\(\frac{x}{3}\)+ \(\frac{y}{5}\)=12(1)
\(\frac{x}{5}\)+ \(\frac{y}{7}\)=8(2)
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\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)
\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
=> x - 29 = 0
=> x = 29.
Giải:
Ta có:
\(\frac{x}{2}=\frac{y}{3};\frac{y}{2}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6};\frac{y}{6}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)
+) \(\frac{x}{4}=2\Rightarrow x=8\)
+) \(\frac{y}{6}=2\Rightarrow y=12\)
+) \(\frac{z}{15}=2\Rightarrow z=30\)
Vậy x = 8
y = 12
z = 30
\(\frac{x}{2}=\frac{y}{3};\frac{y}{2}=\frac{z}{5}\) và x + y + z =50
\(\frac{x}{4}=\frac{y}{6};\frac{y}{6}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}+\frac{y}{6}+\frac{z}{15}=\frac{50}{25}=2\)
=> x = 2.4 = 8
=> y = 2.6 = 12
=> z = 2.15 = 30
Vậy x = 8;y = 12;z = 30.
a) ĐẶT \(\frac{x}{5}=\frac{y}{2}=k;\frac{x}{5}=k\Rightarrow x=5k;\frac{y}{2}=k\Rightarrow y=2k\)
ta có \(x.y=160\)
thay\(5k.2k=160\)
\(k^2.10=160\)
\(k^2=16\)
\(\Rightarrow k=\pm4\)
do đó
\(\frac{x}{5}=\pm4\Rightarrow\hept{\begin{cases}\frac{x}{5}=4\\\frac{x}{5}=-4\end{cases}\Leftrightarrow\hept{\begin{cases}x=5.4=20\\x=5.\left(-4\right)=-20\end{cases}}}\)
\(\frac{y}{2}=\pm4\Rightarrow\hept{\begin{cases}\frac{y}{2}=4\\\frac{y}{2}=-4\end{cases}\Leftrightarrow\hept{\begin{cases}y=2.4=8\\y=2.\left(-4\right)=-8\end{cases}}}\)
vậy các x,y thỏa mãn là \(\left\{x=20;y=8\right\}\left\{x=-20;y=-8\right\}\)
a) X*Y=160
=>X=160/Y (1)
X/5 =Y/2
=> 2x=5y(tính chất tỉ lệ thức)
=>x=5Y/2 (2)
(1),(2)=> 160/y = 5y/2
=> y=8
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
Ta có : \(\frac{x+y}{7}=\frac{x-y}{5}=\frac{xy}{24}\)
BCNN(7,5,24) = 840
=> \(\frac{120\left(x+y\right)}{840}=\frac{168\left(x-y\right)}{840}=\frac{35xy}{840}\)
=> \(120\left(x+y\right)=168\left(x-y\right)=35xy\)
=> \(\frac{x+y}{\frac{1}{120}}=\frac{x-y}{\frac{1}{168}}=\frac{xy}{\frac{1}{35}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
+) \(\frac{x+y}{\frac{1}{120}}=\frac{x-y}{\frac{1}{168}}=\frac{\left(x+y\right)+\left(x-y\right)}{\frac{1}{120}+\frac{1}{168}}=\frac{2x}{\frac{1}{70}}=\left(2:\frac{1}{70}\right)x=140x\)
+) \(\frac{x+y}{\frac{1}{120}}=\frac{x-y}{\frac{1}{168}}=\frac{\left(x+y\right)-\left(x-y\right)}{\frac{1}{120}-\frac{1}{168}}=\frac{2y}{\frac{1}{420}}=\left(2:\frac{1}{420}\right)y=840y\)
=> \(140x=840y\) => \(x=6y\)=> \(\frac{x}{6}=\frac{y}{1}\)
Ta có : \(\frac{xy}{24}=1\Rightarrow xy=24\)
Do đó \(\frac{x}{6}=\frac{y}{1}=\frac{xy}{6}=\frac{24}{6}=4\)
Vậy x = 24,y = 4
\(\hept{\begin{cases}\frac{x}{3}+\frac{y}{5}=12\\\frac{x}{5}+\frac{y}{7}=8\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{15}+\frac{y}{25}=\frac{12}{5}\\\frac{x}{15}+\frac{y}{21}=\frac{8}{3}\end{cases}}}\)
\(\Rightarrow\left(\frac{x}{15}+\frac{y}{21}\right)-\left(\frac{x}{15}+\frac{y}{25}\right)=\frac{12}{5}-\frac{8}{3}\)
\(\Rightarrow\frac{y}{21}-\frac{y}{25}=\frac{-4}{15}\)
\(\Rightarrow\frac{4y}{525}=-\frac{4}{15}\Rightarrow\frac{4y}{525}=\frac{-140}{525}\)
\(\Rightarrow y=-35\Rightarrow x=65\)