f(x)+q(x)=3x\(^2\)+4x\(^6\)-3xyz\(^5\)+9x\(^6\)-5xyz\(^7\)+8x\(^2\)

=-5xyz+13x\(^6\)-3xyz+11x\(^2\)

Áp dụng dãy tỉ số bằng nhau 

\(\frac{12x-15y+20z-12x+15y-20z}{27}=0\) 

rút gọn mới được bằng 0 nha

\(\Rightarrow\frac{12x-15y}{7}=0\Leftrightarrow12x-15y=0\Leftrightarrow12x=15y\Rightarrow\frac{x}{12}=\frac{y}{15}\left(1\right)\)

\(\Rightarrow\frac{15y-20z}{11}=0\Leftrightarrow15y=20z\Rightarrow\frac{y}{15}=\frac{z}{20}\left(2\right)\)

Từ 1 và 2 ta được 

\(\frac{x}{12}=\frac{y}{15}=\frac{z}{20}\)

Rồi đến đây tự làm nha mỏi tay lắm x+y+z=48 dưới mấu số cũng rứa rồi ngủ đây 

a, \(7^6+7^5-7^4=7^4\left[7^2+4-1\right]=7^4\cdot55⋮55\)

b, \(A=1+5+5^2+5^3+...+5^{50}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{51}\)

\(\Rightarrow5A-A=\left[5+5^2+5^3+5^4+...+5^{51}\right]-\left[1+5+5^2+5^3+...+5^{50}\right]\)

\(\Rightarrow4A=5^{51}-1\Leftrightarrow A=\frac{5^{51}-1}{4}\)

bn lam dê ma

Đặng Giang Nam xạo , mới có tiểu học làm dc toán lớp 7 ??????

1 \(-\)\(\frac{1}{3.5}\)\(-\)\(\frac{1}{5.7}\)\(-\)\(\frac{1}{7.9}\)\(-\)..... \(-\)\(\frac{1}{53.55}\)\(-\)\(\frac{1}{55.57}\)

= 1 \(-\)( \(\frac{1}{3.5}\)  + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + ..... + \(\frac{1}{53.55}\)  + \(\frac{1}{55.57}\)  )

= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)+ \(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)+ \(\frac{1}{7}\)\(-\)\(\frac{1}{9}\)+....+ \(\frac{1}{53}\)\(-\)\(\frac{1}{55}\)+ \(\frac{1}{55}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)

= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)

= 1 \(-\) \(\frac{6}{19}\). \(\frac{1}{2}\)= 1 \(-\)\(\frac{3}{19}\)= \(\frac{16}{19}\)

\(1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)

đặt \(A=1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)

\(A=1-\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

đặt \(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{53.55}+\frac{1}{55.57}\)

\(2B=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

\(2B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{53.55}+\frac{2}{55.57}\)

\(2B=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{55-53}{53.55}+\frac{57-55}{55.57}\)

\(2B=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+\frac{9}{7.9}-\frac{7}{7.9}+...+\frac{55}{53.55}-\frac{53}{53.55}+\frac{57}{55.57}-\frac{55}{55.57}\)

\(2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(2B=\frac{1}{3}-\frac{1}{57}\)

\(2B=\frac{54}{171}\)

\(\Rightarrow B=\frac{54}{171}:2\)

\(\Rightarrow B=\frac{9}{57}\)

mà \(A=1-B\)

\(\Rightarrow A=1-\frac{9}{57}\)

\(\Rightarrow A=\frac{48}{57}\)

chúc bạn học giỏi ^^

\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{7}\right]\)

\(Q=\left(\frac{2}{25}-\frac{126}{125}\right):\frac{4}{7}:\frac{-68}{9}\)

\(Q=\frac{-116}{125}:\frac{4}{7}:\frac{-68}{9}\)

\(Q=\frac{-203}{125}:\frac{-68}{9}\)

\(Q=\frac{1827}{8500}\)

Ta có:

\(\frac{x}{2}=\frac{y}{3}\) và \(xy=6\)

\(\Rightarrow x=2k;y=3k\) và \(xy=6\)

\(\Rightarrow2k.3k=6\Leftrightarrow6k^2=6\Leftrightarrow k=\hept{\begin{cases}1\\-1\end{cases}}\)

\(\Rightarrow x=2.1=2;x=2.-1=-2\)

\(\Rightarrow y=3.1=3;y=3.-1=-3\)

Vậy \(x\in\left\{2;-2\right\};y\in\left\{3;-3\right\}\)

A=|x−7| +6-x

ta có \(\left|x-7\right|\).\(\ge\)0 (\(\forall\)x)

=> A đạt giá trị nhỏ nhất khi \(\left|x-7\right|\)= 0

<=> x-7 = 0

<=> x = 7

thay x = 7 vào A ta được A = \(\left|7-7\right|\)+6-7

A = -1

_ Yuki _ Dễ thương _

mk bt rồi (-1)

kết quả bằng 0

\(-\frac{3}{10}-0,125+\frac{-7}{10}+1,125\)

\(=\left(-\frac{3}{10}-\frac{7}{10}\right)+1,125-0,125\)

\(=-\frac{10}{10}+1=-1+1=0\)