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a,\(\frac{x}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2.3}{5}=\frac{6}{5}\)
b,\(\frac{-9}{x}=\frac{\frac{\left(-x\right)}{4}}{49}\)
\(\Rightarrow x=-9.49.\frac{-4}{x}\)
\(\Rightarrow x=\frac{1764}{x}\)
\(\Rightarrow x^2=1764=42^2\)
\(\Rightarrow x=\pm2\)
\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49
Ta có : \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)
Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)
\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186
Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
\(\frac{5}{4}\cdot\left(-\frac{12}{7}\right)=\frac{-60}{28}=\frac{-15}{7}\)
\(\frac{-4}{3}:\frac{13}{9}=-\frac{4}{3}\cdot\frac{9}{13}=-\frac{36}{39}=-\frac{12}{13}\)
\(-\frac{5}{7}\cdot\frac{49}{3}:\frac{7}{-6}=-\frac{5}{7}\cdot\frac{49}{3}\cdot-\frac{6}{7}=\frac{1470}{147}=10\)
\(-\frac{9}{25}:6=-\frac{9}{150}=-\frac{3}{50}\)
2.
\(\frac{7}{4}:\left(\frac{2}{3}-\frac{5}{4}\right)\cdot\left(-\frac{1}{4}\right)\)
\(=\frac{7}{4}:\left(-\frac{7}{12}\right)\cdot\left(-\frac{1}{4}\right)\)
\(=\frac{3}{4}\)
b,giá trị của x thỏa mãn đẳng thức \(-\frac{2}{3}x=\frac{4}{5}\)là
\(-\frac{2}{3}x=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{6}{5}\)
Ta có : \(\frac{x-1}{5}=\frac{y-2}{2}=\frac{z-2}{3}=\frac{2y-4}{4}=\frac{x-1+2y-4-\left(z-2\right)}{5+4-3}=\frac{x-1+2y-4-z+2}{6}\)
\(=\frac{x+2y-z-3}{6}=\frac{3}{6}=\frac{1}{2}\)
Nên : \(\frac{x-1}{5}=\frac{1}{2}\Rightarrow x-1=\frac{5}{2}\Rightarrow x=\frac{7}{2}\)
\(\frac{y-2}{2}=\frac{1}{2}\Rightarrow y-2=1\Rightarrow y=3\)
\(\frac{z-2}{3}=\frac{1}{2}\Rightarrow z-2=\frac{3}{2}\Rightarrow z=\frac{7}{2}\)
Vậy ,,,,,,,,,,,,,,,,,,
Bài 2:
a) \(\frac{x}{-27}=\frac{-3}{x}\Leftrightarrow-\frac{x}{27}=-\frac{3}{x}\Leftrightarrow-x.x=\left(-27\right).\left(-3\right)\Leftrightarrow-x^2=-81\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
b) \(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\Leftrightarrow-\frac{9}{x}=-\frac{49x}{4}\Leftrightarrow-9.4=-x.49x\Leftrightarrow-36=-49x^2\Leftrightarrow\orbr{\begin{cases}x=\frac{6}{7}\\x=-\frac{6}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{7}\\x=-\frac{6}{7}\end{cases}}\)
a) x/-27=-3/x
suy ra: x.x=-3.(-27)
suy ra: x^2=81. Suy ra: x^2=9^2 hoặc (-9)^2
suy ra: x=9 hoặc x=-9
Vậy x=9 hoặc x=-9
\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)
\(\Rightarrow x.x=\left(-9\right).\left(-\frac{4}{49}\right)\)
\(\Rightarrow x^2=\frac{36}{49}\)
\(\Rightarrow x^2=\hept{\begin{cases}\left(\frac{6}{7}\right)^2\\\left(-\frac{6}{7}\right)^2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}+\frac{6}{7}\\-\frac{6}{7}\end{cases}}\)
\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)
=> -x.x = -9.4/49
=> -x2 = \(\frac{-36}{49}\)
=> x2 = \(\frac{36}{49}\)
=> x2 = \(\left(\pm\frac{6}{7}\right)^2\)
=> x = \(\pm\frac{6}{7}\)