a \(\dfrac{25x^3y}{7z}\cdot\dfrac{28z}{15x^2y^5}\)
b \(\dfrac{x^2+3x+9}{2x+10}\cdot\dfrac{x+5}{x^3+27}\)
c \(\dfrac{3x-6}{x-1}\cdot\dfrac{1-x^3}{10-5x}\)
d \(\dfrac{3x-2}{x^2+1}\cdot\dfrac{x-1-x^2}{4-9x^2}\)
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\(a,=\dfrac{2\left(2x^2+1\right).\left(3x+2\right).2\left(2-x\right)}{\left(x-2\right)\left(x-4\right)\left(2x^2+1\right)}=\dfrac{-4.\left(3x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-4\left(3x+2\right)}{x-4}\\ b,=\dfrac{\left(x+3\right).\left(x+2\right)}{x.\left(x+3\right)^2}\times\dfrac{x\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+3\right)\left(x+2\right)x\left(x+3\right)}{x.\left(x+3\right)^2.\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{-1}{x-1}\)
\(=\dfrac{x^3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{-1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4-x+x^3+x+x-1-x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4+x^3}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3}{x-1}\)
b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)-x^2\left(x-1\right)-1\left(x+1\right)+1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^3-x^3+x^2-x-1+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^2-2}{\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{4-2x+x^2}{2+x}-2-x\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2\left(2+x\right)}{2+x}-\dfrac{x\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-4-2x-2x-x^2}{2+x}\)
\(=\dfrac{-6x}{2+x}\)
Còn lại thì dễ rồi, bạn tự làm nha ^^
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 \(x=\dfrac{67}{225}\).
Chúc bạn học tốt!
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NHỚ TRẢ LỜI NHANH GIÙM MK NHA
a) Ta có: \(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)
\(=\dfrac{x-2}{x-1}\)
b) Ta có: \(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)
\(=\dfrac{4+4x}{x\left(3x+6\right)}+\dfrac{x^2}{x\left(3x+6\right)}\)
\(=\dfrac{x^2+4x+4}{3x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{3x\left(x+2\right)}\)
\(=\dfrac{x+2}{3x}\)
c) Ta có: \(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)
\(=\dfrac{x\left(x-2\right)}{x-1}\cdot\dfrac{1}{x}\cdot\dfrac{x^2-2x+1}{x^2-4}\)
\(=\dfrac{x-2}{x-1}\cdot\dfrac{\left(x-1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-1}{x+2}\)
\(B=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3\left(x^2-4\right)}{2x^2-8x+8}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x^2-4x+4\right)}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{1}{3}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)
\(=\dfrac{x-2}{x+2}\cdot\dfrac{3\left(5x+10\right)+7\left(x-2\right)}{21\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)
\(=\dfrac{1}{x+2}\cdot\dfrac{15x+30+7x-14}{21}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{22x+16}{21\left(x+2\right)}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{2\left(x-2\right)\left(22x+16\right)+21\left(x+2\right)\left(3x+6\right)}{42\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{\left(2x-4\right)\left(22x+16\right)+\left(21x+42\right)\left(3x+6\right)}{42\left(x^2-4\right)}\)
\(=\dfrac{44x^2+32x-88x-64+63x^2+126x+126x+252}{42x^2-168}\)
\(=\dfrac{107x^2+196x+188}{42x^2-168}\)
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)
Nguyễn Ngọc Thanh Trúc đề là gì
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