\(\frac{5x}{x-13}=21.25\)
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NT
4
7 tháng 3 2017
\(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+......+\frac{3}{21.25}\)
\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{21.25}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{21}-\frac{1}{25}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{25}\right)\)
\(=\frac{3}{4}.\frac{24}{25}\)
\(=\frac{18}{25}\)
7 tháng 3 2017
\(4A=3-\frac{1}{5}+\frac{3}{5}-\frac{3}{9}+\frac{3}{9}-\frac{3}{13}+...+\frac{3}{21}-\frac{3}{25}\)\(\frac{3}{25}\)
\(4A=3-\frac{3}{25}\)
\(4A=\frac{72}{25}\)
\(A=\frac{18}{25}\)
k minh ha
AH
Akai Haruma
Giáo viên
1 tháng 12 2019
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(4x^2+1\geq 4x\)
\(\Rightarrow \left\{\begin{matrix} 5x^2-x+3\geq x^2+3x+2\\ 5x^2+x+\geq x^2+5x+6\\ 5x^2+3x+13\geq x^2+7x+12\\ 5x^2+5x+21\geq x^2+9x+20\end{matrix}\right.\)
\(\text{VT}\leq \frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)
\(\Leftrightarrow \text{VT}\leq \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}\)
\(\Leftrightarrow \text{VT}\leq \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+\frac{(x+4)-(x+3)}{(x+3)(x+4)}+\frac{(x+5)-(x+4)}{(x+4)(x+5)}\)
\(\Leftrightarrow \text{VT}\leq \frac{1}{x+1}-\frac{1}{x+5}\)
\(\Leftrightarrow \text{VT}\leq \frac{4}{x^2+6x+5}\)
Dấu "=" xảy ra khi $4x^2=1, x>0$ hay $x=\frac{1}{2}$
Vậy $x=\frac{1}{2}$ là nghiệm của PT.
NT
6
N
13 tháng 7 2018
bài này nâng cao lớp 6 mk giải rồi bạn nhờ ai giảng hộ nha nếu bn 5 lên 6
B=1/4.(4/1.5+4/5.9+......+4/25.29)
B=1/4.(1-1/5+1/5-1/9+.....+1/25-1/29)
B=1/4.(1-1/29)
B=1/4.28/29
B=7/29
HN
13 tháng 7 2018
\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)
\(\Rightarrow4B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}+\frac{4}{25.29}\)
\(\Rightarrow4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}+\frac{1}{25}-\frac{1}{29}\)
\(\Rightarrow4B=1-\frac{1}{29}\)
\(\Rightarrow4B=\frac{29}{29}-\frac{1}{29}=\frac{28}{29}\)
\(\Rightarrow B=\frac{28}{29}:4=\frac{28}{29}.\frac{1}{4}=\frac{7}{29}\)
Vậy ....
4 tháng 12 2016
a)\(dk,x\ne7;x\ne0\)
\(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}=\frac{\left(4x+13\right)+\left(x-48\right)}{5x\left(x-7\right)}\\ \)
\(=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}\)
b)
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{1-\left(5x\right)^2}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(\frac{1+5x}{x\left(1-5x\right)\left(1+5x\right)}+\frac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-15x+5x+1}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}\)
19 tháng 9 2019
\(\frac{1-5x}{2}=\frac{13}{36}-\frac{x+1}{3}\)
\(\Leftrightarrow\frac{1-5x}{2}=\frac{13}{36}-\frac{12.\left(x+1\right)}{36}\)
\(\Leftrightarrow\frac{1-5x}{2}=\frac{13-12x-12}{36}\)
\(\Leftrightarrow\frac{1-5x}{2}=\frac{1-12x}{36}\)
\(\Leftrightarrow2.\left(1-12x\right)=36.\left(1-5x\right)\)
\(\Leftrightarrow2-24x=36-180x\)
\(\Leftrightarrow-24x+180x=36-2\)
\(\Leftrightarrow156x=34\)
\(\Leftrightarrow x=\frac{17}{78}\)
S
10 tháng 5 2017
\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)
\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(=2100\left(1-\frac{1}{25}\right)\)
\(=2100\cdot\frac{24}{25}\)
\(=2016\)
10 tháng 5 2017
\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\frac{24}{25}\)
\(A=2016\)
HT
23 tháng 5 2016
A=\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\) ( với \(x^4-8x^2-9=x^4-9x^2+x^2-9=x^2\left(x^2-9\right)+\left(x^2-9\right)=\left(x^2-9\right)\left(x^2+1\right)=\left(x-3\right)\left(x+3\right)\left(x^2+1\right)\)
A= \(\frac{13-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-\frac{2}{x-3}=0\) \(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\) \(\Leftrightarrow\left(10x-30\right)\left(x-3\right)+6-2\left(x+3\right)=0\Leftrightarrow-x^2+11x-30=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=5\end{array}\right.\)
T
17 tháng 8 2016
pt đã cho có dạng \(\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{4}{13}\Leftrightarrow....\)
17 tháng 8 2016
bạn tuấn mình thấy vậy nè
Gỉa sử cho x=1 ta thấy \(\frac{1}{1\times4}\ne\frac{1}{1}-\frac{1}{4}\)
Bạn bấm máy tính thử xem dấu bằng chỉ áp dụng với 2 số tự nhiên liên tiếp thôi còn cái này cách 3 lận
giải thích giúp mình với