\(6x^4+7x^3-37x^2-8x+12\\ =6x^4-3x^3+10x^3-5x^2-32x^2+16x-24x+12\\ =3x^3\left(2x-1\right)+5x^2\left(2x-1\right)-16x\left(2x-1\right)-12\left(2x-1\right)\\ =\left(2x-1\right)\left(3x^3+5x^2-16x-12\right)\\ =\left(2x-1\right)\left(3x^3-6x^2+11x^2-22x+6x-12\right)\\ =\left(2x-1\right)\left[3x^2\left(x-2\right)+11x\left(x-2\right)+6\left(x-2\right)\right]\\ =\left(2x-1\right)\left(x-2\right)\left(3x^2+11x+6\right)\\ =\left(2x-1\right)\left(x-2\right)\left(3x^2+9x+2x+6\right)\\ =\left(2x-1\right)\left(x-2\right)\left[3x\left(x+3\right)+2\left(x+3\right)\right]\\ =\left(2x-1\right)\left(x-2\right)\left(x+3\right)\left(3x+2\right)\)

Chón ý B 5/8. Mình trả lời nhanh nhất nè bạn. Tick cho mình đi!

Bạn xem lại các đáp án nhé, mình tính không ra bạn ạ!

 a : b = 9 (dư 8)

a = b x 9 + 8

a - b = 88

b x 9 + 8 - b = 88

b x 8 = 80

b = 80 : 8

b = 10

⇒ a = b x 9 + 8 = 10 x 9 + 8 = 98

Vậy số bị chia là: 98

       số chia là: 10

Gọi số bị chia và số chia lần lượt là a và b 

Ta có: 

\(a=b\times9+8\) 

\(a-b-8=b\times8\)

Mà hiệu số bị chia và số chia là 88 nên:

\(88-8=b\times8\)

\(b\times8=80\)

\(b=10\)

hay số chia là \(10\); số bị chia là \(88+10=98\) 

Đáp số:...

\(a.\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{8}{9}\right)^2\\ =\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{\left(2\sqrt{2}\right)^2}{3^2}\right)^2\\ =\left(\dfrac{3}{2}\right)^4\cdot\left(\dfrac{2\sqrt{2}}{3}\right)^4\\ =\left(\dfrac{3}{2}\cdot\dfrac{2\sqrt{2}}{3}\right)^4\\ =\left(\sqrt{2}\right)^4\\ =4\\ b.\left(\dfrac{-3}{5}\right)^6\cdot\left(\dfrac{-5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot\left(-\dfrac{3}{5}\right)^5\cdot\left(\dfrac{-5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot\left(-\dfrac{3}{5}\cdot-\dfrac{5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot1\\ =-\dfrac{3}{5}\\ c.\left(\dfrac{4}{7}\right)^3\cdot\left(\dfrac{4}{7}\right)^5\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)^8\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)\cdot\left(\dfrac{4}{7}\right)^7\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)\cdot\left(\dfrac{4}{7}\cdot\dfrac{7}{4}\right)^7\\ =\dfrac{4}{7}\)

a: \(\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{8}{9}\right)^2=\dfrac{3^4}{4^4}\cdot\dfrac{8^2}{9^2}\)

\(=\dfrac{3^4}{3^4}\cdot\dfrac{2^6}{2^8}=\dfrac{1}{2^2}=\dfrac{1}{4}\)

b: \(\left(-\dfrac{3}{5}\right)^6\cdot\left(-\dfrac{5}{3}\right)^5\)

\(=\left(-\dfrac{3}{5}\right)^5\cdot\left(-\dfrac{5}{3}\right)^5\cdot\dfrac{-3}{5}=\left(-\dfrac{3}{5}\cdot\dfrac{-5}{3}\right)^5\cdot\dfrac{-3}{5}\)

\(=1^5\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\)

c: \(\left(\dfrac{4}{7}\right)^3\cdot\left(\dfrac{4}{7}\right)^5\cdot\left(\dfrac{7}{4}\right)^7=\left(\dfrac{4}{7}\right)^8\cdot\left(\dfrac{7}{4}\right)^7\)

\(=\left(\dfrac{4}{7}\cdot\dfrac{7}{4}\right)^7\cdot\dfrac{4}{7}=1^7\cdot\dfrac{4}{7}=\dfrac{4}{7}\)

d: \(\dfrac{8^{14}}{4^4\cdot64^5}=\dfrac{2^{42}}{2^8\cdot2^{30}}=2^4=16\)

e: \(\dfrac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\dfrac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\dfrac{3^{41}}{3^{43}}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

\(x^4+x^3+2x^2+x+1\\= x^4+x^3+x^2+x^2+x+1\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+1\right)\left(x^2+x+1\right)\)

