a, 1076 x 204

b, 196168 : 217

11678523096

11,678,519,868

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

\(TH1:x-5-x=3\)

           \(-5=3\)(ko xảy ra)

            \(xkoTM\)

\(TH2:-\left(x-5\right)-x=3\)

            \(5-x-x=3\)

            \(5-2x=3\)

             \(2x=2\)

             x=1

Vậy x=1

x = 1

ai tk mình mình tk lại cho

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

Sửa đề: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2\)

\(=x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)\)

\(=x^2-25-x^2-6x-9+3x^2-12x+12\)

\(=3x^2-18x-22\)

Ta có: \(\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

\(=x^2+2x+1-\left(x^2-16\right)+3x^2\)

\(=4x^2+2x+1-x^2+16=3x^2+2x+17\)

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

=>\(3x^2-18x-22=3x^2+2x+17\)

=>-20x=39

=>\(x=-\frac{39}{20}\)

Sửa đề: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2\)

\(=x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)\)

\(=x^2-25-x^2-6x-9+3x^2-12x+12\)

\(=3x^2-18x-22\)

Ta có: \(\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

\(=x^2+2x+1-\left(x^2-16\right)+3x^2\)

\(=4x^2+2x+1-x^2+16=3x^2+2x+17\)

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

=>\(3x^2-18x-22=3x^2+2x+17\)

=>-20x=39

=>\(x=-\frac{39}{20}\)

cậu chia từng câu ra cho mình nhé

Câu 1:

[x - 3] + [x - 2] + [x - 1] + ... + [x + 5] = 0

x - 3 + x - 2 + x - 1 + ...+ x + 4 + x + 5 = 0

[x + x + x + x +...+ x] - [3 + 2 + 1+ .. - 4 - 5] = 0

Xét dãy số: 3; 2; 1;...; -5

Dãy số trên có số số hạng là: (-5 - 3) : (-1) + 1 = 9

Tổng dãy số trên là:[-5 + 3] x 9 : 2 = - 9

9x - (-9) = 0

9x = 9

x = 9 : 9

x = 1

Vậy x = 1