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\(a,\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{23}{20}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=-\dfrac{23}{20}\end{matrix}\right.\\ b,\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{2}{9}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{9}\\x+\dfrac{1}{3}=-\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=-\dfrac{5}{9}\end{matrix}\right.\\ c,\Leftrightarrow3^x\cdot6=54\Leftrightarrow3^x=9=3^2\Leftrightarrow x=2\)
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ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(3)=(y)/(4)=(x+y)/(3+4)=(90)/(7)`
`->` $\begin{cases}x=\dfrac{90}{7}.3=\dfrac{30}{7} \\ y=\dfrac{90}{7}.4=\dfrac{360}{7} \end{cases}$
1)\(\dfrac{x}{5}=\dfrac{y}{3}\) áp dụng...ta đc:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{20}{2}=10\)
x=50
y=30
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a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
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1. Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)
+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)
+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)
+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)
Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)
2,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)
+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)
+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)
+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)
Vậy \(x=-2;y=-3;c=-4\)
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a) Ta có :\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)\(=\dfrac{5x+2y+z}{555+444+333}=\dfrac{1100}{1332}=\dfrac{275}{333}\)
Từ đó tìm được x;y;z
b) Từ \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)
Đặt \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4k\\y^2=9k\end{matrix}\right.\)
\(\Rightarrow x^2\cdot y^2=4k\cdot9k=52\)
\(\Rightarrow36k^2=52\)
\(\Rightarrow k^2=\dfrac{13}{9}\) (sai đề)
b: Sửa đề: x^2+y^2=52
Đặt x/2=y/3=k
=>x=2k; y=3k
x^2+y^2=52
=>4k^2+9k^2=52
=>k^2=4
TH1: k=2
=>x=4; y=6
TH2: k=-2
=>x=-4; y=-6
c: Đặt x/5=y/3=k
=>x=5k; y=3k
x^2-y^2=16
=>25k^2-9k^2=16
=>k^2=1
TH1: k=1
=>x=5; y=3
TH2: k=-1
=>x=-5; y=-3
d: Đặt x/2=y/3=k
=>x=2k; y=3k
Ta có: xy=54
=>2k*3k=54
=>6k^2=54
=>k^2=9
TH1: k=3
=>x=6; y=9
TH2: k=-3
=>x=-6; y=-9
e: Đặt x/4=y/3=k
=>x=4k; y=3k
Ta có: xy=12
=>4k*3k=12
=>k^2=1
TH1: k=1
=>x=4; y=3
TH2: k=-1
=>x=-4; y=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=k\)
Theo đề bài, ta có :
\(xy=54\Rightarrow2k.3k=54\)
\(\Rightarrow5k=54\Rightarrow k=10,8\)
Ta thấy :
\(\dfrac{x}{2}=10,8\Rightarrow x=10,8.2=21,6\)
\(\dfrac{y}{3}=10,8\Rightarrow y=10,8.3=32,4\)
Đặt :\(\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
mà \(xy=54\)
hay 2k . 3k = 54
\(\Rightarrow6.k^2=54\)
\(\Rightarrow k^2=9=\left(\pm3\right)^2\)
Với k = 3 \(\Rightarrow\) \(x=2.3=6;y=3.3=9\)
Với k = -3 \(\Rightarrow x=2.\left(-3\right)=-6;y=3.\left(-3\right)=-9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(\Rightarrow xy=2k\cdot3k=6k^2=54\Rightarrow k^2=9\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)
+) Với k = 3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=9\end{matrix}\right.\)
+) Với k = -3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=3k=3\cdot\left(-3\right)=-9\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;9\right);\left(-6;-9\right)\)
\(\dfrac{x}{2}=\dfrac{y}{3}\&xy=54\)
Áp dụng tc dãy tỉ số BN ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2.3}=\dfrac{54}{6}=9\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=9\\\dfrac{y}{3}=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=27\end{matrix}\right.\)
\(\Leftrightarrow x^3=-4\cdot54=-216=\left(-6\right)^3\\ \Leftrightarrow x=-6\)
\(\Leftrightarrow x^3=-216\\ \Leftrightarrow x=-6\)