a: x+(x+1)+(x+2)+...+(x+30)=496

=>(x+x+...+x)+(1+2+3+...+30)=496

=>\(31x+30\times\dfrac{31}{2}=496\)

=>\(31x+465=496\)

=>31x=31

=>x=1

b: \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)

=>\(51x-\left(1+2+3+...+50\right)=1530\)

=>\(51x-\dfrac{50\times51}{2}=1530\)

=>\(51x-1275=1530\)

=>51x=1275+1530=2805

=>x=2805:51=55

a, \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=496\)

\(\Leftrightarrow31x+1+2+...+30=496\Leftrightarrow31x+\dfrac{\left(30+1\right).30}{2}=496\)

\(\Leftrightarrow31x+465=496\Leftrightarrow31x=31\Leftrightarrow x=1\)

b, \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)

\(\Leftrightarrow51x+\dfrac{\left(-1-50\right).50}{2}=1530\Leftrightarrow51x-1275=1530\Leftrightarrow51x=2805\Leftrightarrow x=55\)

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

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

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

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

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

=>\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

=>\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

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

=>x+2004=0

=>x=-2004

b, \(275-5\left(2x-1\right)=200\Leftrightarrow5\left(2x-1\right)=75\Leftrightarrow2x-1=15\Leftrightarrow x=8\)

c, \(3x-38:2=206\Leftrightarrow3x-19=206\Leftrightarrow3x=225\Leftrightarrow x=75\)

d, \(2x+x+5x=400\Leftrightarrow8x=400\Leftrightarrow x=50\)

\(sin^210+sin^220+sin^245+sin^270+sin^280\)

\(=sin^210+sin^220+sin^245+cos^220+cos^210=1+1+sin^245=2+\dfrac{1}{2}=\dfrac{5}{2}\)

(x-15):12=79 dư 8

=>\(x-15=79\times12+8=956\)

=>x=956+15=971

(x - 15) : 12 = 79 dư 8

x - 15 = 79 x 12 + 8

x - 15 = 956

x = 956 + 15

x= 971 

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}.2021\)

\(=\left\{47-\left[736:2^4\right]\right\}.2021\)

\(=\left\{47-\left[736:16\right]\right\}.2021\)

\(=\left\{47-46\right\}.2021\)

\(=1.2021\)

\(=2021\)

\(\left\{47-\left[736:\left(5-3\right)^4\right]\right\}\cdot2021\)

\(=\left\{47-736:16\right\}\cdot2021\)

\(=\left(47-46\right)\cdot2021=2021\)

2(x-1)+3(x-2)=x-4

=> 2x-2+3x-6=x-4

=> 5x-8=x-4

=> 5x-x=8-4

=> 4x=4

=> x=4:4

=> x=1

Vậy: x=1

\(2\left(x-1\right)+3\left(x-2\right)=x-4\)

\(2x-2+3x-6=x-4\)

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

\(5x-8=x-4\)

\(5x-x=-4+8\)

\(4x=4\)

\(x=1\)

\(\left(2345-45\right)+2345\)

\(=2300+2345\)

\(=4645\)

\(\left(2345-45\right)+2345\)

\(=2300+2345\)

\(=4645\)

\(\left(7+x\right)-\left(21-13\right)=32\)

\(\left(7+x\right)-8=32\)

\(7+x=32+8\)

\(7+x=40\)

\(x=40-7\)

\(x=33\)

Vậy \(x=33\)

\(\left(7+x\right)-\left(21-13\right)=32\)

\(7+x-21+13=32\)

\(x+\left(7+13-21\right)=32\)

\(x-1=32\)

\(x=33\)