Bài làm:

Đặt \(\frac{x}{7}=\frac{y}{5}=\frac{z}{3}=k\)

=> \(\hept{\begin{cases}x=7k\\y=5k\\z=3k\end{cases}}\)

Mà \(yz=135\Leftrightarrow15k^2=135\Leftrightarrow k^2=9\Rightarrow k=\pm3\)

=> \(\hept{\begin{cases}x=\pm21\\y=\pm15\\z=\pm9\end{cases}}\)

Đặt \(\frac{x}{7}=\frac{y}{5}=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=5k\\z=3k\end{cases}}\)

Khi đó yz = 135

<=> 5k.3k = 135

=> 15.k² = 135

=> k² = 9

=> k = \(\pm\)3

Nếu k = 3 => x = 21 ; y = 15 ; z = 9

Nếu k = -3 => x = -21 ; y = -15 ; z = -9

Vậy các cặp (x;y;z) thỏa mãn bài toán là (21 ; 15 ; 9) ; (-21 ; - 15 ; -9)

|x + 4/15| - |-3,75| = -|- 2,15|

=> |x + 4/15| - 3,75 = 2,15

=> |x + 4/15| = 59/10

=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{59}{10}\\x+\frac{4}{15}=-\frac{59}{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{169}{30}\\x=\frac{-185}{30}\end{cases}}\)

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\frac{4}{15}\right|=1,6\)

\(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)

\(\Rightarrow\hept{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{8}{5}-\frac{4}{15}=\frac{4}{3}\\x=\frac{-8}{5}-\frac{4}{15}=\frac{-28}{15}\end{cases}}}\)

\(-2,5+\left|3x+5\right|=-1,5\)

\(\Leftrightarrow\left|3x+5\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)

vậy x=-4/3 hoặc x=-2

\(-2,5+\left|3x+5\right|=-1,5\)

                  \(\left|3x+5\right|=-1,5+2,5\)

                  \(\left|3x+5\right|=1\)

Ta có : \(\left|3x+5\right|\ge0\forall x\)

\(\Rightarrow\left|3x+5\right|=3x+5\)

\(\Rightarrow3x+5=1\)

               \(3x=1-5\)

               \(3x=-4\)

                  \(x=-4\div3\)

                  \(x=\frac{-4}{3}\)

\(\left|x-3,5\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-3,5=5\\x-3,5=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8,5\\x=-2,5\end{cases}}}\)

|x-3,5|=5

x-3,5=5  hoặc    x-3,5=-5

x=5+3,5  hoặc  x=-5+3,5

x=8,5   hoặc    x=-1,5

vậy x=8,5  hoặc      x=-1,5

các bạn vào trang cá nhân của mik đi, có cái này hay lắm!!!

bạn Giang Trà ko được trả lời linh tinh vớ vẩn như thế

a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)

b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)

c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)

\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)

\(\left|\frac{-2}{3}+\frac{3}{2}\right|.\left(\frac{-4}{5}-\left|-\frac{3}{-2}-1\right|\right)\)

\(=\left|\frac{5}{6}\right|.\left(\frac{-4}{5}-\left|\frac{1}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\frac{1}{2}\right)\)

\(=\frac{5}{6}.\frac{-13}{10}\)

\(=\frac{-13}{12}\)

\(\left|\frac{-2}{3}+\frac{3}{2}\right|.\left(\frac{-4}{5}-\left|-\frac{-3}{2}-1\right|\right)\)

\(=\left|\frac{-4}{6}+\frac{9}{6}\right|.\left(\frac{-4}{5}-\left|\frac{3}{2}-1\right|\right)\)

\(=\left|\frac{5}{6}\right|.\left(\frac{-4}{5}-\left|\frac{3}{2}-\frac{2}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\left|\frac{1}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\frac{1}{2}\right)\)

\(=\frac{5}{6}.\left(\frac{-8}{10}-\frac{5}{10}\right)\)

\(=\frac{5}{6}.\frac{-13}{10}\)

\(=\frac{5.\left(-13\right)}{6.10}=\frac{-13}{6.2}=\frac{-13}{12}=-1\frac{1}{12}\)

Chúc bạn học tốt

a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)

b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)

c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)

Các bạn vào trang cá nhân của mik đi, có cái này hay lắm!!!

\(\hept{\begin{cases}4x=5y\\3x-2y=35\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{5}}\\3x-2y=35\end{cases}}\Rightarrow\hept{\begin{cases}\frac{3x}{\frac{3}{4}}=\frac{2y}{\frac{2}{5}}\\3x-2y=35\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{3x}{\frac{3}{4}}=\frac{2y}{\frac{2}{5}}=\frac{3x-2y}{\frac{3}{4}-\frac{2}{5}}=\frac{35}{\frac{7}{20}}=100\)

\(\frac{3x}{\frac{3}{4}}=100\Rightarrow3x=75\Rightarrow x=25\)

\(\frac{2y}{\frac{2}{5}}=100\Rightarrow2y=40\Rightarrow y=20\)

Vậy x = 25 , y = 20

Ta có : 4x = 5y

\(\Rightarrow\frac{x}{5}=\frac{y}{4}\)

\(\Rightarrow\frac{3x}{15}=\frac{2y}{8}\)

và \(3x-2y=35\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{3x}{15}=\frac{2y}{8}=\frac{3x-2y}{15-8}=\frac{35}{7}=5\)

Khi đó : \(\frac{x}{5}=5\Rightarrow x=25\)

              \(\frac{y}{4}=5\Rightarrow y=20\)

Vậy \(x=25;y=20\)

mấy cái dấu kẻ dọc ngắn là giá trị tuyệt đối

\(\left(\frac{-2}{5}+\frac{3}{10}\right)\div\left(\frac{2}{-5}+\left|-\frac{4}{3}-2\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\left|\frac{-10}{3}\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\frac{10}{3}\right)\)

\(=\frac{-1}{10}\div\frac{44}{15}\)

\(=\frac{-3}{88}\)