\(\text{Giải :}\)

\(\frac{\left(\text{0,9}\right)^{\text{6}}}{\left(\text{0,3}\right)^{\text{5}}}=\frac{\left(\text{0,9}\right)^{\text{5}}}{\left(\text{0,3}\right)^{\text{5}}}.\left(\text{0,9}\right)=\text{3 . 0,9}=\text{2,7}\)

\(\frac{0,9^6}{0,3^5}\)=\(\frac{\left(0,3\right)^{2^6}}{0,3^5}\)

=\(\frac{0,3^{12}}{0,3^5}\)=(0,3)⁷

\(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)

\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)

Nhận thấy 19¹⁷ + 19 > 19¹⁶ + 19

=> \(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)

=> \(-\frac{18}{19^{17}+19}>-\frac{18}{19^{16}+19}\)

=> \(1-\frac{18}{19^{17}+19}>1-\frac{18}{19^{16}+19}\)

=> \(\frac{x}{19}>\frac{y}{19}\)

=> x > y

Vậy x > y

Ta có : \(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)

\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)

Vì\(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)\(\Rightarrow\frac{x}{19}>\frac{y}{19}\)

mà \(x,y>0\)

\(\Rightarrow x>y\)

Ta có x^2-x-2=0

=>x(x-1)=2

=>x(x-1)=1.2=(-1).(-2)

=>x= 2 hoặc  x = -1

Ta có : x² - x - 2 = 0

<=> x² - 2x + x - 2 = 0

<=> x(x - 2) + (x - 2) = 0

<=> (x + 1)(x - 2) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

Vậy x \(\in\left\{-1;2\right\}\)

Ta có 13x = \(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

13y = \(\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

Vì 13¹⁷ + 1 > 13¹⁶ + 1

=> \(\frac{1}{13^{17}+1}< \frac{1}{13^{16}+1}\)

=> \(\frac{12}{13^{17}+1}< \frac{12}{13^{16}+1}\)

=> \(1+\frac{12}{13^{17}+1}< 1+\frac{12}{13^{16}+1}\)

=> 13x < 13y 

=> x < y

Vậy x < y

\(\frac{5}{9}\div\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}\div\left(\frac{1}{15}-\frac{2}{3}\right).\)

= \(\frac{5}{9}\div\left(\frac{-3}{22}\right)+\frac{5}{9}\div\left(\frac{-3}{5}\right)\)

= \(\frac{-110}{27}+\frac{-25}{27}\)

= \(-\frac{135}{27}=-5\)

\(\frac{\text{5}}{\text{9}}:\left(\frac{\text{1}}{\text{11}}-\frac{\text{5}}{\text{22}}\right)+\frac{\text{5}}{\text{9}}:\left(\frac{\text{1}}{\text{15}}-\frac{\text{2}}{\text{3}}\right)\)

\(=\frac{\text{5}}{\text{9}}:\frac{-\text{3}}{\text{22}}+\frac{\text{5}}{\text{9}}:\frac{-\text{9}}{\text{15}}\)

\(=\frac{\text{5}}{\text{9}}.\frac{\text{22}}{-\text{3}}+\frac{\text{5}}{\text{9}}.\frac{\text{15}}{-\text{9}}\)

\(=\frac{-\text{5}}{\text{27}}.\left(\text{22 + 5}\right)=\frac{-\text{5}}{\text{27}}.\text{27 = -5}\)

x³ - 27 = 0

x³ = 0 + 27

x³ = 27

x³ = 3³

\(\Rightarrow\)x = 3

Hok tốt

a) Ta có: Ot là tia phân giác của \(\widehat{xOy}\) => \(\widehat{O_1}=\widehat{O_2}=\frac{\widehat{xOy}}{2}\) (1)

On là tia phân giác của \(\widehat{xAm}\) => \(\widehat{A_1}=\widehat{A_2}=\frac{1}{2}\widehat{xAm}\) (2)

Mà Am // Oy (gt) => \(\widehat{xAm}=\widehat{xOy}\) (đồng vị) (3)

