\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2008}-1\right).\left(\dfrac{1}{2009}-1\right)\)

\(=-\dfrac{1}{2}.-\dfrac{2}{3}.-\dfrac{3}{4}...-\dfrac{2007}{2008}.-\dfrac{2008}{2009}\)

\(=-\left(\dfrac{1.2.3...2007.2008}{2.3.4...2008.2009}\right)\)

\(=-\dfrac{1}{2009}\)

ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

=>\(\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

=>\(\dfrac{-8x^2\cdot4}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x\cdot4\left(2x+1\right)}{12\left(2x-1\right)\left(2x+1\right)}-\dfrac{3\left(2x-1\right)\left(8x+1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

=>\(-32x^2=8x\left(2x+1\right)-3\left(16x^2-6x-1\right)\)

=>\(-32x^2=16x^2+8x-48x^2+18x+3\)

=>26x+3=0

=>\(x=-\dfrac{3}{26}\left(nhận\right)\)

Câu 4:

a: 2x+y=6

=>y=-2x+6

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=-2x+6\end{matrix}\right.\)

b: x+3y=2

=>x=2-3y

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y\in R\\x=2-3y\end{matrix}\right.\)

c: 3x-2y=1

=>3x=2y+1

=>\(x=\dfrac{2}{3}y+\dfrac{1}{3}\)

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{2}{3}y+\dfrac{1}{3}\end{matrix}\right.\)

d: 2x+0y=4

=>2x=4

=>x=2

vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y\in R\\x=2\end{matrix}\right.\)

e: \(0x-3y=3\)

=>-3y=3

=>y=-1

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y=-1\\x\in R\end{matrix}\right.\)

Câu 5:

a: 3x-y-2=0

=>3x-2-y=0

=>y=3x-2

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=3x-2\end{matrix}\right.\)

b: 0x+2y=3

=>2y=3

=>\(y=\dfrac{3}{2}\)

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=\dfrac{3}{2}\end{matrix}\right.\)

mik cảm ơn b nha

ĐKXĐ: \(x\notin\left\{5;-5;0;\dfrac{5}{2}\right\}\)

\(\dfrac{3}{4\left(x-5\right)}+\dfrac{5}{5x-2x^2}=\dfrac{-7}{6\left(x+5\right)}\)

=>\(\dfrac{3}{4\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=\dfrac{-5}{5x-2x^2}\)

=>\(\dfrac{9\left(x+5\right)+14\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}=\dfrac{5}{x\left(2x-5\right)}\)

=>\(\dfrac{23x-25}{12\left(x-5\right)\left(x+5\right)}=\dfrac{5}{x\left(2x-5\right)}\)

=>\(60\left(x-5\right)\left(x+5\right)=\left(2x^2-5x\right)\left(23x-25\right)\)

=>\(46x^3-50x^2-115x^2+125x=60\left(x^2-25\right)\)

=>\(46x^3-165x^2+125x-60x^2+1500=0\)

=>\(46x^3-225x^2+125x+1500=0\)

=>\(x\simeq-1,99\left(nhận\right)\)

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

\(\dfrac{9x-0,7}{4}-\dfrac{5x-1,5}{7}=\dfrac{7x-1,1}{3}-\dfrac{5\left(0,4-2x\right)}{6}\)

=>\(\dfrac{7\left(9x-0,7\right)-4\left(5x-1,5\right)}{28}=\dfrac{2\left(7x-1,1\right)-5\left(0,4-2x\right)}{6}\)

=>\(\dfrac{63x-4,9-20x+6}{28}=\dfrac{14x-2,2-2+10x}{6}\)

=>\(\dfrac{43x+1,1}{28}=\dfrac{24x-4,2}{6}\)

=>\(28\left(24x-4,2\right)=6\left(43x+1,1\right)\)

=>672x-117,6=258x+6,6

=>414x=124,2

=>\(x=124,2:414=0,3\)

Em cảm ơn nhìu ạ 😍💞

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