Tìm x
a, \(\frac{13x-16}{15}+\frac{x-32}{35}< \frac{x-6}{21}\)
b, \(\frac{5x+12}{14}+\frac{11x+28}{3}>\frac{4x+9}{17}\)
c, \(\left(2x-5\right).\left(x^2+4\right)< 0\)
trình bày cách làm nữa nha . lm dc 1 câu cx lm giúp mk nha
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\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=x+\frac{7}{12}\)
\(\frac{3.3\left(2x+1\right)}{12}-\frac{2\left(5x+3\right)}{12}+\frac{4\left(x+1\right)}{12}=\frac{12x+7}{12}\)
\(18x+9-10x-6+4x+4=12x+7\)
\(0x=0\) ( vô số nghiệm )
Vậy x \(\in\)R
b) ĐKXĐ : x \(\ne\)-1;-3;-5;-7
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)
\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)
\(\left(x+1\right)\left(x+7\right)=16\)
Ta thấy x+1 và x+7 là 2 số cách nhau 6 đơn vị . Mà x + 1 < x + 7
\(\Rightarrow\)\(\hept{\begin{cases}x+1=2\\x+7=8\end{cases}\Rightarrow x=1}\)
hoặc \(\hept{\begin{cases}x+1=-2\\x+7=-8\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=-15\end{cases}}\)( loại )
Vậy x = 1
\(a.\)\(\frac{13x-16}{15}+\frac{x-32}{35}< \frac{x-6}{21}\)\(MC:105\)
\(\Leftrightarrow\frac{7\left(13x-16\right)}{105}+\frac{3\left(x-2\right)}{105}< \frac{5\left(x-6\right)}{105}\)
\(\text{Khử mẫu ta dc pt tương đương vs pt:}\)
\(\Leftrightarrow7\left(13x-16\right)+3\left(x-2\right)< 5\left(x-6\right)\)
\(\Leftrightarrow91x-112+3x-6< 5x-30\)
\(\Leftrightarrow94x-118< 5x-30\)
\(\Leftrightarrow94x-5x< 118-30\)
\(\Leftrightarrow89x< 88\)
\(\Leftrightarrow x< \frac{88}{89}\)
.\(b.\)\(\frac{5x+12}{14}+\frac{11x+28}{3}>\frac{4x+9}{17}\)\(MC:714\)
\(\text{Khi khử mẫu pt ta dc pt tương đương}:\):
\(\Leftrightarrow51\left(5x+12\right)+238\left(11x+28\right)>42\left(4x+9\right)\)
\(\Leftrightarrow255x+612+2618x+6664>168x+378\)
\(\Leftrightarrow2873x+7276>168x+378\)
\(\Leftrightarrow2873x-168x>-7276+378\)
\(\Leftrightarrow2705x>-6898\)
\(\Leftrightarrow x>-\frac{6898}{2705}\)