địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(\dfrac{3}{5}\)+4-\(\dfrac{6}{7}\)=\(\dfrac{3}{5}\)+\(\dfrac{20}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{23}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{161}{35}\)-\(\dfrac{30}{35}\)=\(\dfrac{131}{35}\)

m.ng oiii giải giúp mình với ạ mình đang cần gấp!!!

a) Thay \(x=-1\) và \(y=\dfrac{1}{4}\) vào, ta được:

       \(2\cdot\left(-1\right)^2\cdot\dfrac{1}{4}\)

    = \(\dfrac{1}{2}\)

b) Thay \(x=-\dfrac{1}{2}\) và \(y=-4\) vào, ta được:

        \(-\dfrac{1}{2}\cdot\left(-\dfrac{1}{2}\right)^3\cdot\left(-4\right)^2\)

     =  \(\left(-\dfrac{1}{2}\right)^4\cdot16\)

     =  1

[545-(45+100)]:50-8+2^2

=(545-145):50-4

= 400:50-4

=8-4=4

\(\text{Tính giá trị của biểu thức :}\)

\(a,\text{ }\left[545-\left(45+4\cdot25\right)\right]\text{ : }50-2000\text{ : }250+2^{15}\text{ : }2^{13}\)

\(=\left[545-145\right]\text{ : }50-8+2^2\)

\(=400\text{ : }50-8+4\)

\(=8-8+4\)

\(=4\)

\(\left(\frac{2}{5}\right)^2+5\frac{1}{2}:\left(4,5-2\right)-0,2\)

\(=\frac{4}{25}+\frac{11}{2}:\frac{5}{2}-\frac{1}{5}\)

\(=\frac{4}{25}+\frac{11}{2}.\frac{2}{5}-\frac{1}{5}\)

\(=\frac{4}{25}+\frac{11}{5}-\frac{1}{5}\)

\(=\frac{4}{25}+\frac{55}{25}-\frac{5}{25}\)

\(=\frac{54}{25}\)

a) Đề sai

b) \(\left|x+\frac{4}{5}\right|=\frac{1}{7}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{5}=\frac{1}{7}\\x+\frac{4}{5}=\frac{-1}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{7}-\frac{4}{5}\\x=\frac{-1}{7}-\frac{4}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{35}-\frac{28}{35}\\x=\frac{-5}{35}-\frac{28}{35}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-23}{35}\\x=\frac{-33}{35}\end{cases}}}\)

Vậy \(x=\frac{-23}{35}\)hoặc \(x=\frac{-33}{35}\)

( Hỗn số có thể viết thành a + b/c ( đối với hỗn số âm thì không). VD: \(1\frac{1}{2}=1+\frac{1}{2}\)) mà chỗ và là chỉ hỗn số phải chứ, xem đầu bài mk viết như vậy có đúng không nha bn, nếu ko thì nhắn tin cho mk, để mk sửa nhé!

\(B=1+[\frac{13}{15.\left(0,5\right)^2.3}+\left(\frac{8}{15-1}+\frac{19}{60}\right):1]+\frac{23}{24}\)

\(B=1+\left[\frac{13}{11,25}+\left(\frac{4}{7}+\frac{19}{60}\right)\right]+\frac{23}{24}\)

\(B=1+\left[\frac{52}{45}+\frac{373}{420}\right]+\frac{23}{24}\)

\(B=1+2+\frac{11}{252}+\frac{23}{24}\)

\(B=3+1+\frac{1}{504}\)

\(B=4+\frac{1}{504}\)

\(B=4\frac{1}{504}\)

Bài 2

\(a,\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{4;5;6\right\}\)

\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)

\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)

Mà \(\frac{15}{2}\notin n\)

\(\Rightarrow x\in\left\{7;8\right\}\)

#)Giải :

Bài 1 :

a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)

b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)

Bài 2 : 

a) \(\left(x-5\right)^2=\left(x-5\right)^6\)

\(\Leftrightarrow x^4-625=x^6-15625\)

\(\Leftrightarrow x^6-x^4=15000\)

\(\Leftrightarrow x^6-x^4=5^6-5^4\)

\(\Leftrightarrow x=5\)

b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow2x-15=1\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Bài 1 :

\(A=\frac{2^{13}+2^5}{2^{10}+2^2}\)\(=\frac{2^{10}.2^3+2^2.2^3}{2^{10}+2^2}\) \(=\frac{2^3.\left(2^{10}+2^2\right)}{2^{10}+2^2}=2^3=8\)

Vậy A = 8

A=\(1+2^1+2^2+2^3+2^4+...+2^{2015}\)

2A=\(2^1+2^2+2^3+2^4+...+2^{2015}+2^{2016}\)

2A-A=\(2^{2016}-1\)

Vậy A=\(2^{2016}-1\)

\(A=1+2^1+2^2+2^3+....+2^{2015}\)

\(2A=2\left(1+2+2^2+.....+2^{2015}\right)\)

\(2A=2+2^2+2^3+....+2^{2016}\)

\(2A-A=\left(2+2^2+.....+2^{2016}\right)-1-2-2^2-...-2^{2015}\)

\(A=2^{2016}-1\)

Vậy\(A=---\)

Cách làm của mình gần giống bạn Đoàn Khánh Linh nhưng mình sẽ viết cho ngắn gọn:

Ta có:

\(A=1+2^1+2^2+2^3+........+2^{2015}\)

\(2A=2+2^2+2^3+2^4+.........+2^{2016}\)

\(2A-A=\left(2+2^2+2^3+2^4+.......+2^{2016}\right)-\left(1+2^1+2^2+2^3+.........+2^{2015}\right)\)

\(A=2^{2016}-1\)

Vậy A = \(2^{2016}-1\)

\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)

\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)

\(=0\)

Ta có :                  

                ^{82 - 576 : 32}

^{= 64 - 576 : 9}

^{= 64 - 64}

^{=  0}

 (1¹ + 2² + 3³ + 4⁴ +...+ 2022²⁰²²) . 0

^{= 0}