a) 4.25-12.5+170:10

=100-60+17

=40+17

=57

b) (7+3³:3²).4-3

=(7+3).4-3

=10.4-3

=40-3

=37

c) 12:{400:[500-(125+25.7)]}

=12:{400:[500-(125+175)]}

=12:{400:[500-300]}

=12:{400:200}

=12:2

=6

d) 168+{[2.(2⁴+3²)-256⁰]:7²}

=168+{[2.(16+9)-1]:49}

=168+{[2.25-1]:49}

=168+{[50-1]:49}

=168+{49:49}

=168+1

=169

\(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{3}-125\frac{3}{5}\)

\(=\left|-\frac{419}{15}\right|+\left(-\frac{419}{15}\right)\)

\(=\frac{419}{15}+\left(-\frac{419}{15}\right)=0\)

học tốt ~~

a) 5^x=125

=> 5^x=5^2

=>x=2

b) 3^2*x=81

=>3^2*x=9^2

=>3^2*x=(3^2)^2

=>x=2

câu c chưa hiểu chỗ 5^2x3 hay 5^2*3 vậy ?

a) \(5^x=125\)

\(5^x=5^3\)

\(\Rightarrow x=3\)

b)\(3^{2x}=81\)

\(3^{2x}=3^4\)

\(\Rightarrow2x=4\)

\(x=4:2\)

\(x=2\)

c)\(5^{2x-3}-2.5^2=5^2.3\)

\(5^{2x-3}=5^2.3+5^2.2\)

\(5^{2x-3}=5^2.\left(3+2\right)\)

\(5^{2x-3}=5^3\)

\(\Rightarrow2x-3=3\)

\(2x=3+3\)

\(2x=6\)

\(x=6:2\)

\(x=3\)

5^2x3 nha

a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)

b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)

c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)

a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) =  \(\dfrac{6}{7}\)  

b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)          

c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)

1. \(A=\left(2^{2017}\cdot3+2^{2017}\cdot5\right):2^{2018}\)

\(A=\left[2^{2017}.\left(3+5\right)\right]:\left(2^{2018}\right)\)

\(A=\left[2^{2017}.2^3\right]:\left(2^{2018}\right)\)

\(A=2^{2020}:2^{2018}=2^2=4\)

2. a) 2 + x : 5 = 6

=> x : 5 = 4

=> x = 20

b) 5x(7 + 48:x) = 45

=> x(7 + 48:x) = 9

=> 7x + 48 = 9

=> 7x = -39

=> x = -39/7.

c) Không hiểu đề câu này cho lắm.

3. \(25^{30}=\left(5^2\right)^{30}=5^{60};125^{19}=\left(5^3\right)^{19}=5^{57}\)

Vì 60 > 57 => \(25^{30}>125^{19}\)

4. \(S=1+7^1+...+7^{100}\)

\(\Rightarrow7S=7+7^2+...+7^{101}\)

\(\Rightarrow7S-S=7+7^2+...+7^{101}-1-7-...-7^{100}\)

\(\Rightarrow6S=7^{101}-1\)

\(\Rightarrow S=\frac{7^{101}-1}{6}\)

5. \(Q=1+2+2^2+...+2^{49}\)

\(\Rightarrow2Q=2+2^2+...+2^{50}\)

\(\Rightarrow2Q-Q=2+2^2+...+2^{50}-1-2-...-2^{49}\)

\(\Rightarrow Q=2^{50}-1\)

\(\Rightarrow2^{50}-1+1=2^n\)

\(\Rightarrow2^{50}=2^n\Rightarrow n=50\)

1/\(\left(x-6\right)^2=9\Rightarrow\orbr{\begin{cases}x-6=-3\\x-6=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=9\end{cases}}}\)
2/\(5^{x+1}=125\Leftrightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)
3/ đề thiếu
 

x = 9

x = 2

câu 3 mình ko hỉu

à xin lỗi mình nghi thiếu

5^2x-3-2x5²=5²x3

a: =46-16+35-5=30+30=60

b: =32-12+34-14+36-16+38-18-10

=20+20+20+20-10

=80-10=70

c: \(=125-125-170+120=-50\)

d: =(-1)+(-1)+...+(-1)

=-50

#\(N\)

`a, 4573 + 46 - 4573 + 35 - 16 - 5`

`= (4573-4573) + (46 - 16)+(35-5)`

`= 0 +30+30 = 60`

`b, 32+34+36+38-10-12-14-16-18`

`= (32-12)+(34-14)+(38-18)+10`

`= 20+20+20+10 = 70`

`c, 125-170+120+(-125)`

`= (125 + -125)-170+120`

`= 0-170+120`

`=-170 + 120 = -50`

`d, 1-2+3-4+...-98+99-100`

Các số hạng có trong biểu thức: \(\left(100-1\right)\div1+1=100\) `(` số hạng `)`

`=> (1-2)+(3-4)+...+(97-98)+(99-100)`

Các cặp mà trong bthuc có là: \(100\div2=50\) 

`=> (-1)+(-1)+...(-1)+(-1) = (-50)`

`e,`

*Mình xp sửa đề phải là `1-5 + 7-11+...+997-1001` nhỉ? Vì để như vậy nó lẻ á ._.

`1-5+7-11+...+997-1001`

Các số hạng có trong biểu thức là: \(\left(1001-1\right)\div4+1=251\) `(` số `)`

`-> (1-5)+(7-11)+...+(997-1001)`

Các cặp được ghép ở trong bthuc là: \(251\div2=125,5\) 

`-> (-4)+(-4)+...+(-4) = (-4)*125,5 = -502`

\(\frac{3}{2\times5}+\frac{2}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{602\times605}\)

\(=\frac{5-2}{2\times5}+\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+...+\frac{605-602}{602\times605}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\frac{1}{2}-\frac{1}{605}=\frac{603}{1210}\)

\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{2}{13\times15}\)

\(=\frac{7-4}{3\times7}+\frac{12-7}{7\times12}+\frac{13-12}{12\times13}+\frac{15-13}{13\times15}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)