\(1.\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)\\ =\dfrac{1}{3}\left(\dfrac{24}{20}-\dfrac{45}{20}\right)\\ =\dfrac{1}{3}\cdot\dfrac{-21}{20}\\ =\dfrac{-7}{20}\\ 2.-\dfrac{7}{5}\cdot\left(\dfrac{15}{14}+\dfrac{5}{7}\right)\\ =-\dfrac{7}{5}\cdot\left(\dfrac{15}{14}+\dfrac{10}{14}\right)\\ =-\dfrac{7}{5}\cdot\dfrac{25}{14}\\ =\dfrac{-5}{2}\\ 3.\dfrac{1}{5}:\dfrac{3}{10}+\dfrac{5}{6}\\ =\dfrac{1}{5}\cdot\dfrac{10}{3}+\dfrac{5}{6}\\ =\dfrac{2}{3}+\dfrac{5}{6}\\ =\dfrac{4}{6}+\dfrac{5}{6}\\ =\dfrac{3}{2}\)

1: \(\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)=\dfrac{1}{3}\cdot\dfrac{24-45}{20}\)

\(=\dfrac{1}{3}\cdot\dfrac{-21}{20}=\dfrac{-7}{20}\)

2: \(\dfrac{-7}{5}\left(\dfrac{15}{14}+\dfrac{5}{7}\right)=-\dfrac{7}{5}\cdot\left(\dfrac{15}{14}+\dfrac{10}{14}\right)\)

\(=-\dfrac{7}{5}\cdot\dfrac{25}{14}=\dfrac{-5}{2}\)

3: \(\dfrac{1}{5}:\dfrac{3}{10}+\dfrac{5}{6}=\dfrac{1}{5}\cdot\dfrac{10}{3}+\dfrac{5}{6}=\dfrac{2}{3}+\dfrac{5}{6}=\dfrac{4}{6}+\dfrac{5}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)

4: \(-\dfrac{4}{5}:\left(\dfrac{20}{9}-\dfrac{8}{3}\right)=\dfrac{-4}{5}:\left(\dfrac{20}{9}-\dfrac{24}{9}\right)\)

\(=-\dfrac{4}{5}:\dfrac{-4}{9}=\dfrac{4}{5}\cdot\dfrac{9}{4}=\dfrac{9}{5}\)

5: \(\dfrac{10}{7}:\dfrac{5}{14}-\dfrac{2}{3}=\dfrac{10}{7}\cdot\dfrac{14}{5}-\dfrac{2}{3}\)

\(=\dfrac{140}{35}-\dfrac{2}{3}=4-\dfrac{2}{3}=\dfrac{12}{3}-\dfrac{2}{3}=\dfrac{10}{3}\)

6: \(-\dfrac{3}{4}:\left(\dfrac{1}{4}-\dfrac{5}{8}\right)=\dfrac{-3}{4}:\left(\dfrac{2}{8}-\dfrac{5}{8}\right)=\dfrac{-3}{4}:\dfrac{-3}{8}\)

\(=\dfrac{3}{4}:\dfrac{3}{8}=\dfrac{3}{4}\cdot\dfrac{8}{3}=\dfrac{8}{4}=2\)

7: \(\dfrac{5}{26}-\dfrac{5}{7}:\dfrac{2}{7}=\dfrac{5}{26}-\dfrac{5}{7}\cdot\dfrac{7}{2}=\dfrac{5}{26}-\dfrac{5}{2}\)

\(=\dfrac{5}{26}-\dfrac{65}{26}=\dfrac{-60}{26}=\dfrac{-30}{13}\)

8: \(\dfrac{3}{4}:\dfrac{-3}{5}+\dfrac{1}{2}=\dfrac{3}{4}\cdot\dfrac{5}{-3}+\dfrac{1}{2}=-\dfrac{5}{4}+\dfrac{1}{2}\)

\(=-\dfrac{5}{4}+\dfrac{2}{4}=-\dfrac{3}{4}\)

9: \(\dfrac{1}{3}\cdot\left(\dfrac{2}{15}-\dfrac{4}{9}\right):\dfrac{1}{9}\)

\(=\dfrac{1}{3}\cdot9\cdot\left(\dfrac{6}{45}-\dfrac{20}{45}\right)\)

\(=3\cdot\dfrac{-14}{45}=\dfrac{-14}{15}\)

a: \(\dfrac{14}{-27}\cdot x=\dfrac{7}{9}\)

=>\(x=\dfrac{-7}{9}:\dfrac{14}{27}=\dfrac{-7}{9}\cdot\dfrac{27}{14}=\dfrac{-1}{2}\cdot3=-\dfrac{3}{2}\)

b: \(\left(2x-1\right):\dfrac{8}{9}=\dfrac{15}{4}\)

=>\(2x-1=\dfrac{15}{4}\cdot\dfrac{8}{9}=\dfrac{120}{36}=\dfrac{10}{3}\)

=>\(2x=\dfrac{10}{3}+1=\dfrac{13}{3}\)

=>\(x=\dfrac{13}{3}:2=\dfrac{13}{6}\)

c: \(\dfrac{2}{5}:x=\dfrac{3}{16}\)

=>\(x=\dfrac{2}{5}:\dfrac{3}{16}=\dfrac{2}{5}\cdot\dfrac{16}{3}=\dfrac{32}{15}\)

d: \(\dfrac{11}{12}-\left(\dfrac{2}{5}-3x\right)=\dfrac{2}{3}\)

