-13/6 nha bạn

TL

-13/6

HT nha bn

\(\left(3^2\right)^2+2^x=5\left(5+2^2.3\right)\)

\(\Rightarrow81+2^x=5.17\)

\(\Rightarrow81+2^x=85\)

\(\Rightarrow2^x=85-81=4=2^2\Rightarrow x=2\)

\(5^{x-1}+5=150\Rightarrow5^{x-1}=150-5=145\)

Mà không có lũy thừa nào có cơ số là 5 mà kết quả bằng 145 nên x không thỏa mãn điều kiện trên.

\(\left(3-x\right)^6=\left(3-x\right)^4\)

\(\Rightarrow3-x=\left\{-1;0;1\right\}\)

\(\Rightarrow x=\left\{4;3;2\right\}\)

9^2+2^x=5(5+4.3)

81+2^x=5(5+12)

81+2^x=5.17

81+2^x=85

2^x=85-81

2^x=4

vi 4= 2^2 . x=2

mk nhác tính lắm nên ns sơ qua thui nha

bạn bấm máy tính cầm tay các số mũ (lũy thừa) bấm giống như trên

sau đó bạn giải biểu thức như bình thường thôi

ok

\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)

=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)

=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)

=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)

=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)

=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)

=\(\frac{29}{25600}\)

Đoạn \(\frac{2^6\cdot\left(27+2\right)}{2^6\cdot5^2\cdot2^{10}}\)là sai rồi bn ơi!!!

Bn phải lm như mục trên là:

\(\frac{2^6\left(2+3^3\right)}{2^7\left(5^2+2^2\right)}\)\(=\frac{2^6\cdot29}{2^7\cdot29}=\frac{1}{2}\)

Nhưng dù sao cx c.ơn bn vì đã giúp mk,mk sẽ cho bn 1 ths nka!!!

\(14:\left(4\frac{2}{3}-1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)

\(=14:\left(\frac{14}{3}-\frac{14}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)

\(=14:\left(\frac{14}{3}-\frac{14}{9}+\frac{2}{3}+\frac{8}{9}\right)\)

\(=14:\left(\frac{16}{3}-\frac{6}{9}\right)\)

\(=14:\left(\frac{16}{3}-\frac{2}{3}\right)\)

\(=14:\frac{14}{3}\)

\(=14.\frac{3}{14}=3\)

1. Tính hợp lí

a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)

\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)

\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)

\(=1+\dfrac{-7}{19}\)

\(=\dfrac{12}{19}\)

b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)

\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)

\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)

\(=2.\dfrac{-5}{2}\)

\(=-5\)

c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)

\(=\dfrac{3}{5}.-1\)

\(=\dfrac{-3}{5}\)

d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)

\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)

\(=\dfrac{7}{4}.1\)

\(=\dfrac{7}{4}\)

Chúc bạn học tốt

a) $0,7+\genfrac{}{}{0ex}{}{-7}{19}-\left(-0,3\right)$

$=\genfrac{}{}{0ex}{}{7}{10}+\genfrac{}{}{0ex}{}{-7}{19}+\genfrac{}{}{0ex}{}{3}{10}$

$=\left(\genfrac{}{}{0ex}{}{7}{10}+\genfrac{}{}{0ex}{}{3}{10}\right)+\genfrac{}{}{0ex}{}{-7}{19}$

$=1+\genfrac{}{}{0ex}{}{-7}{19}$

$=\genfrac{}{}{0ex}{}{12}{19}$

b) $\genfrac{}{}{0ex}{}{5}{3}.\left(-2,5\right):\genfrac{}{}{0ex}{}{5}{6}$

$=\genfrac{}{}{0ex}{}{5}{3}.\genfrac{}{}{0ex}{}{-5}{2}.\genfrac{}{}{0ex}{}{6}{5}$

$=\left(\genfrac{}{}{0ex}{}{5}{3}.\genfrac{}{}{0ex}{}{6}{5}\right).\genfrac{}{}{0ex}{}{-5}{2}$

$=2.\genfrac{}{}{0ex}{}{-5}{2}$

$=-5$

c) $0,6.\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{12}{17}$

$=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{12}{17}$

$=\genfrac{}{}{0ex}{}{3}{5}.\left(\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{12}{17}\right)$

$=\genfrac{}{}{0ex}{}{3}{5}.-1$

$=\genfrac{}{}{0ex}{}{-3}{5}$

d) $\genfrac{}{}{0ex}{}{7}{4}.\genfrac{}{}{0ex}{}{5}{2}-\genfrac{}{}{0ex}{}{7}{4}.\genfrac{}{}{0ex}{}{3}{2}$

$=\genfrac{}{}{0ex}{}{7}{4}.\left(\genfrac{}{}{0ex}{}{5}{2}-\genfrac{}{}{0ex}{}{3}{2}\right)$

$=\genfrac{}{}{0ex}{}{7}{4}.1$

$=\genfrac{}{}{0ex}{}{7}{4}$

mình giúp rùi đó nhớ tick mình nha

Xét 3 số tự nhiên liên tiếp \(2005^n,2005^n+1,2005^n+2\) luôn có ít nhất 1 số chia hết cho 3

Mà:\(2005\equiv1\)(mod 3)

 \(\Rightarrow2005^n\equiv1^n=1\)(mod 3)

\(\Rightarrow2005^n\) không chia hết cho 3

Nên trong 2 số  \(2005^n+1,2005^n+2\) luôn có 1 số chia hết cho 3

\(\Rightarrow\left(2005^n+1\right)\left(2005^n+2\right)⋮3\)

Xét \(n=2k\left(k\in N\right)\)Ta có :

\(\left(2005^n+1\right)\left(2005^n+2\right)=\left(2005^{2k}+1\right)\left(2005^{2k}+2\right)\)

\(=\left(2005^{2k}+1\right)\left(2005^{2k}-1+3\right)\)

Vì \(2005^{2k}-1⋮2004⋮3\) do đó \(\left(2005^n+1\right)\left(2005^n+2\right)⋮3\)

Xét \(n=2k+1\) thì \(2005^n+1=2005^{2k+1}+1⋮2007⋮3\)

Ta có ngay ĐPCM