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\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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NHỚ TRẢ LỜI NHANH GIÙM MK NHA
1,
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)
Hoàng Ngọc Anh bài này nè bn giúp mk nha!!! ngày mai mk phải nộp bài rùi =.=
a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)
\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)
\(\Rightarrow x=\dfrac{-79}{210}\)
b) Tương tự câu a)
a,
\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)
b,
\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
c,
\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)
\(\Rightarrow5\left|x+1\right|^2=180\)
\(\Rightarrow\left|x+1\right|^2=36\)
Mà \(\left|x+1\right|\ge0\)
=> x + 1 = 6 <=> x = 7