\(\left|x^2-2x\right|=2x-x^2\\ =>\left|x^2-2x\right|=-\left(x^2-2x\right)\\ =>x^2-2x\le0\\ =>x\left(x-2\right)\le0\\ =>0\le x\le2\)

`S = 1.4+2.5 + 3.6 +...+ 100.103`

`S = 1 . (2+2) + 2 (3+2) + 3 . (4+2) + ... + 100 . (101 + 2) `

`S = 1.2 + 2 . 1 + 2.3 + 2.2 + 3.4 + 2.3 + ... + 100 . 101 + 2. 100`

`S = 1.2 + 2.3+ 3.4 +... + 100.101 + 2.1 + 2.2 + 2.3 + ... + 2.100`

`S =  1.2 + 2.3+ 3.4 +... + 100.101 + 2(1+2+3+...+100) `

`S =  1.2 + 2.3+ 3.4 +... + 100.101 + 2 (100+1) . [(100-1):1+1] : 2`

`S =  1.2 + 2.3+ 3.4 +... + 100.101 + 10100`

Đặt `A = 1.2 + 2.3+ 3.4 +... + 100.101`

`3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .... + 100 . 101. (102 - 99)`

`3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102 -99.100.101`

`3A = 100.101.102`

`A = 343400`

`S = A + 10100`

`= 343400 + 10100`

`= 353500`

\(7\cdot4^{x-1}+4^{x+1}=4^x\cdot7\cdot\dfrac{1}{4}+4^x\cdot4\)

\(=4^x\left(7,25+4\right)=11,25\cdot4^x\)

\(\left(2x-3\right)^9-2x+3=0\)

=>\(\left(2x-3\right)^9-\left(2x-3\right)=0\)

=>\(\left(2x-3\right)\left[\left(2x-3\right)^8-1\right]=0\)

=>\(\left[{}\begin{matrix}2x-3=0\\\left(2x-3\right)^8-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\\left(2x-3\right)^8=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x-3=1\\2x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

Số số hạng là \(\dfrac{\left(2x+1-3\right)}{2}+1=\dfrac{2x-2}{2}+1=x-1+1=x\left(số\right)\)

Tổng của dãy số là \(\dfrac{x\left(2x+1+3\right)}{2}=x\left(x+2\right)\)

Theo đề, ta có: x(x+2)=624

=>\(x^2+2x-624=0\)

=>(x+26)(x-24)=0

=>\(\left[{}\begin{matrix}x=-26\\x=24\end{matrix}\right.\)

8

2³.6 - 72 : 3²

= 8.6 - 72 : 9

= 48 - 8

= 40

Xét tổng: S = 1 + 2 + 3 + ... + x 

Số lượng số hạng: (x - 1) :  1 + 1 = x (số hạng)

`=>S=((x+1)*x)/2` 

\(=>\dfrac{\left(x+1\right)x}{2}=820\\ =>x\left(x+1\right)=1640\\ =>x^2+x-1640=0\\ =>\left(x^2-40x\right)+\left(41x-1640\right)=0\\ =>x\left(x-40\right)+41\left(x-40\right)=0\\ =>\left(x+41\right)\left(x-40\right)=0\\ =>\left[{}\begin{matrix}x=40\\x=-41\end{matrix}\right.\)

Mà x > 0 => x = 40 

a: \(\left|0,5x-2\right|-\left|x+\dfrac{2}{3}\right|=0\)

=>\(\left|\dfrac{1}{2}x-2\right|=\left|x+\dfrac{2}{3}\right|\)

=>\(\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+\dfrac{2}{3}\\\dfrac{1}{2}x-2=-x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x=2+\dfrac{2}{3}=\dfrac{8}{3}\\\dfrac{3}{2}x=-\dfrac{2}{3}+2=\dfrac{4}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{8}{3}:\dfrac{-1}{2}=\dfrac{8}{3}\cdot\left(-2\right)=-\dfrac{16}{3}\\x=\dfrac{4}{3}:\dfrac{3}{2}=\dfrac{8}{9}\end{matrix}\right.\)

b: 

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

=>\(\left|x+1\right|=2x-\dfrac{1}{4}\)

=>\(\left\{{}\begin{matrix}2x-\dfrac{1}{4}>=0\\\left(2x-\dfrac{1}{4}\right)^2=\left(x+1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(2x-\dfrac{1}{4}-x-1\right)\left(2x-\dfrac{1}{4}+x+1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(x-\dfrac{5}{4}\right)\left(3x+\dfrac{3}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{4}\)

c: \(3x-\left|x+15\right|=\dfrac{5}{4}\)

=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)

=>\(\left\{{}\begin{matrix}3x-\dfrac{5}{4}>=0\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16,25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=8,125\)

d: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

=>\(\left|3x+\dfrac{5}{4}\right|=\dfrac{3}{2}-\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

e: \(\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\)

=>\(\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=-3x+\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-3x=-\dfrac{1}{2}+1\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

f: \(\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)

=>\(\left[{}\begin{matrix}2x-1=x+\dfrac{1}{3}\\2x-1=-x-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=\dfrac{1}{3}+1\\2x+x=-\dfrac{1}{3}+1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{4}{3}\\3x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{2}{9}\end{matrix}\right.\)

\(\left(\dfrac{7}{30}\right)-\left(\dfrac{11}{36}\right)+\left(\dfrac{23}{30}\right)-\left(\dfrac{25}{36}\right)\)

\(=\dfrac{7}{30}+\dfrac{23}{30}-\left(\dfrac{11}{36}+\dfrac{25}{36}\right)\)

\(=\dfrac{30}{30}-\dfrac{36}{36}\)

\(=1-1\)

\(=0\)

`(7x - 11)^3 = 2^5 . 5^2 + 100`

`=> (7x - 11)^3 = 32 . 25 + 100`

`=> (7x - 11)^3 = 800 + 100`

`=> (7x - 11)^3 = 900`

`=> 7x - 11 =` \(\sqrt[3]{900}\)

`=> 7x = 11 +` \(\sqrt[3]{900}\)

`=> x =` \(\dfrac{11+\sqrt[3]{900}}{7}\)

Vậy ...

Sửa đề:

\(\left(7x-11\right)^2=2^5.5^2+100\)

\(\left(7x-11\right)^2=32.25+100\)

\(\left(7x-11\right)^2=800+100\)

\(\left(7x-11\right)^2=900\)

\(\left(7x-11\right)^2=30^2\)

\(\Rightarrow7x-11=30\)

\(7x=30+11\)

\(7x=41\)

\(x=41\div7\)

\(x=\dfrac{41}{7}\)

Vậy \(x=\dfrac{41}{7}\)