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a)\(-\dfrac{2}{3}+1=-\dfrac{7}{9}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{9}-1\)
\(-\dfrac{2}{3}x=-\dfrac{16}{9}\)
\(x=-\dfrac{16}{9}:-\dfrac{2}{3}\)
\(x=-\dfrac{16}{9}.-\dfrac{3}{2}\)
\(x=\dfrac{8}{3}\)
b)\(\left|x-\dfrac{5}{3}\right|=\dfrac{1}{2}\)
\(x-\dfrac{5}{6}=\dfrac{-1}{2}\) hoặc \(x-\dfrac{5}{6}=\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}+\dfrac{5}{6}\) hoặc \(x=\dfrac{1}{2}+\dfrac{5}{6}\)
\(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{4}{3}\)
c)\(\left|x+\dfrac{4}{9}\right|-\dfrac{1}{2}=\dfrac{3}{2}\)
\(\left|x+\dfrac{4}{9}\right|=\dfrac{3}{2}+\dfrac{1}{2}\)
\(\left|x+\dfrac{4}{9}\right|=2\)
\(x+\dfrac{4}{9}=2\) hoặc \(x+\dfrac{4}{9}=-2\)
\(x=2-\dfrac{4}{9}\) hoặc \(x=\left(-2\right)-\dfrac{4}{9}\)
\(x=\dfrac{14}{9}\) hoặc \(x=\dfrac{-22}{9}\)
a, \(-\dfrac{2}{3}x+1=-\dfrac{7}{9}\)
\(\dfrac{\left(-2\right)}{3}x=\dfrac{-7}{9}+\dfrac{-9}{9}\)
\(x=\dfrac{-16}{9}.\dfrac{-3}{2}=\dfrac{-8}{3}.\dfrac{-1}{1}\)
\(x=\dfrac{8}{3}\)
a, \(\dfrac{x}{y}=\dfrac{4}{9}\Rightarrow\dfrac{x}{4}=\dfrac{y}{9}\)
Theo tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x+y}{4+9}=\dfrac{-30}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-\dfrac{30}{13}\right)=\dfrac{-120}{13}\\y=9.\left(-\dfrac{30}{13}\right)=\dfrac{-270}{13}\end{matrix}\right.\)
Vậy....
b, \(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{4}=\dfrac{y}{7}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{2x-y}{2.4-7}=\dfrac{10}{1}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.10=40\\y=7.10=70\end{matrix}\right.\)
Vậy......
c, Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{9}=\dfrac{x-zy+z}{4-9.6+9}=\dfrac{-30}{-41}=\dfrac{30}{41}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\dfrac{30}{41}=\dfrac{120}{41}\\y=6.\dfrac{30}{41}=\dfrac{180}{41}\\z=9.\dfrac{30}{41}=\dfrac{270}{41}\end{matrix}\right.\)
Vậy....
1
\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)
\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)
2
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
Có \(a\left(b+1\right)< b\left(a+1\right)\Leftrightarrow ab+a< ab+b\)
\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)
Áp dụng \(\frac{2^{2018}}{3^{2019}}< \frac{2^{2018}+1}{3^{2019}+1}\)
Ta có:
\(1-\frac{a}{b}=\frac{b-a}{b}\)
\(1-\frac{a+1}{b+1}=\frac{b+1-a-1}{b+1}=\frac{b-a}{b+1}\)
Vì b < b + 1 và a < b; a, b nguyên dương => b - a > 0 nên \(\frac{b-a}{b}>\frac{b-a}{b+1}\)
Do đó \(1-\frac{a}{b}>1-\frac{a+1}{b+1}\)
\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)
Áp dụng chứng minh tương tự nhé bạn
mk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước
a)
\(\dfrac{2}{3}-\dfrac{5}{12}x=\dfrac{-8}{3}\)\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}-\left(-\dfrac{8}{3}\right)\)
\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\)
\(\Rightarrow x=\dfrac{10}{3}:\dfrac{5}{12}=8\)
b) \(3x-2\left(2x-1\right)=1\dfrac{1}{3}\)\(\Rightarrow3x-4x+2=\dfrac{4}{3}\)
\(\Rightarrow3x-4x=\dfrac{4}{3}-2\)
\(\Rightarrow-x=-\dfrac{2}{3}\)\(\Rightarrow x=\dfrac{2}{3}\)
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\Rightarrow\left(x+4\right)\left(x+4\right)=20.5\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=10^2\) hoặc \(\left(x+4\right)^2=\left(-10\right)^2\)
=> x+4=10 => x+4=-10
=> x=6 => x=-14
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