Tính giá trị của biểu thức :
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-3.\frac{1}{297}.\frac{1}{358}\)
Tính giá trị của biểu thức :
A=\(\frac{1}{358}\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right)-7.\frac{1}{358}-3.\frac{1}{297}.\frac{1}{358}\)
Tính giá trị của biểu thức :
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-3.\frac{1}{297}.\frac{1}{358}\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-4.2.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{358}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}+2.\dfrac{1}{297}.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-4.2.\dfrac{1}{297}\)
\(A=0+0+\dfrac{-8}{297}\)
\(A=\dfrac{-8}{297}\)
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= \(-\dfrac{8}{297}\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{359}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-2.4.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{359}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}-2.\dfrac{1}{297}.\dfrac{1}{359}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-8.\dfrac{1}{297}\)
\(A=-4.\dfrac{1}{297}.\dfrac{1}{359}-8.\dfrac{1}{297}\)
\(A=-4.\dfrac{1}{297}\left(\dfrac{1}{359}+2\right)\)
\(A=\dfrac{-4}{297}.\dfrac{719}{359}=\dfrac{-2876}{106623}\)
Chúc bạn học tốt!!!
Tính giá trị của biểu thức \(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
Tính giá trị biểu thức D biết :\(D=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
Cho biểu thức: \(A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\)
Hãy tính giá trị của A theo hai cách:
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
Bài 1: Rút gọn biểu thức
a) \(5^{n+1}-2.5^5\)
b) \(2x^{n-1}.\left(x^{n+1}-y^{n+1}\right)+y^{n+1}.\left(2x^{n-1}-y^{n-1}\right)\)
Bài 2: Chứng minh rằng
a) \(8^5+16^4\)chia hết cho 3
b) \(2^8+2^9+2^{10}\)chia hết cho 7
Bài 3: Tính giá trị biểu thức:
\(x^8-2005.x^7+2005.x^6-2005.x^5....-2005.x+2005vớix=2004\)
Bài 4: Tính giá trị biểu thức
\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
Đăng từng bài một rồi tui làm cho~
Nhìn như này hoa mắt lắm :(
Bài 4:
Đặt\(a=\frac{1}{3589}\)và\(b=\frac{1}{297}\)
\(A=a\left(7+b\right)-\left(4-a\right)2b-7a-3ab\)
\(A=7a+ab-8b+2ab-7a-3ab\)
\(A=-8b=\frac{-8}{297}\)
Tính giá trị biểu thức
\(A=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)...\left(1-1\frac{2}{7}\right)\left(1-1\frac{3}{7}\right)\)
\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)
\(=0\)
Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0
Tính giá trị của các biểu thức sau:
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\end{array}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
Giải phương trình:
a) (x - 1)(x - 3)(x + 5)(x + 7) - 297 = 0
b) \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2=\left(x+4\right)^2+4\left(x+\frac{1}{x}\right)^2\left(x^2+\frac{1}{x^2}\right)\)
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 :)