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 

a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

\(\Rightarrow5^x=5^5\)

\(\Rightarrow x=5\)

Ta có \(x\left(4x+2\right)-\left(2x-5\right)\left(2x+5\right)=1\)

\(\Leftrightarrow4x^2+2x-\left(4x^2-25\right)=1\)

\(\Leftrightarrow4x^2+2x-4x^2+25=1\)

\(\Leftrightarrow2x=-24\)

\(\Leftrightarrow x=-12\)

Vậy x=-12

\(x\left(4x+2\right)-\left(2x-5\right)\left(2x+5\right)=1\)

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

\(4x^2+2x-4x^2+25=1\)

\(2x+25=1\)

\(2x=-24\)

\(x=-12\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\\ \Leftrightarrow4x=-38\Leftrightarrow x=-\dfrac{19}{2}\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)

\(\Rightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(\Rightarrow4x=-38\Rightarrow x=-\dfrac{19}{2}\)

x(5 – 2x) + 2x(x – 1) = 15

(x.5 – x.2x) + (2x.x – 2x.1) = 15

5x – 2x2 + 2x2 – 2x = 15

(2x2 – 2x2) + (5x – 2x) = 15

3x = 15

x = 5.

Vậy x = 5.

TH1: \(x\ge5\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=x-5\\\left|2x-1\right|=2x-1\end{matrix}\right.\)

PT <=> \(x-5+2x-1=2x+3\)

<=> x = 9 (Tm)

TH2: \(\dfrac{1}{2}\le x< 5\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=2x-1\end{matrix}\right.\)

PT <=> 5 - x + 2x -1 = 2x + 3

<=> x = 1(Tm)

TH3: \(x< \dfrac{1}{2}\)

<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=1-2x\end{matrix}\right.\)

PT <=> \(5-x+1-2x=2x+3\)

<=> \(5x=3< =>x=\dfrac{3}{5}\left(l\right)\)

KL: x \(\in\left\{1;9\right\}\)

\(\Leftrightarrow\left(2x-1\right)^4\left(2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)