\(x-91=9\)
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\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
\(\Leftrightarrow-10\dfrac{1}{9}< =x< =-3\)
hay \(x\in\left\{-10;-9;-8;...;-4;-3\right\}\)
100 x 9 + 100 x 91 x 30 - 21
= 100 x ( 9 + 91 x 30 ) - 21
= 273900 - 21
= 273879
100 x 9 + 100 x 91 x 30 - 21
= 100 x ( 9 + 91 ) x 30 - 21
= 100 x 100 x 30 - 21
= 10 000 x 30 - 21
= 300 000 - 21
= 299 979
_Học _ tốt_
\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)
Áp dụng BDT: Cô-si:
\(\Rightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)\ge2\sqrt{\dfrac{9}{\sqrt{x-19}}\cdot\sqrt{x-19}}+2\sqrt{\dfrac{16}{\sqrt{y-5}}\cdot\sqrt{y-5}}+2\sqrt{\dfrac{25}{\sqrt{z-91}}\cdot\sqrt{z-91}}\\ =2\cdot3+2\cdot4+2\cdot5=24\)Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{x-19}}=\sqrt{x-19}\\\dfrac{16}{\sqrt{y-5}}=\sqrt{y-5}\\\dfrac{25}{\sqrt{z-91}}=\sqrt{z-91}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-19=9\\y-5=16\\z-91=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=21\\z=116\end{matrix}\right.\)
Vậy các số \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)
Theo đề bài ta có:
a : 7 (dư 5)
a : 13 (dư 4)
=> a + 9 chia hết cho 7 và 13.
7 và 13 đều là số nguyên tố => a + 9 chia hết cho 7 x 13 = 91.
=> a chia cho 91 dư 91-9 = 82.
Vậy số tự nhiên đó chia cho 7 dư 5, chia cho 13 dư 4. Nếu đem chia số đó cho 91 dư 82.
Các bạn ơi mình ko hiểu cách giải tí nào luôn ý, giảng cho mình cái chỗ sao lại ra a + 9 chia hết cho 7 và 13.
7 và 13 đều là số nguyên tố => a + 9 chia hết cho 7 x 13 = 91.
=> a chia cho 91 dư 91-9 = 82.
x-91=9
x=9+91
x=100
ai k mình k lại
\(x-91=9\)
\(x=91+9\)
\(x=100\)