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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\left(1+1+1\right)!=6\)
\(2+2+2=6\)
\(\left(3+3-3\right)!=6\)
\(\sqrt{4}+\sqrt{4}+\sqrt{4}=6\)
\(5+5:5=6\)
\(7-7:7=6\)
\(\sqrt{8+\left(8:8\right)}!=6\)
\(\left(9-9\right)+\sqrt{9}!=6\)
\(\sqrt{10-\left(10:10\right)}!=6\)
Tất cả các phép đều sai
~~ tk mk nhé....~~
Ai tk mk mk tk lại ~~~
Kb lun nha....n_n
\(S=\frac{1}{x\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{2}\left[\frac{2}{x\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{x+2-x}{x\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)