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1: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x-y}{7-13}=\dfrac{42}{-6}=-7\)
=>x=-48; y=-91
2: x/y=3/4
=>4x=3y
=>4x-3y=0
mà 2x+y=10
nên x=3 và y=4
3: =>7x-3y=0 và x-y=-24
=>x=18 và y=42
4: =>7x-5y=0 và x+y=24
=>x=10 và y=14
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1: f(1)=3 nên a+5=3
hay a=-2
2: f(-3)=-2 nên -3a+5=-2
=>-3a=-7
hay a=7/3
3: f(-1)=4 nên -a+5=4
hay a=1
4: f(1/2)=4 nên 1/2a+5=4
=>1/2a=-1
hay a=-2
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\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\)
\(\Rightarrow\frac{11}{10}x=\frac{-33}{25}\)
\(\Rightarrow x=\frac{-33}{25}\div\frac{11}{10}\)
\(\Rightarrow x=\frac{-6}{5}\)
Vậy \(x=\frac{-6}{5}.\)
\(\frac{1}{2}.x+\frac{3}{5}.x=\frac{-33}{25}\)
\(\Leftrightarrow x.\left(\frac{1}{2}+\frac{3}{5}\right)=\frac{-33}{25}\)
\(\Leftrightarrow x.\left(\frac{5}{10}+\frac{6}{10}\right)=\frac{-33}{25}\)
\(\Leftrightarrow x.\frac{11}{10}=\frac{-33}{25}\)
\(\Leftrightarrow x=\frac{-33}{25}:\frac{11}{10}=\frac{-33}{25}.\frac{10}{11}\)
\(\Leftrightarrow x=\frac{-330}{275}=\frac{-6}{5}\)
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a. -3/4 x 12/-5 x (-25/6)=-15/2
b. -2 x -38/21 x -7/4 x (-3/8)=-19/8
c. (11/12: 33/16) x 3/5=4/15
d. 7/23 x [(-8/6)- 45/18]=-7/6
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\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
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a. \(\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{-33}{25}\)
\(\Rightarrow\dfrac{11}{10}x=\dfrac{-33}{25}\)
\(\Rightarrow x=\dfrac{-33}{25}:\dfrac{11}{10}=\dfrac{-6}{5}\)
Vậy.........
b. \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+\dfrac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}+\dfrac{-3}{7}:x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{-3}{7}:x=\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)
Vậy................
\(\frac{1}{2}.x+\frac{3}{5}.x=-\frac{33}{25}\)
\(\Leftrightarrow x.\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)
\(\Leftrightarrow x.\frac{11}{10}=-\frac{33}{25}\)
\(\Leftrightarrow x=-\frac{33}{25}:\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{6}{5}\)
Vậy \(x=\frac{6}{5}\)
\(\frac{1}{2}.x+\frac{3}{5}.x=-\frac{33}{25}\)
\(x.\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)
\(x.\left(\frac{5}{10}+\frac{6}{10}\right)=-\frac{33}{25}\)
\(x.\frac{11}{10}=-\frac{33}{25}\)
\(x=-\frac{33}{25}:\frac{11}{10}\)
\(x=-\frac{33}{25}.\frac{10}{11}\)
\(x=-\frac{6}{5}\)
Vậy \(x=-\frac{6}{5}\)
Chúc bạn học tốt !!!