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Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{5}+\frac{y}{5}=\frac{x+y}{5+5}=\frac{24}{10}=2,4\)
=> \(\begin{cases}x=12\\y=12\end{cases}\)
a) \(\frac{16}{2^x}=1\Leftrightarrow2^x=16\Leftrightarrow2^x=2^4\Leftrightarrow x=4\)
b)\(5^{x+2}=625\Leftrightarrow5^{x+2}=5^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c)\(\frac{x+3}{8}=\frac{2}{x-3}\left(đk:x\ne3\right)\Leftrightarrow\left(x+3\right).\left(x-3\right)=2.8\Leftrightarrow x^2-9=16\Leftrightarrow x^2=25\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
d)\(\frac{x^2}{6}=\frac{24}{25}\Leftrightarrow25x^2=24.6\Leftrightarrow\left(5x\right)^2=144\Leftrightarrow\orbr{\begin{cases}5x=12\\5x=-12\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{5}\\x=-\frac{12}{5}\end{cases}}\)
a) 16=2^x \(\Leftrightarrow\)x=4
b)5^x+2=5^4\(\Leftrightarrow\)x+2=4\(\Leftrightarrow\)x=2
k đi, mk làm tiếp cho
Áp dụng tc của dãy tỉ số bằng nhau ta có :
\(\frac{x}{5}=\frac{y}{7}=\frac{x+y}{5+7}=\frac{24}{12}=2\)
\(\Rightarrow\begin{cases}x=60\\y=84\end{cases}\)
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
\(\left(-5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
\(< =>\left(\frac{-43}{8}+x-\frac{173}{24}\right):\left(\frac{-50}{3}\right)=0\)
\(< =>\frac{-43}{8}+x-\frac{173}{24}=0\)
\(< =>\frac{-43}{8}+x=\frac{173}{24}\)
\(< =>x=\frac{173}{24}+\frac{43}{8}\)
\(< =>x=\frac{151}{12}\)
Tìm x:
\(\dfrac{4-x}{6-x}\)=\(\dfrac{x-3}{x-8}\)\(\Rightarrow\)(4-x)(x-8)=(6-x)(x-3)
\(\Rightarrow\)12x-x2-32=9x-x2-18
\(\Rightarrow\)3x=14\(\Rightarrow\)x=\(\dfrac{14}{3}\).
\(\dfrac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)
=5.(1-7.22) = 5.(1-28) = 5.(-27) = -135
x + 3/16 = -5/24
=> x = -5/24 - 3/16
=> x = -19/48
Vậy x = -19/48
\(x+\frac{3}{16}=-\frac{5}{24}\)
\(x=-\frac{5}{24}-\frac{3}{16}\)
\(x=-\frac{19}{48}\)