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

bạn nào có thể biến đổi cái này để mình dùng vi-ét cái

\(x^2_1+2x^2_2=3\) biến đổi hộ cái 

\(\left\{{}\begin{matrix}x+2y=2\\mx-y=m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\2mx-2y=2m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2mx+x=2+2m\\x+2y=2\end{matrix}\right.\\ \left\{{}\begin{matrix}x\left(2m+1\right)=2\left(m+1\right)\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\\dfrac{2\left(m+1\right)}{2m+1}+2y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\2m+2+4my+2y=4m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\y\left(4m+2\right)=2m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\y=\dfrac{2m}{4m+2}\end{matrix}\right.\\ thay.....x,y....vào....ta.....được\\ \dfrac{2\left(m+1\right)}{2m+1}+\dfrac{2m}{4m+2}=1\\ \Leftrightarrow\dfrac{4\left(m+1\right)}{4m+2}+\dfrac{2m}{4m+2}=\dfrac{4m+2}{4m+2}\\ \Rightarrow4m+4+2m=4m+2\\ \Leftrightarrow2m=-2\\ \Leftrightarrow m=-1\\ vậy...m=-1...thì...tm\)                         \(thay....m=3...vào...ta...có...hpt:\\ \left\{{}\begin{matrix}x+2y=2\\3x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\6x-2y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7x=8\\x+2y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8}{7}\\y=\dfrac{3}{7}\end{matrix}\right.\) 

\(thay...m=3....ta...có:\\ \left\{{}\begin{matrix}x+2y=2\\3x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\6x-2y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7x=8\\x+2y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8}{7}\\y=\dfrac{3}{7}\end{matrix}\right.\\ vậy...với..m=3...thì...hệ....phương....trình....có...nghiệm...duy...nhất\left\{x=\dfrac{8}{7};y=\dfrac{3}{7}\right\}\)

\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}\\x^2-y^2=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x^2-6y^2=5xy\\x^2-y^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x^2-6y^2=5xy\\6x^2-6y^2=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}0=5xy-30\\6x^2-6y^2=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5xy=30\\6x^2-6y^2=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{5y}\\6\left(\dfrac{30}{5y}\right)^2-6y^2=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{5y}\\\dfrac{216}{y^2}-6y^2=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{5y}\\216-6y^4=30y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{5y}\\y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

                     \(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\\ \Leftrightarrow\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x\sqrt{x}+1}{\sqrt[]{x}\left(\sqrt{x}+1\right)}\right):\dfrac{2\left(\sqrt{x}-1\right)^2}{x-1}\\ \Leftrightarrow\dfrac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}-\dfrac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)}:\dfrac{2\sqrt[]{x}-1}{\sqrt{x}+1}\\ \Leftrightarrow\dfrac{x^2+x\sqrt{x}-\sqrt{x}-1-x^2+x\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(x-1\right)}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ \Leftrightarrow\dfrac{2x\sqrt{x}-2\sqrt{x}}{\sqrt{x}\left(x-1\right)}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

giúp em 2 câu này gấp với ạ >.<

a) \(\left(x+7\right)\left(3x+1\right)=49-x^2\\ \Rightarrow3x^2+x+21x+7=49-x^2\\ \Rightarrow4x^2+22x-42=0\\ \Rightarrow4x^2+28x-6x-42=0\\ \Leftrightarrow4x\left(x+7\right)-6\left(x+7\right)=0\\ \Leftrightarrow\left(4x-6\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+7=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=\dfrac{3}{2}\end{matrix}\right.\)                                                         b)\(\left(2x+1\right)^2=\left(x-1\right)^2\\ \Rightarrow4x^2+4x+1=x^2-2x+1\\ \Leftrightarrow3x^2+6x=0\\ \Leftrightarrow3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c)\(x^3-5x^2+6x=0\\ \Rightarrow\\ x^3-3x^2-2x^2+6x=0\Leftrightarrow x^2\left(x-3\right)-2x\left(x-3\right)=0\\ \Leftrightarrow\left(x^2-2x\right)\left(x-3\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)                      d)\(3x^2+5x+2=0\\3x^2+3x+2x +2=0\Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\end{matrix}\right.\)

hãy cho tôi biết cách làm 1 bài văn nghị luận về tư tưởng đạo lí chỉ cần phần thân bài TB: Lđ 1: giải thích câu tục ngữ Lđ 2 ........................... Lđ n là j mình nhớ là có bểu hiện, có câu hỏi và mở rộng vấn đề nhanh nha chỉ cần phần thân bài

\(\begin{cases} 5x-4y=32\\ 2x+3y=-1 \end{cases} \)

➞\(\begin{cases} 10x-8y=64\\ 10x+15y=-5 \end{cases} \)

➜\(\begin{cases} -23y=69\\ 10x+15y=-5 \end{cases} \)

➜\(\begin{cases} x=-3\\ y=4 \end{cases} \)

`vậy hpt có nghiệm duy nhất (x;y)=(-3;4)`