Tính nhanh:
4/3x9/8x16/15x25/24x29/30x....x10000/9999.
Các bạn làm ơn giúp mik vs.ai nhanh mik tick cho nha.thank you very much
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
a)1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72+89/90
=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10 = 81/10
b)4/3.7 + 4/7.11 + 4/11.15 + 4/15.19 + 4/19.23 + 4/23.27
=4.(4/3.7 + 4/7.11 + ........+ 4/23.27 )
=1.( 1/3.7 + 1/7.11 + ......+ 1/23.27 )
=1.(1/3 - 1/7 + 1/7 - 1/11 +............ + 1/23 - 1/27 )
=1.(1/3 - 1/27 )
=1.(9/27 - 1/27)
=1.8/27
=8/27
c)1/10+1/40+1/88+1/154+1/138+1/340
=1/2.5 + 1/5.8 + 1/11.8 + 1/11.14 + 1/14.17 + 1/17.20
=1/3. (3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 + 3/14.17 + 3/17.20 )
=1/3. ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20 )
=1/3. ( 1/2 - 1/20 )
=1/3. 9/20
=3/20
P/S: CHÚC HOK TỐT !
\(a,\frac{57}{300}=\frac{19}{100}=19\%\)
\(b,\frac{24}{400}=\frac{6}{100}=6\%\)
\(c,\frac{12}{500}=\frac{2.4}{100}=2,4\%\)
\(d,\frac{750}{600}=\frac{125}{100}=125\%\)
~Study well~
#๖ۣۜNamiko#
GIẢI
a) \(\frac{57}{300}=\frac{19}{100}=19\%\)
b) \(\frac{24}{400}=\frac{6}{100}=6\%\)
c)\(\frac{12}{500}=\frac{6}{250}=2,4\%\)
d )\(\frac{750}{600}=\frac{125}{100}=125\%\)
P/S: HOK TỐT !
Ta có \(\Delta'=1-m\ge0\)=>\(m\le1\)
Theo viet ta có
\(x_1+x_2=2\)
Vì x1 là nghiệm của phương trình
=> \(x_1^2=2x_1-m\)
Khi đó
\(P=\frac{m^3-m^2+4m}{2\left(x_1+x_2\right)+m^2-m}+m^2+1\)
\(=\frac{m\left(m^2-m+4\right)}{m^2-m+4}+m^2+1=m^2+m+1=\left(m+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(MinP=\frac{3}{4}\)khi \(m=-\frac{1}{2}\)(thỏa mãn \(x\le1\))
\(\frac{2}{x^2+y^2}+\frac{2}{y^2+z^2}+\frac{2}{z^2+x^2}=3+\frac{z^2}{x^2+y^2}+\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}\le3+\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2zx}\)
\(=3+\frac{x^3+y^3+z^3}{2xyz}\)
\(\Rightarrow\)\(A\le3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\sqrt{\frac{2}{3}}\)
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
ghi lại đầu bài đi em