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a;\(\dfrac{17}{24}\) < \(\dfrac{17}{34}\) ⇒ \(\dfrac{-17}{24}\) > \(\dfrac{-17}{34}\) = - \(\dfrac{1}{2}\)
\(\dfrac{25}{31}\) > \(\dfrac{25}{50}\) ⇒ - \(\dfrac{25}{31}\) < \(\dfrac{-25}{50}\) = - \(\dfrac{1}{2}\)
Vậy - \(\dfrac{17}{34}\) > - \(\dfrac{25}{31}\)
b; \(\dfrac{27}{38}\) > \(\dfrac{27}{39}\) > \(\dfrac{25}{39}\)
⇒ - \(\dfrac{27}{38}\) < - \(\dfrac{25}{39}\) = \(\dfrac{-125}{195}\)
Vậy - \(\dfrac{27}{38}\) < - \(\dfrac{125}{195}\)
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)
\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)
\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=3(a+b+c+d)/a+b+c+d=3
suy ra k=3
taco:\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)=>\(\dfrac{b+c+d}{a}+1=\dfrac{c+d+a}{b}+1=\dfrac{a+b+d}{c}+1=\dfrac{a+b+c}{d}+1=k+1\)=>\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=k+1=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)
=>k+1=4
=>k=3
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{4}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ..
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)
Thay vào ta được:
\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)
Vậy...................
Chúc bạn học tốt!!!
1) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (1)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
2) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Cô làm rồi em nhé:
https://olm.vn/cau-hoi/giup-em-voiii.8161766187032
- \(\dfrac{11}{24}\) : \(\dfrac{17}{23}\) - \(\dfrac{11}{24}\) : \(\dfrac{17}{11}\) - \(\dfrac{1}{12}\)
= - \(\dfrac{11}{24}\) x \(\dfrac{23}{17}\) - \(\dfrac{11}{24}\) x \(\)\(\dfrac{11}{17}\) - \(\dfrac{1}{12}\)
= - \(\dfrac{11}{24}\) x (\(\dfrac{23}{17}\) + \(\dfrac{11}{17}\)) - \(\dfrac{1}{12}\)
= - \(\dfrac{11}{24}\) x 2 - \(\dfrac{1}{12}\)
= - \(\dfrac{11}{12}\) - \(\dfrac{1}{12}\)
=- \(\dfrac{12}{12}\)
= - 1