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\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)
\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)
\(\Leftrightarrow-16x=-14\)
\(\Rightarrow x=\dfrac{7}{8}\)
\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)
\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)
\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)
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
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)
\(=\dfrac{\dfrac{2^3\cdot3^2}{3^3\cdot4^2}\cdot\left(-1\right)}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot2^6\cdot3^3}}=\dfrac{-\dfrac{1}{2}\cdot\dfrac{1}{3}}{-\dfrac{1}{2^4}\cdot5\cdot\dfrac{1}{3^3}}=\dfrac{1}{6}:\dfrac{5}{2^4\cdot3^3}\)
\(=\dfrac{1}{6}\cdot\dfrac{2^4\cdot3^3}{5}=\dfrac{2^3\cdot3^2}{5}=\dfrac{72}{5}\)
\(B=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3\left(x^2-4\right)}{2x^2-8x+8}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x^2-4x+4\right)}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{1}{3}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)
\(=\dfrac{x-2}{x+2}\cdot\dfrac{3\left(5x+10\right)+7\left(x-2\right)}{21\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)
\(=\dfrac{1}{x+2}\cdot\dfrac{15x+30+7x-14}{21}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{22x+16}{21\left(x+2\right)}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{2\left(x-2\right)\left(22x+16\right)+21\left(x+2\right)\left(3x+6\right)}{42\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{\left(2x-4\right)\left(22x+16\right)+\left(21x+42\right)\left(3x+6\right)}{42\left(x^2-4\right)}\)
\(=\dfrac{44x^2+32x-88x-64+63x^2+126x+126x+252}{42x^2-168}\)
\(=\dfrac{107x^2+196x+188}{42x^2-168}\)
\(M=\left(\dfrac{x}{y}+1-5\right)^3=\left(\dfrac{x}{y}-4\right)^3\)
\(=\left(\dfrac{12}{2}-4\right)^3=\left(6-4\right)^3=2^3=8\)
b) \(\dfrac{x^2+2\cdot x+2}{x+1}>\dfrac{x^2+4\cdot x+5}{x+2}-1\)
\(\Leftrightarrow\dfrac{x^2+2\cdot x+2}{x+1}-\dfrac{x^2+4\cdot x+5}{x+2}+1>0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x+1\right)\left(x^2+4x+5\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+4x^2+5x+x^2+4x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+5x^2+9x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-x^3-5x^2-9x-5+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{0+0+1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)>0\)
\(\left\{{}\begin{matrix}x+1>0\\x+2>0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+1< 0\\x+2< 0\end{matrix}\right.\)
↓
\(\left\{{}\begin{matrix}x>-1\\x>-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< -1\\x< -2\end{matrix}\right.\)
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2