= 40 nha, đúng thì tk mình nha bạn

Bài 3: 

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

a) Ta có: \(A=\left(\dfrac{x+1}{x-2}+\dfrac{x}{x+2}+\dfrac{2x^2+3}{x^2-4}\right):\left(1-\dfrac{x-3}{x+2}\right)\)

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

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

\(=\dfrac{4x^2+x+5}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{5}\)

\(=\dfrac{4x^2+x+5}{5\left(x-2\right)}=\dfrac{4x^2+x+5}{5x-10}\)

b) Vì x=-1 thỏa mãn ĐKXĐ nên Thay x=-1 vào biểu thức \(A=\dfrac{4x^2+x+5}{5x-10}\), ta được:

\(A=\dfrac{4\cdot\left(-1\right)^2-1+5}{5\cdot\left(-1\right)-10}=\dfrac{4-1+5}{-5-10}=\dfrac{-8}{15}\)

Vậy: Khi x=-1 thì \(A=-\dfrac{8}{15}\)

c) Để A=-3 thì \(\dfrac{4x^2+x+5}{5x-10}=-3\)

\(\Leftrightarrow4x^2+x+5=-3\left(5x-10\right)\)

\(\Leftrightarrow4x^2+x+5=-15x+30\)

\(\Leftrightarrow4x^2+16x-25=0\)

\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot4+16-41=0\)

\(\Leftrightarrow\left(2x+4\right)^2=41\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=\sqrt{41}\\2x+4=-\sqrt{41}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{41}-4\\2x=-\sqrt{41}-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}-4}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}-4}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: Khi A=-3 thì \(x\in\left\{\dfrac{\sqrt{41}-4}{2};\dfrac{-\sqrt{41}-4}{2}\right\}\)

Camr ơn bạn nhiều ạ

Hình nhỏ quá. Bạn nên gõ đề để được hỗ trợ tốt hơn.

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: Xét tứ giác AMBQ có

P là trung điểm chung của AB và MQ

MA=MB

=>AMBQ là hình thoi

b: BC=2*AM=20cm

\(AC=\sqrt{20^2-18^2}=\sqrt{76}=2\sqrt{19}\left(cm\right)\)

\(S_{ABC}=\dfrac{1}{2}\cdot2\sqrt{19}\cdot18=18\sqrt{19}\left(cm^2\right)\)

c: Để AMBQ là hình vuông thì góc ABM=45 độ

=>góc ABC=45 độ

d: \(MP=\dfrac{AC}{2}=\dfrac{2\sqrt{19}}{2}=\sqrt{19}\left(cm\right)\)

=>MQ=2 căn 19(cm)

\(S_{AMBQ}=\dfrac{1}{2}\cdot2\sqrt{19}\cdot18=18\sqrt{19}\left(cm^2\right)\)