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Lời giải:
$8^{2x}=27.3^x$
$\Rightarrow (2^3)^{2x}=3^3.3^x$
$\Rightarrow 2^{6x}=3^{x+3}$
Dạng toán này không phù hợp với lớp 6 bạn nhé. Bạn xem lại khi đã viết đúng đề chưa vậy?
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\(\Rightarrow8C=8^2+8^3+8^4+...+8^{100}\\ \Rightarrow8C-C=8^2+8^3+...+8^{100}-8-8^2-...-8^{99}\\ \Rightarrow7C=8^{100}-8\\ \Rightarrow7C+8=8^{100}=8^{2x}\\ \Rightarrow2x=100\Rightarrow x=50\)
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Tính 25 . ( -3) = - 75
Suy ra kết quả:
(+25) . ( +3) = 75;
(-25) . ( -3) = 75;
(-25) . ( +3) = -75;
(+3) . ( -25) = -75
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính 25 . ( -3) = - 75
Suy ra kết quả:
(+25) . ( +3) = 75;
(-25) . ( -3) = 75;
(-25) . ( +3) = -75;
(+3) . ( -25) = -75
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(A=\dfrac{19}{25}\left(\dfrac{3}{7}+\dfrac{25}{19}\right)-\dfrac{3}{7}\left(\dfrac{19}{25}-\dfrac{7}{3}\right)\\ A=\dfrac{19}{25}.\dfrac{3}{7}+\dfrac{19}{25}.\dfrac{25}{19}-\dfrac{3}{7}.\dfrac{19}{25}+\dfrac{3}{7}.\dfrac{7}{3}\\ A=\dfrac{19}{25}.\dfrac{3}{7}-\dfrac{19}{25}.\dfrac{3}{7}+\dfrac{25}{9}.\dfrac{19}{25}+\dfrac{3}{7}.\dfrac{7}{3}\\ A=0+1+1\\ A=2\)
Vậy A= 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{19}{25}.\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}.\left(\frac{19}{25}-\frac{7}{3}\right)\)
=> \(A=\frac{19}{25}.\frac{232}{133}-\frac{3}{7}.\frac{232}{75}\)
=> \(A=\frac{232}{175}-\frac{232}{175}\)
=> \(A=0\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
19/25 . 3/7 . 25/19 - 3/7 . 19/25 . 7/3
( 19/25 . 25/19 ) . 3/7 - ( 3/7 . 7/3 ) . 19/25
1 . 3/7 - 1 . /19/25
3/7 - 19/25
-58/175
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A = - 2 + (-1)\\A = - 3\end{array}\)
b,
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)
8^2x = 3 = 2^5