Từ (1), (2) và (3) => \(\widehat{O_1}=\widehat{O_2}=\widehat{A_1}=\widehat{A_2}\)

mà \(\widehat{A_2}\) và \(\widehat{O_2}\)ở vị trí đồng vị => An // Ot

b) Ta có: \(\hept{\begin{cases}AH\perp Ot\left(gt\right)\\Ot//On\left(cmt\right)\end{cases}}\Rightarrow AH\perp An\)

Xét tam giác OAH vuông tại H có: \(\widehat{O_2}+\widehat{A_3}=90^0\)

Lại có: \(\widehat{A_1}+\widehat{A_4}=90^0\)(phụ nhau)

mà \(\widehat{O_2}=\widehat{A_1}\) (cm câu a)

=> \(\widehat{A_3}=\widehat{A_4}\) -> AH là tia phân giác của \(\widehat{OAm}\)

Có chắc là đúng ko hả bạn?

9. ta có: a/3b=b/3c=c/3d=d/3a=a+b+c+d/3(a+b+c+d)=1/3(theo t/c tỉ lệ thức)

suy ra a/3b=1/3 suy ra a=b (1)

b/3c=1/3 suy ra b=c (2)

c/3d=1/3 suy ra c=d (3)

d/3a=1/3 suy ra d=a(4)

từ (1,2,3,4) suy ra a=b=c=d

11. a/b=b/c=c/d=a+b+c/b+c+d(theo t/c tỉ lệ thức)

ta có: a/b.b/c.c/d=a/d (1)

a/b.b/c.c/d mà a/b=b/c=c/d=a+b+c/b+c+d nên a/b.b/c.c/d=a+b+c/b+c+d.a+b+c/b+c+d.a+b+c/b+c+d=(a+b+c/b+c+d)³  (2)

từ (1),(2) suy ra a/d=(a+b+c/b+c+d)³ 

12. a/b+c=b/c+a=c/a+b=a+b+c/a+b+c+a+b+c=1/2

Gọi x,y,z,t lần lượt là số học sinh các khối:6,7,8,9

Theo đề bài ta có: \(\frac{x}{9}\) =\(\frac{y}{8}\)=\(\frac{z}{7}\)=\(\frac{t}{6}\)và y - t=70

Theo tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{y}{8}\)=\(\frac{t}{6}\)=\(\frac{y-t}{8-6}\)=\(\frac{70}{2}\)=35

Do đó:

x=315

y=280

z=245

t=210

Bài 2 :

a) \(\left(x-3\right)\left(x+7\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x-3>0\\x+7< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x-3< 0\\x+7>0\end{cases}}\)

mà \(x+3< x+7\forall x\)

nên \(\hept{\begin{cases}x-3< 0\\x+7>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x< 3\\x>-7\end{cases}}\)

\(\Leftrightarrow-7< x< 3\)

b) \(\left(2x+1\right)\left(x+5\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}2x+1>0\\x+5>0\end{cases}}\)hoặc \(\hept{\begin{cases}2x+1< 0\\x+5< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>-\frac{1}{2}\\x>-5\end{cases}}\)hoặc \(\hept{\begin{cases}x< -\frac{1}{2}\\x< -5\end{cases}}\)

hay \(\orbr{\begin{cases}x>-\frac{1}{2}\\x< -5\end{cases}}\)

Bài 3: 

\(A=\frac{\frac{3}{7}-\frac{3}{5}+\frac{3}{11}}{-\frac{2}{7}+\frac{2}{5}-\frac{2}{11}}.\frac{\frac{1}{3}-0,25+0,2}{1\frac{1}{6}-0,875+0,7}+\frac{6}{7}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{5}+\frac{1}{11}\right)}{-2\left(\frac{1}{7}-\frac{1}{5}+\frac{1}{11}\right)}.\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}+\frac{6}{7}\)

\(=\frac{-3}{2}.\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}+\frac{6}{7}\)

\(=-\frac{3}{2}.\frac{2}{7}+\frac{6}{7}\)

\(=-\frac{3}{7}+\frac{6}{7}=\frac{3}{7}\)

\(B=\left(-19,75\right)+\left(-45,95\right)+\left(4,75+5,95\right)\)

\(=\left(-19,75+4,75\right)+\left(-45,95+5,95\right)\)

\(=\left(-15\right)+\left(-40\right)=-55\)