\(\frac{5}{x}+\frac{4}{x+1}=\frac{3}{x+2}+\frac{2}{x+3}\)

\(\Leftrightarrow\frac{5\left(x+1\right)+4x}{x\left(x+1\right)}=\frac{3\left(x+3\right)+2\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{5x+5+4x}{x^2+x}=\frac{3x+9+2x+4}{x^2+5x+6}\)

\(\Leftrightarrow\frac{9x+5}{x^2+x}=\frac{5x+13}{x^2+5x+6}\)

\(\Leftrightarrow\left(9x+5\right)\left(x^2+5x+6\right)=\left(5x+13\right)\left(x^2+x\right)\)

\(\Leftrightarrow9x^3+45x^2+54x+5x^2+25x+30=5x^3+5x^2+13x^2+13x\)

\(\Leftrightarrow9x^3+50x^2+79x+30=5x^3+18x^2+13x\)

\(\Leftrightarrow9x^3-5x^3+50x^2-18x^2+79x-13x+30=0\)

\(\Leftrightarrow4x^3+32x^2+66x+30=0\)

\(\Leftrightarrow2x^3+16x^2+33x+15=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2,3660\right)\left(x+0,6340\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x\approx2,3660\end{cases}or_{ }x\approx0,6340}\)

Nghiệm là -5 thôi nha, phần còn lại khác 0 nên loại

\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)

\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)

Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)

Nhưng \(2\sqrt{x}+1\ge1\)

\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

Vậy \(x\in\left\{0;4\right\}\)

30 đơn vị là j hả bạn

Mong bn xem lại giúp VNM

Hội con 🐄 chúc bạn học tốt!!!

Lấy điểm F sao cho DF // AM và F thuộc BC

Theo quy tắc hình bình hành ( AM//DF ; AD //MF)

\(\overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{AM}\)

Vì AMFD là hình bình hành nên \(\left|\overrightarrow{AD}\right|=\left|\overrightarrow{MF}\right|\Rightarrow BF=\frac{a}{2}+a=\frac{3a}{2}\)

Theo định lý Pytago ta có:

\(\left|\overrightarrow{AF}\right|^2=a^2+\left(\frac{3a}{2}\right)^2=a^2+\frac{9a^2}{4}=\frac{13a^2}{4}\)

\(\Rightarrow\left|\overrightarrow{AF}\right|=\sqrt{\frac{13a^2}{4}}=\frac{a\sqrt{13}}{2}\)

Dễ tính được \(AM=\frac{\sqrt{5}a}{2}\)

Ta thấy M là trung điểm của BC tức \(MB=MC=\frac{1}{2}BC=\frac{1}{2}AB\Rightarrow\widehat{AMB}=60^0\) 

\(AD//BC\Rightarrow\widehat{DAC}=\widehat{AMB}=60^0\)

\(\Rightarrow\overrightarrow{AD}+\overrightarrow{AM}=\sqrt{a^2+\frac{5a^2}{4}-2\cdot a\cdot\frac{\sqrt{5}a}{2}\cdot\cos120}\)

\(\Rightarrow\overrightarrow{AD}+\overrightarrow{AM}=\sqrt{\frac{9a^2}{4}+\frac{\sqrt{5}a^2}{2}}=\sqrt{\frac{9a^2+2\sqrt{5}a^2}{4}}=\frac{a}{2}\sqrt{9+2\sqrt{5}}\)

Chắc vậy ạ 

Sai thì thông cảm mk nha