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1, \(A=\dfrac{-2}{4}+\dfrac{2}{7}-\dfrac{5}{28}\)
\(A=\dfrac{-1}{2}+\dfrac{2}{7}-\dfrac{5}{28}\)
\(A=\dfrac{-14+8-5}{28}=\dfrac{-11}{28}\)
2, \(B=\left(\dfrac{5}{7}.0,6-5:3\dfrac{1}{2}\right).\left(40\%-1,4\right).\left(-2\right)^3\)
\(B=\dfrac{-13}{14}.\left(-1\right).8=\dfrac{52}{7}\)
\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)
\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)
\(\Leftrightarrow-16x=-14\)
\(\Rightarrow x=\dfrac{7}{8}\)
\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)
\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)
\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)
Cái này không biết tách kiểu gì cho vừa nên bạn nhấn máy tính nhé
Mode 5 3 rồi lần lượt điền vào theo thứ tự trên thì
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
\(C=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{a+1}{a}\right):\dfrac{a-1}{a^3}\)
\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right):\dfrac{a-1}{a^3}\)
\(=\dfrac{a+2\cdot\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)
\(=\dfrac{2a\left(a+1\right)}{\left(a^2+1\right)\cdot\left(a+1\right)^3}\cdot\dfrac{a^2}{a-1}\)
\(=\dfrac{2a^3}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2
\(=\dfrac{\dfrac{2^3\cdot3^2}{3^3\cdot4^2}\cdot\left(-1\right)}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot2^6\cdot3^3}}=\dfrac{-\dfrac{1}{2}\cdot\dfrac{1}{3}}{-\dfrac{1}{2^4}\cdot5\cdot\dfrac{1}{3^3}}=\dfrac{1}{6}:\dfrac{5}{2^4\cdot3^3}\)
\(=\dfrac{1}{6}\cdot\dfrac{2^4\cdot3^3}{5}=\dfrac{2^3\cdot3^2}{5}=\dfrac{72}{5}\)