4(x-2)-3(x+1)=5

=>\(4x-8-3x-3=5\)

=>\(x-11=5\)

=>x=11+5=16

\(4\left(x-2\right)-3\left(x+1\right)=5\)

\(\Leftrightarrow4x-8-3x-3=5\)

\(\Leftrightarrow\left(4x-3x\right)=5+8+3\)

\(\Leftrightarrow x=16\)

Vậy \(x=16\)

Ta có:

\(\dfrac{1}{2}=\dfrac{1\times2}{2\times2}=\dfrac{2}{4};\dfrac{1}{4}=\dfrac{1}{4}\)

Vì \(\dfrac{2}{4}>\dfrac{1}{4}\) nên \(\dfrac{1}{2}>\dfrac{1}{4}\)

Mọi người ơi, cuộc thi Bài văn số 326 đâu rồi nhỉ? 

= > Cậu vô olm.vn rồi lướt xuống là sẽ thấy.

\(\dfrac{1}{2}>\dfrac{1}{4}\left(\dfrac{1}{4}< \dfrac{1}{2}\right)\)

\(\dfrac{29}{2}=\dfrac{28+1}{2}=14+\dfrac{1}{2}=14\dfrac{1}{2}\)

\(\dfrac{15}{4}=\dfrac{12+3}{4}=3+\dfrac{3}{4}=3\dfrac{3}{4}\)

\(\dfrac{31}{2}=\dfrac{30+1}{2}=15\dfrac{1}{2}\)

\(\dfrac{29}{3}=\dfrac{27+2}{3}=9\dfrac{2}{3}\)

\(\dfrac{125}{8}=\dfrac{120+5}{8}=15+\dfrac{5}{8}=15\dfrac{5}{8}\)

\(\dfrac{36}{27}=\dfrac{27+9}{27}=1+\dfrac{9}{27}=1\dfrac{9}{27}\)

\(\dfrac{124}{15}=\dfrac{120+4}{15}=8+\dfrac{4}{15}=8\dfrac{4}{15}\)

\(\dfrac{96}{3}=\dfrac{93+3}{3}=31\dfrac{3}{3}\)

\(\dfrac{129}{24}=\dfrac{120+9}{24}=5+\dfrac{9}{24}=5\dfrac{9}{24}\)

\(\dfrac{78}{13}=\dfrac{65+13}{13}=5+\dfrac{13}{13}=5\dfrac{13}{13}\)

\(\dfrac{91}{4}=\dfrac{88+3}{4}=22+\dfrac{3}{4}=22\dfrac{3}{4}\)

\(\dfrac{115}{8}=\dfrac{112+3}{8}=14+\dfrac{3}{8}=14\dfrac{3}{8}\)

a: \(B=2021\times2025=\left(2023-2\right)\times\left(2023+2\right)=2023\times2023-2\times2\)

=>\(B=A-4\)

=>A lớn hơn B 4 đơn vị

b: \(C=35\times53-18=35\times35+35\times18-18\)

\(=35\times35+18\times\left(35-1\right)\)

\(=35\times35+18\times34\)

\(D=35+53\times34\)

\(=35+\left(35-1\right)\times\left(35+18\right)\)

\(=35+35\times35+35\times18-35\times1-18\)

\(=35\times35+35\times17+17=35\times35+36\times17\)

\(=35\times35+18\times34\)

=C

=>C=D

Gọi số cần tìm có dạng là \(X=\overline{ab}\)

Vì viết thêm số 7 vào bên trái số đó thì sẽ được số mới gấp 36 lần số cần tìm nên ta có: \(\overline{7ab}=36\times\overline{ab}\)

=>\(700+\overline{ab}=36\times\overline{ab}\)

=>\(35\times X=700\)

=>X=20

Vậy: Số cần tìm là 20

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012-1}{4048}=\dfrac{1011}{4048}\)

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+\dfrac{1}{4\cdot10}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2023\cdot2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1012-1}{2024}\)

\(=\dfrac{1011}{4048}\)

Có đáp án luôn rồi!

@456 troll thật

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