ĐKXĐ: \(x\notin\left\{-7;3;-3\right\}\)

a) Ta có: \(B=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{x-3}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)

\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)

\(=\dfrac{x^2+1-x^2+3x+5x+15}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+10-x-3}{x+3}\)

\(=\dfrac{8x+16}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+7}\)

\(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)

b) Ta có: |x-1|=2

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Thay x=-1 vào biểu thức \(B=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\), ta được:

\(B=\dfrac{8\cdot\left(-1\right)+16}{\left(-1-3\right)\left(-1+7\right)}=\dfrac{-8+16}{-4\cdot6}=\dfrac{8}{-24}=\dfrac{-1}{3}\)

Vậy: Khi x=-1 thì \(B=\dfrac{-1}{3}\)

c) Để \(B=\dfrac{x+5}{6}\) thì \(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}=\dfrac{x+5}{6}\)

\(\Leftrightarrow6\left(8x+16\right)=\left(x+5\right)\left(x-3\right)\left(x+7\right)\)

\(\Leftrightarrow48x+96=\left(x^2-3x+5x-15\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x^2+2x-15\right)\left(x+7\right)=48x+96\)

\(\Leftrightarrow x^3+7x^2+2x^2+14x-15x-105-48x-96=0\)

\(\Leftrightarrow x^3+9x^2-49x-201=0\)

\(\Leftrightarrow x^3+3x^2+6x^2+18x-67x-201=0\)

\(\Leftrightarrow x^2\left(x+3\right)+6x\left(x+3\right)-67\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+6x-67\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+6x+9-76\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-76\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+3-2\sqrt{19}\right)\left(x+3+2\sqrt{19}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+3-2\sqrt{19}=0\\x+3+2\sqrt{19}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=2\sqrt{19}-3\left(nhận\right)\\x=-2\sqrt{19}-3\left(nhận\right)\end{matrix}\right.\)

Vậy: Để \(B=\dfrac{x+5}{6}\) thì \(x\in\left\{2\sqrt{19}-3;-2\sqrt{19}-3\right\}\)

a: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)

\(=\dfrac{35}{6}:\dfrac{95-126}{30}-\dfrac{11}{31}\)

\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)

\(=\dfrac{-35\cdot5}{31}-\dfrac{11}{31}=\dfrac{-186}{31}=-6\)

b: \(=\left(-8\right)\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-4:\dfrac{27-14}{12}=\dfrac{-4\cdot12}{13}=\dfrac{-48}{13}\)

phép tính đầu kết quả là -6

phếp tính thứ 2 kết quả là-48/13

câu 1: 
UCLN(15;19)=1

1Yes, i have. 2 I do exercises everyday. 3 I’ll help them to find a new house or give them to someone can do that 

1: \(100-x^2=\left(10-x\right)\left(10+x\right)\)

2: \(b^2-a^2=\left(b-a\right)\left(b+a\right)\)

3: \(\left(3y\right)^2-\left(4x\right)^2=\left(3y-4x\right)\left(3y+4x\right)\)

4. (x - 2)(x + 2) = x² - 4

5. (2x - y)(2x + y) = 4x² - y²

6. \(\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)=\dfrac{1}{4}x^2-y^2\)

mờ quá

bạn chụp dọc đc ko ạ , khó nhìn và mờ quá

mink kiểm tra rồi , nhưng tụi mink khác trường mà làm sao các thầy cô giáo cho đề giống nhau được

Lớp mấy vậy bn

lớp 6 nha các bạn 

nhưng khác trưởng cũng được cho mk xin nha