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Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
nhìn căng nhể :))
a) ( x - 1 )( x - 3 )( x + 5 )( x + 7 ) - 297 = 0
<=> [ ( x - 1 )( x + 5 ) ][ ( x - 3 )( x + 7 ) ] - 297 = 0
<=> ( x2 + 4x - 5 )( x2 + 4x - 21 ) - 297 = 0
Đặt t = x2 + 4x - 5
pt <=> t( t - 16 ) - 297 = 0
<=> t2 - 16t - 297 = 0
<=> t2 - 27t + 11t - 297 = 0
<=> t( t - 27 ) + 11( t - 27 ) = 0
<=> ( t - 27 )( t + 11 ) = 0
<=> ( x2 + 4x - 5 - 27 )( x2 + 4x - 5 + 11 ) = 0
<=> ( x2 + 4x - 32 )( x2 + 4x + 6 ) = 0
<=> ( x2 - 4x + 8x - 32 )( x2 + 4x + 6 ) = 0
<=> [ x( x - 4 ) + 8( x - 4 ) ]( x2 + 4x + 6 ) = 0
<=> ( x - 4 )( x + 8 )( x2 + 4x + 6 ) = 0
Đến đây dễ rồi :)
Ta có công thức :
\(1-\frac{1}{k^2}=\frac{k^2-1^2}{k^2}=\frac{\left(k+1\right)\left(k-1\right)}{k^2}\)
Áp dụng công thức trên ta được :
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\frac{2^2-1^2}{2^2}.\frac{3^2-1^2}{3^2}.\frac{4^2-1^2}{4^2}....\frac{n^2-1^2}{n^2}\)
\(=\frac{\left(2+1\right)\left(2-1\right)}{2.2}.\frac{\left(3+1\right)\left(3-1\right)}{3.3}.\frac{\left(4+1\right)\left(4-1\right)}{4.4}...\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\frac{\left[1.2.3.....\left(n+1\right)\right].\left[3.4.5...\left(n-1\right)\right]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\frac{1}{2n}=\frac{n+1}{2n}\)
mới lớp 6 mà giải đc toan lớp 8 , anh đây thuông minh quá ))
\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
\(A=\frac{1}{3589}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{3589}\right).2.\frac{1}{297}-7.\frac{1}{3589}-3.\frac{1}{3689}.\frac{1}{297}\)
\(A=7.\frac{1}{3689}+\frac{1}{3589}.\frac{1}{297}-8.\frac{1}{297}+2.\frac{1}{3589}.\frac{1}{297}-7.\frac{1}{3589}\)
\(A=-8.\frac{1}{297}\)
\(A=\frac{-8}{297}\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-4.2.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{358}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}+2.\dfrac{1}{297}.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-4.2.\dfrac{1}{297}\)
\(A=0+0+\dfrac{-8}{297}\)
\(A=\dfrac{-8}{297}\)
Chúc bạn học tốt!!!
A= \(\dfrac{1}{358}\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
A= \(\dfrac{7}{358}+\dfrac{1}{358.297}-\dfrac{8}{297}+\dfrac{2}{358.297}-\dfrac{7}{358}-\dfrac{3}{358.297}\)
A= \(-\dfrac{8}{297}\)