có:

t(t + 1) - 12

= t² + t + 1/4 - 49/4

= (t + 1/2)² - (7/2)²

= (t + 1/2 + 7/2)(t + 1/2 - 7/2)

= (t + 4)(t - 3)

nhân váo như bình thường sau đó bấm máy tính shift solve =? rồi chia hoocne 

2x+2x+1+2x+2+...+2x+2015=22019−82x+2x+1+2x+2+...+2x+2015=22019−8

⇒2x(1+2+22+...+22015)=22019−8⇒2x(1+2+22+...+22015)=22019−8

⇒2x(22016−1)=22019−8⇒2x(22016−1)=22019−8

⇒2x=22019−822016−1⇒2x=22019−822016−1

⇒2x=⇒2x=23.(22016−1)22016−123.(22016−1)22016−1 

⇒2x=23⇒2x=23

⇒x=3⇒x=3

Vậy x=3

`Answer:`

\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)

\(\Leftrightarrow-6x-53=80\)

\(\Leftrightarrow-6x=133\)

\(\Leftrightarrow x=-\frac{133}{6}\)

\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)

\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)

\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)

\(\Leftrightarrow22x^2-9x+30=22x^2\)

\(\Leftrightarrow-9x+30=0\)

\(\Leftrightarrow-9x=-30\)

\(\Leftrightarrow x=\frac{10}{3}\)

Xét khai triển:

\(\left(x+1\right)^n=C_n^0+C_n^1x+C_n^2x^n+C_n^3x^3+...+C_n^nx^n\)

Đạo hàm 2 vế:

\(n\left(x+1\right)^{n-1}=C_n^1+2C_n^2x+3C_n^3x^2+...+nC_n^nx^{n-1}\)

Thay \(x=1\) vào ta được:

\(n.2^{n-1}=C_n^1+2C_n^2+3C_n^3+...+nC_n^2=256n\)

\(\Rightarrow2^{n-1}=256=2^8\Rightarrow n=9\)

Câu 2:

\(\left(x-2\right)^{80}=a_0+a_1x+a_2x^2+a_3x^3+...+a_{80}x^{80}\)

Đạo hàm 2 vế:

\(80\left(x-2\right)^{79}=a_1+2a_2x+3a_3x^2+...+80a_{80}x^{79}\)

Thay \(x=1\) ta được:

\(80\left(1-2\right)^{79}=a_1+2a_2+3a_3+...+80a_{80}\)

\(\Rightarrow S=80.\left(-1\right)^{79}=-80\)

cảm ơn anh