\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)

\(=\frac{2.6}{3.5}\)

\(=\frac{4}{5}\)

Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}\)

\(=\frac{2\cdot6}{3\cdot5}=\frac{4}{5}\)

giúp mk vs mk cần rất gấp

a) Ta có:

\(\frac{x}{2}=\frac{y}{5}\Leftrightarrow\frac{x}{6}=\frac{y}{15}\)

\(\frac{y}{3}=\frac{z}{2}\Leftrightarrow\frac{y}{15}=\frac{z}{10}\)

=> \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=\frac{x+y+z}{6+15+10}=-\frac{12}{31}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{72}{31}\\y=-\frac{180}{31}\\z=-\frac{120}{31}\end{cases}}\)

b) \(2x=3y=5z\Leftrightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x-y+z}{15-10+6}=\frac{33}{11}=3\)

=> \(\hept{\begin{cases}x=45\\y=30\\z=18\end{cases}}\)

mn trả lời cho mình ạ mình cần gấppp

a) \(\left|2,3-x\right|=1,2\)

\(\Leftrightarrow\orbr{\begin{cases}2,3-x=1,2\\2,3-x=-1,2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1,1\\x=3,5\end{cases}}\)

Ta có: \(7\left(x-1\right)+2x\left(1-x\right)=0\)

\(\Leftrightarrow7\left(x-1\right)-2x\left(x-1\right)=0\)

\(\Leftrightarrow\left(7-2x\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7-2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}\)

a) Vì \(A=2-\left|x+\frac{5}{6}\right|\le2-0=2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x+\frac{5}{6}\right|=0\Rightarrow x=-\frac{5}{6}\)

Vậy Max(A) = 2 khi \(x=-\frac{5}{6}\)

b) Vì \(B=5-\left|\frac{2}{3}-x\right|\le5-0=5\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|\frac{2}{3}-x\right|=0\Rightarrow x=\frac{2}{3}\)

Vậy Max(B) = 5 khi \(x=\frac{2}{3}\)

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

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

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

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

| x + \(\frac{4}{5}\)| - | - 3,75 | = - | - 2,15 |

=> | x + \(\frac{4}{5}\)| - 3,75 = - 2,15

=> | x + \(\frac{4}{5}\)| = 1,6

=> \(\orbr{\begin{cases}x+\frac{4}{5}=1,6\\x+\frac{4}{5}=-1,6\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=-\frac{12}{5}\end{cases}}\)

Ta có \(\frac{5a+9c}{5b+9d}=\frac{3a+8c}{3b+8d}\)\(\left(a;b;c;d\ge0\right)\)

Thỏa mãn điều kiện \(a=b=c=d\)

\(\frac{5a+9c}{5b+9d}=\frac{3a+8c}{3a+8d}\)

\(\Rightarrow\frac{x\left(5+9\right)}{x\left(5+9\right)}=\frac{x\left(3+8\right)}{x\left(3+8\right)}\Leftrightarrow1=1\)

Vậy \(\frac{5a+9c}{5b+9d}=\frac{3a+8c}{3b+8d}\).

Thiếu đk: \(\frac{a}{b}=\frac{c}{d}\)

Ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{9c}{9d}=\frac{5a+9c}{5b+9d}\)

mà \(\frac{a}{b}=\frac{c}{d}=\frac{3a}{3b}=\frac{8c}{8d}=\frac{3a+8c}{3b+8d}\)

\(\Rightarrow\frac{5a+9c}{5b+9d}=\frac{3a+8c}{3b+8d}\)

        Bài làm :

Ta có ;

• 40=2³.5
• 48=2⁴.3
• 42=2.3.7

Vậy BCNN(40;48;42)=2⁴.3.5.7=1680

\(40=2^3.5\) .

\(48=2^4.3\)

\(42=2.3.7\)

BCNN (40,48,42) = 2⁴.3.5.7=1680

#Satan_Dilys

#9:23_27/9/2020