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Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=140\)
\(\Leftrightarrow\left(4x+720\right):x=140\)
\(\Leftrightarrow4x+720=140.x\)
\(\Leftrightarrow4x-140x=-720\)
\(\Leftrightarrow x.\left(-136\right)=-720\)
\(\Leftrightarrow x=-720:\left(-136\right)\)
\(\Leftrightarrow x=\frac{90}{17}\)
\(2\)) Mình đang nghĩ
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)
\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)
\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)
\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)
\(\Rightarrow-480\le x\le480\)
\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)
\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)
\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)
\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)
\(\Leftrightarrow x-3=\frac{90}{7}\)
\(\Rightarrow x=\frac{111}{7}\)