tìm x biết
\(\left(x-\frac{2}{3}\right)\times\left(x+\frac{1}{4}\right)=0\)
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\(\Leftrightarrow\frac{4x^2}{5}\times\frac{2x-3}{6}-\frac{3x-10}{15}\times\frac{4x^2+3}{3}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{4x^2\left(2x-3\right)}{30}-\frac{\left(3x-10\right)\left(4x^2+3\right)}{45}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{12x^2\left(2x-3\right)}{90}-\frac{2\left(3x-10\right)\left(4x^2+3\right)}{90}=\frac{44x^2}{90}\)
\(\Leftrightarrow12x^2\left(2x-3\right)-2\left(3x-10\right)\left(4x^2+3\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-2\left(12x^3+9x-40x^2-30\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-24x^3-18x+80x^2+60=44x^2\)
\(\Leftrightarrow24x^3-36x^2-24x^3-18x+80x^2-44x^2=-60\)
\(\Leftrightarrow\left(24x^3-24x^3\right)+\left(-36x^2+80x^2-44x^2\right)-18x=-60\)
\(\Leftrightarrow-18x=-60\)
\(\Leftrightarrow x=\frac{-60}{-18}\)
\(\Leftrightarrow x=\frac{10}{3}\)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
\(\left(x-\frac{2}{3}\right)\times\left(x+\frac{1}{4}\right)=0\)
Để đầu bài = 0 thì :
\(\orbr{\begin{cases}\left(x-\frac{2}{3}\right)=0\Rightarrow x=\frac{2}{3}\\\left(x+\frac{1}{4}\right)=0\Rightarrow x=-\frac{1}{4}\end{cases}}\)
Vậy \(x=\frac{2}{3}\)hoặc \(x=-\frac{1}{4}\)