c) (x+1) + (x+2) + ... + (x+5) = 90

=> 5x + ( 1 + 2 + ... + 5 ) = 90

5x + 15 = 90

5x = 90 - 15

5x = 75

x = 75 : 5

x  = 15

d) (x+1) + (x+2) + .... + (x+100) = 20150

=> 100x + ( 1+2+...+100 ) = 20150

100x + 5050 = 20150

100x = 20150 - 5050

100x = 15100

x = 15100 : 100

x = 151

Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90

<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90

<=> 5x + 15 = 90

=> 5x = 75

=> x = 15 

c)   \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)

 \(\Leftrightarrow x+1+x+2+x+3+x+4+x+5=90\)

\(\Leftrightarrow5x+\left(1+2+3+4+5\right)=90\)

\(\Leftrightarrow5x+15=90\)

\(\Leftrightarrow5x=75\)

\(\Leftrightarrow x=15\)

d)    \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+......+\left(x+99\right)+\left(x+100\right)=20150\)

\(\Leftrightarrow x+1+x+2+x+3+......+x+99+x+100=20150\)

\(\Leftrightarrow100x+\left(1+2+3+.....+99+100\right)=20150\)

\(\Leftrightarrow100x+5050=20150\)

\(\Leftrightarrow100x=15100\)

\(\Leftrightarrow x=151\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\left(x\cdot100\right)+\left(1+2+...+100\right)=5750\)

\(\left(x\cdot100\right)+\left(100+1\right)\cdot\frac{100}{2}=5750\)

\(\left(x\cdot100\right)+101\cdot50=5750\)

\(\left(x\cdot100\right)+5050=5750\)

\(x\cdot100=5750-5050\)

\(x\cdot100=700\)

\(x=700\div100\)

\(x=7\)

Ta có: ( x+1)+(x+2)+(x+3)+.....+(x+99)+(x+100)=5750

<=>(x+x+x+....+x+x)+(1+2+3+..+99+100)=5750

<=> 100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

Vậy x=7

 hoặc mở câu hỏi tương tự tham khảo.

thank you

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)

\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)

\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)

\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)

\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)

\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2018}=\dfrac{2018}{x\left(x+2018\right)}\)

\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}+...+\dfrac{1}{\left(x+96\right)\left(x+98\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)\)

\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)=\dfrac{1}{4}\left(\dfrac{x^2+198x+9800-x^2-2x}{x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\right)\)

\(B=\dfrac{196x+9800}{4x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\)

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

\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)

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

\(=\left(x+1\right)\cdot\left(-1\right)\)

\(=-1\left(x+1\right)\)

b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)

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

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

\(=x^3-1-x^3-8+12x-12\)

\(=-21+12x\)

c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)

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

\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)

\(=0\)

thực hiện phép tính

Phân thức cuối hình như mẫu sai rồi bạn

Phải là (x+9)(x+10) mới đúng chứ

Nếu đề bài đúng thì sẽ làm như sau:

A = \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\) (ĐKXĐ tự tìm nhé, chứ viết dài lắm)

A = \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+9}-\frac{1}{x+100}\)

A = \(\frac{1}{x}-\frac{1}{x+100}\)

A = \(\frac{x+100-x}{x\left(x+100\right)}\)

A = \(\frac{100}{x\left(x+100\right)}\)

Vậy A = \(\frac{100}{x\left(x+100\right)}\)

Chúc bn học tốt!!

a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)

\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)

=2