ĐKXĐ: \(\left\{{}\begin{matrix}u\ne\frac{1}{3}\\u\ne-\frac{11}{3}\end{matrix}\right.\)

\(\frac{1}{\left(3u-1\right)^2}-\frac{3}{\left(3u+11\right)^2}+\frac{2}{\left(3u-1\right)\left(3u+11\right)}=0\)

\(\Leftrightarrow\left(3u+11\right)^2-3\left(3u-1\right)^2+2\left(3u-1\right)\left(3u+11\right)=0\)

\(\Leftrightarrow\left(3u+11\right)^2-\left(3u-1\right)\left(3u+11\right)+3\left[\left(3u-1\right)\left(3u+11\right)-\left(3u-1\right)^2\right]=0\)

\(\Leftrightarrow12\left(3u+11\right)-36\left(3u-1\right)=0\)

\(\Leftrightarrow3u=7\Rightarrow u=\frac{7}{3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}1-3u\ne0\\3u+11\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3u\ne1\\3u\ne-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u\ne\frac{1}{3}\\u\ne-\frac{11}{3}\end{matrix}\right.\)

Ta có: \(\frac{2}{\left(1-3u\right)\left(3u+11\right)}=\frac{1}{9u^2-6u+1}-\frac{3}{\left(3u+11\right)^2}\)

\(\Leftrightarrow\frac{2}{\left(1-3u\right)\left(3u+11\right)}-\frac{1}{\left(3u-1\right)^2}+\frac{3}{\left(3u+11\right)^2}=0\)

\(\Leftrightarrow\frac{2\cdot\left(1-3u\right)\cdot\left(3u+11\right)}{\left(1-3u\right)^2\left(3u+11\right)^2}-\frac{\left(3u+11\right)^2}{\left(1-3u\right)^2\left(3u+11\right)^2}+\frac{\left(1-3u\right)^2\cdot3}{\left(3u+11\right)^2\left(1-3u\right)^2}=0\)

\(\Leftrightarrow\left(2-6u\right)\left(3u+11\right)-\left(9u^2+66u+121\right)+\left(1-6u+9u^2\right)\cdot3=0\)

\(\Leftrightarrow6u+22-18u^2-66u-9u^2-66u-121+3-18u+27u^2=0\)

\(\Leftrightarrow-144u-96=0\)

\(\Leftrightarrow-144u=96\)

\(\Leftrightarrow u=-\frac{96}{144}=-\frac{2}{3}\)(thỏa mãn)

Vậy: \(u=-\frac{2}{3}\)

\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}.\left(6x+1\right)\)

\(\Rightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}.6x+\frac{2}{3}\)

\(\Rightarrow\frac{55-24}{60}-x=4x+\frac{2}{3}\)

\(\Rightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)

\(\Rightarrow\frac{31}{60}-\frac{2}{3}=4x+x=5x\)

\(\Rightarrow5x=-\frac{11}{60}\)

\(\Rightarrow x=\frac{-11}{300}\)

vô tù ko e

WTF Toán Lớp 1

thấy mẹ nhầm rồi,  quy đồng quên nhân:(( mai rảnh check lại:((

Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)

\(=0\)

Do đó biểu thức trên đầu bài bằng 0

bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu

c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)

a.211,0465116

Sai thôi nha

A=\(-\frac{155}{284}\)

B=\(-\frac{43}{3780}\)

Nhân 2 cả 2 vế lên:

\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243

\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)

\(24x+\frac{24}{25}=22x+\frac{224}{243}\)

\(2x=\frac{224}{243}-\frac{24}{25}\)

\(2x=-\frac{232}{6025}\)

\(x=\frac{-116}{6075}\)

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)

\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)

\(12x+\frac{12}{25}=11x+\frac{112}{243}\)

\(11x-12x=\frac{112}{243}-\frac{12}{25}\)

\(-1x=-\frac{116}{6075}\)

\(x=-\frac{116}{6075}\div\left(-1\right)\)

\(x=\frac{116}{6075}\)