\(A=3\left(2x-1\right)-\left|x-5\right|\)

\(=6x-3-\left|x-5\right|\)

TH1 : \(x-5\ge0\Rightarrow x\ge5\Rightarrow\left|x-5\right|=x-5\)

\(A=6x-3-x+5\)

\(=5x+2\)

TH2 : \(x-5< 0\Rightarrow x< 5\Rightarrow\left|x-5\right|=5-x\)

\(A=6x-3-5+x\)

\(=7x-8\)

Vậy ....

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

     \(=x^3-8-x^3-3x^2-3x-1+3x^2-3\)

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

        \(=-3x-11\)

\(\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{15.16.17}\)

=\(\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{3.5.16.17}\)

=\(\frac{1}{1.2}+\frac{1}{2.4}+\frac{1}{4.5}+...+\frac{1}{5.16.17}\)

sai rồi bạn ơi!!! mk biết ko phải là như vậy!

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

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

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

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

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

tíc mình nha

Minh ánh, bn viết sai đầu bài rồi

a) 3(x-1)-2|x+3| = 3x-3-2(x+3) = 3x-3-2x+6 = (3x-2x)+(6-3)=x+3

b) 2|x-3|-|4x-1| = 2(x-3)-(4x-1) = 2x-6-4x+1 = (2x-4x)-(6-1) = -2x-5

 (x-3)^2 *(x+3) *(x-3) =

 (x+4)^2 -(x-2)^2 -2(x+2) *(x-2) =

a) \(\left(x-3\right)^2.\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right).\left(x-3\right).\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right)^3.\left(x+3\right)\)

\(=\left(3x-9\right).\left(x+3\right)\)

Phần b tương tự

`A=1+4+4^2+4^3+....+4^99+4^100`

`=>4A=4+4^2+4^3+4^4+...+4^100+4^101`

`=>4A-A=4^101-1`

`=>3A=4^101-1`

`=>A=(4^101-1)/3`

Ta có: \(A=1+4+4^2+...+4^{99}+4^{100}\)

\(\Leftrightarrow4\cdot A=4+4^2+4^3+...+4^{100}+4^{101}\)

\(\Leftrightarrow4\cdot A-A=4^{101}-1\)

hay \(A=\dfrac{4^{101}-1}{3}\)