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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
Giải:
a) \(\frac{1}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\Leftrightarrow2x=\frac{1}{2}-\left(\frac{1}{5}-\frac{2}{3}\right)\)
\(\Leftrightarrow2x=\frac{1}{2}-\frac{-7}{15}\)
\(\Leftrightarrow2x=\frac{11}{15}\)
\(\Leftrightarrow x=\frac{11}{15}:2\)
\(\Leftrightarrow x=\frac{11}{30}\)
b) \(4\left(\frac{1}{3}-3\right)+\frac{1}{2}=\frac{5}{6}+x\)
\(\Leftrightarrow\frac{-61}{6}=\frac{5}{6}+x\)
\(\Leftrightarrow x=\frac{-61}{6}-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{-66}{6}=-11\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=140\)
\(\Leftrightarrow\left(4x+720\right):x=140\)
\(\Leftrightarrow4x+720=140.x\)
\(\Leftrightarrow4x-140x=-720\)
\(\Leftrightarrow x.\left(-136\right)=-720\)
\(\Leftrightarrow x=-720:\left(-136\right)\)
\(\Leftrightarrow x=\frac{90}{17}\)
\(2\)) Mình đang nghĩ
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
\(\left(1-\frac{3}{10}-x\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1.\)
\(\left(1-\frac{3}{10}-x\right):\left(\frac{9}{10}-\frac{2}{5}\right)+\frac{4}{5}=1\)
\(\left(1-\frac{3}{10}-x\right):\frac{1}{2}+\frac{4}{5}=1\)
\(\left(\frac{7}{10}-x\right):\frac{1}{2}=1-\frac{4}{5}\)
\(\left(\frac{7}{10}-x\right):\frac{1}{2}=\frac{1}{5}\)
\(\frac{7}{10}-x=\frac{1}{5}\times\frac{1}{2}\)
\(\frac{7}{10}-x=\frac{1}{10}\)
\(x=\frac{7}{10}-\frac{1}{10}\)
\(x=\frac{3}{5}\)
\(\left(\text{1}-\frac{\text{3}}{\text{1}\text{0}}-\text{x}\right):\left(\frac{\text{19}}{\text{1}\text{0}}-\text{1}-\frac{\text{2}}{\text{5}}\right)+\frac{\text{4}}{\text{5}}=\text{1}\)
=> \(\left(\text{1}-\frac{\text{3}}{\text{1}\text{0}}-\text{x}\right):\left(\frac{\text{19}}{\text{1}\text{0}}-\text{ }\frac{\text{10}}{\text{10}}-\frac{\text{4}\text{ }}{\text{ }10}\right)=\text{1}-\frac{\text{4}}{\text{5}}=\frac{\text{1}}{\text{5}}\)
=> \(\left(\text{1}-\frac{\text{3}}{\text{1}\text{0}}-\text{x}\right):\frac{\text{1}}{\text{2}}=\frac{\text{1}}{\text{5}}\)
=> \(\text{1}-\frac{\text{3}}{\text{1}\text{0}}-\text{x}=\frac{\text{1}}{\text{5}}\text{x}\frac{\text{1}}{\text{2}}=\frac{\text{1}}{\text{1}\text{0}}\)
=> \(\text{x}=\text{1}-\frac{\text{3}}{\text{1}\text{0}}-\frac{\text{1}}{\text{10}}=\frac{\text{3}}{\text{5}}\)