\(t=\frac{5}{x}-\frac{x}{4}\)
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Câu hỏi của Nguyễn Thành Đăng - Toán lớp 8 - Học toán với OnlineMath
\(B=\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}\div\dfrac{x+4}{x+5}.\dfrac{\left(x+4\right)^2}{x+5}\)
\(B=\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+5}{x+4}.\dfrac{\left(x+4\right)^2}{x+5}\)
\(B=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+4\right)^2}{\left(x+3\right)\left(x+4\right)\left(x+4\right)\left(x+5\right)}\)
\(B=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)^2\left(x+5\right)}{\left(x+3\right)\left(x+4\right)^2\left(x+5\right)}\)
\(B=x+2\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Rightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329+20}{5}=0\)
\(\Rightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Rightarrow x+329=0\).Do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x=-329\)
\(\frac{x-5}{1}+\frac{x-5}{2}+\frac{x-5}{3}+\frac{x-5}{4}=0\)
\(\Rightarrow\left(x-5\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=0\)
Mà \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\ne0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
\(\frac{x-5}{1}+\frac{x-5}{2}+\frac{x-5}{3}+\frac{x-5}{4}=0\)
\(\Rightarrow\left(x-5\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=0\)
\(\Rightarrow x-5=0\). Do \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\ne0\)
\(\Rightarrow x=5\)
a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)
=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)
\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(48;12\right).\)
b)
\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
⇒ \(\frac{5}{3}x=\frac{1}{21}\)
⇒ \(x=\frac{1}{21}:\frac{5}{3}\)
⇒ \(x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}.\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
⇒ \(x-\frac{1}{2}=\frac{1}{3}\)
⇒ \(x=\frac{1}{3}+\frac{1}{2}\)
⇒ \(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}.\)
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a)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\)x/12=4 suy ra x=12.4=48
y/3=4 suy ra y=3.4 =12
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=2\)
suy ra x-2/5=2 hoac x-2/5=-2
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
\(\Leftrightarrow\dfrac{x+2}{327}+1+\dfrac{x+3}{326}+1+\dfrac{x+4}{325}+1+\dfrac{x+5}{324}+1+\dfrac{x+349}{5}-4=0\)
=>x+329=0
hay x=-329
Tìm x, biết:
a, \(\frac{x}{28}=\frac{-4}{7}\)
\(\Leftrightarrow x.7=\left(-4\right).28\)
\(\Leftrightarrow7x=-112\)
\(\Leftrightarrow x=-16.\)
Vậy x = - 16.
b, \(\left|x+\frac{4}{5}\right|-\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}+\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{4}{5}=1\\x+\frac{4}{5}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-\frac{4}{5}\\x=-1-\frac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{5}-\frac{1}{5}\\x=\frac{-5}{5}-\frac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{-9}{5}\end{matrix}\right.\)
Vậy \(x=\frac{1}{5}\)hoặc \(x=\frac{-9}{5}\).
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