=>\(\dfrac{2}{5}-3x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

=>\(3x=\dfrac{2}{5}-\dfrac{1}{4}=\dfrac{3}{20}\)

=>\(x=\dfrac{3}{20}:3=\dfrac{1}{20}\)

\(a)\dfrac{14}{-27}\cdot x=\dfrac{7}{9}\\ x=\dfrac{7}{9}:\dfrac{14}{-27}\\ x=\dfrac{7}{9}\cdot\dfrac{-27}{14}\\x =\dfrac{-3}{2}\\ b)\left(2x-1\right):\dfrac{8}{9}=\dfrac{15}{4}\\ 2x-1=\dfrac{15}{4}\cdot\dfrac{8}{9}\\ 2x-1=\dfrac{10}{3}\\ 2x=\dfrac{10}{3}+1\\ 2x=\dfrac{13}{3}\\ x=\dfrac{13}{3}:2=\dfrac{13}{6}\\ c)\dfrac{2}{5}:x=\dfrac{3}{16}\\ x=\dfrac{2}{5}:\dfrac{3}{16}\\ x=\dfrac{2}{5}\cdot\dfrac{16}{3}\\ x=\dfrac{32}{15}\\ d)\dfrac{11}{12}-\left(\dfrac{2}{5}-3x\right)=\dfrac{2}{3}\\ \dfrac{2}{5}-3x=\dfrac{11}{12}-\dfrac{2}{3}\\ \dfrac{2}{5}-3x=\dfrac{3}{12}\\ \dfrac{2}{5}-3x=\dfrac{1}{4}\\ 3x=\dfrac{2}{5}-\dfrac{1}{4}\\ 3x=\dfrac{3}{20}\\ x=\dfrac{3}{20}:3\\ x=\dfrac{1}{20}\)

a: \(0,25\in Q\)

=>Đúng

b: \(-\dfrac{6}{7}\in Q\)

=>Đúng

c: \(-235\notin Q\)

=>Sai

b) Xét pt hoành độ giao điểm của hàm số đã cho và Ox là \(2x^3+2\left(6m-1\right)x^2-3\left(2m-1\right)x-3\left(1+2m\right)=0\)    (*)

Ta thấy \(x=1\) là nghiệm của pt trên. Lập sơ đồ Horner:

Do đó pt (*) 

\(\Leftrightarrow\left(x-1\right)\left(2x^2+12mx+6m+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2+12mx+6m+3=0\end{matrix}\right.\)

 Xét pt \(2x^2+12mx+6m+3=0\)      (1)

 Ycbt \(\Leftrightarrow\) pt (1) có 2 nghiệm phân biệt \(x_1,x_2\) khác 1 và thỏa mãn \(x_1^2+x_2^2=27\)

 Có \(\Delta'=\left(6m\right)^2-2\left(6m+3\right)=36m^2-12m-6>0\) 

 \(\Leftrightarrow\left[{}\begin{matrix}m>\dfrac{1+\sqrt{7}}{6}\\m< \dfrac{1-\sqrt{7}}{6}\end{matrix}\right.\)

Có 2 nghiệm khác 1 \(\Leftrightarrow2.1^2+12m.1+6m+3\ne0\) 

\(\Leftrightarrow18m+5\ne0\)

\(\Leftrightarrow m\ne-\dfrac{5}{18}\)

Theo định lý Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-6m\\x_1x_2=\dfrac{6m+3}{2}\end{matrix}\right.\)

Để \(x_1^2+x_2^2=27\) 

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=27\)

\(\Leftrightarrow\left(-6m\right)^2-2.\dfrac{6m+3}{2}=27\)

\(\Leftrightarrow36m^2-6m-3=27\)

\(\Leftrightarrow6m^2-m-5=0\)

\(\Leftrightarrow6m^2-6m+5m-5=0\)

\(\Leftrightarrow6m\left(m-1\right)+5\left(m-1\right)=0\)

\(\Leftrightarrow\left(m-1\right)\left(6m+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\left(nhận\right)\\m=-\dfrac{5}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy \(m=1\) hoặc \(m=-\dfrac{5}{6}\) thỏa ycbt.

c) Xét pt \(x^3-3mx^2+\left(3m-1\right)x+6m=0\)   (*)

Ta thấy (*) có nghiệm \(x=-1\). Lập sơ đồ Horner:

Vậy (*) \(\Leftrightarrow\left(x+1\right)\left(x^2-\left(3m+1\right)x+6m\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2-\left(3m+1\right)x+6m=0\end{matrix}\right.\)

Tới đây thì làm tương tự câu b) nhé.

Mỗi kí tự có 10 cách chọn số, 26 cách chọn chữ in hoa và 26 cách chọn chữ in thường. Do đó mỗi kí tự có \(10+2.26=62\) cách chọn. Khi đó số mật khẩu có thể là \(62^{10}\) 

Trong trường hợp xấu nhất, kẻ gian sẽ mất \(62^{10}\) giây, để cho gọn hơn thì là \(62^{10}:60:60:24:365:100=266140083\) thể kỷ

 P/S: Đó là khi kẻ gian không chết trước khi phá được mật khẩu.