\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)

\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)

\(100x-5050=50\)

\(100x=50+5050\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=7\)

\(b,x+\left(1+2+3+...+50\right)=2000\)

 \(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)

\(x+1275=2000\)

\(\Rightarrow x=2000-1275=725\)

c , ( x - 1 ) + ( x - 2 ) + ...... + ( x - 100 ) = 50

( x + x + ........ + ) - ( 1 + 2 + ..... + 100 ) = 50

100x - 5050 = 50

100x = 50 + 5050

100x = 5100

=> x = 5100/100 = 51

\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)

\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

\(=>x+2001=0\)

\(x=-2001\)

\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)

\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(=>x+1998=0\)

\(x=-1998\)

dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)

\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)

\(\Rightarrow x+2001=0\)

\(\Leftrightarrow x=-2001\)

x+1250=9978-6540

x+1250=93438

x+1250=93438-6540

x=92188

x + 1250 = 93438

x            = 93438 - 1250

x            = 92188

x+1250=99978-6540

x+1250=93438

x+1250=93438-1250

x=92188

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

<=>  \(\frac{x+4}{2000}+1+\frac{x+3}{2001}=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

<=>  \(\frac{x+4}{2000}+\frac{x+4}{2001}=\frac{x+4}{2002}+\frac{x+4}{2003}\)

<=>  \(\left(x+4\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

<=>  \(x+4=0\)   do 1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0

<=>  \(x=-4\)

Vậy...

Ta có :

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy ...

Ta có: \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

x = -2014

Thế x = 3 , y = -5 vào biểu thức ta được :

a.3[ 3 - ( -5 ) ] + ( -5 )⁴( 3 - 5 )

= a.3.8 + 625.( -2 )

= 24a - 1250

Bài 1: Tìm x:

\(\text{a, x.(-26)+26:(-2)=(-39)}\)

\(x.\left(-26\right)+\left(-13\right)=\left(-39\right)\)

\(\Rightarrow x.\left(-26\right)=52\)

\(x=52:\left(-26\right)\)

\(x=-2\)

\(\text{b, 41.(-79)-179.(-41)}\)

\(=41.\left(-79+179\right)\)

\(=41.100\)

\(=4100\)

\(\text{c, -19.(73-219)-219.(19-73)}\)

\(=-19.73+19.219-219.19+219.73\)

\(=73.\left(-19+219\right)+19.\left(219-219\right)\)

\(=73.200+19.0\)

\(=14600\)

Bài 2:Tìm số tự nhiên x:

\(a,18⋮x+4\)

\(\Rightarrow\left(x+4\right)\inƯ\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)

Ta có bảng sau :

Vậy \(x\in\left\{...\right\}\)

\(b,3x+4⋮x-3\)

ta có \(\frac{3x+4}{x-3}=3\left(x-3\right)+\frac{13}{x-3}\)
\(=3+\frac{13}{x-3}\)
để \(3x+4⋮x-3\Rightarrow13⋮x-3\)
=> x-3 thuộc ước của 13={1;-1;13;-13}
+x-3=1=>x=(tm)
+x-3=-1=>x=2(tm)
+x-3=-13=>x=-10(tm)
+x-3=13=>x=16(tm)

học tốt

Bài 1: Tìm x:

a, x.(-26)+26:(-2)=(-39)

    x.(-26)+(-13)=-39

    x.(-26)=-39+13

    x.(-26)=-26

            x=-26:(-26)

            x=1

Vậy x=1

b, 19-|x+3|=10

         |x+3|=19-10

         |x+3|=9

\(\Rightarrow\orbr{\begin{cases}x+3=9\Rightarrow x=9-3=6\\x+3=-9\Rightarrow x=-9-3=-12\end{cases}}\)

Vậy x=6 hoặc x=-12

 Bài 3: Tính nhanh

19.(73-219)-219.(19-73)

=19.73-19.219-219.19+219.73

=(19.73+219.73)-(19,219-219.19)

=[73.(19+219)]-0

=[73.238]-0

=17374

Bài 1

a)-26x-13=-39

-26x=-39+13

-26x=-26

x=1

b)-|x-3|=10-19

-|x-3|=-9

|x-3|=9

x-3=9 hoặc x-3=-9

x=9+3.        x=-9+3

x=12.          x=-6