\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)

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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)

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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)

\(\sqrt{4x^2-4x+1}=2x+1\\ \Leftrightarrow\sqrt{\left(2x+1\right)^2}=2x+1\\ \Leftrightarrow\left|2x+1\right|=2x+1\)

\(TH_1:2x+1>0\Leftrightarrow x>-\dfrac{1}{2}\\ 2x+1=2x+1\left(LD\forall x\right)\)

\(TH_2:2x+1\le0\Leftrightarrow x\le-\dfrac{1}{2}\\ -2x-1=2x+1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)

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

\(a,A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(dk:x>0,x\ne1\right)\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)

\(b,x=3+2\sqrt{2}=\sqrt{2}^2+2\sqrt{2}.1+1=\left(\sqrt{2}+1\right)^2\)

\(A=\sqrt{x}-1=\sqrt{\left(\sqrt{2}+1\right)}^2-1=\sqrt{2}+1-1=\sqrt{2}\)

Dựa vào lý thuyết.

\(\left(a\right)\dfrac{x^2+3x}{x+3}=\dfrac{x\left(x+3\right)}{x+3}=x\\ \left(b\right)\dfrac{x-2}{x^2-5x+6}=\dfrac{x-2}{x^2-2x-3x+6}=\dfrac{x-2}{x\left(x-2\right)-3\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\\ \left(c\right)\dfrac{1}{x+y}+\dfrac{y}{x^2-y^2}=\dfrac{x-y+y}{x^2-y^2}=\dfrac{x}{x^2-y^2}\)

\(TXD\) \(D=R\backslash\left\{0\right\}\) là tập đối xứng.

\(\forall x\in D\Rightarrow-x\in D\)

Có \(f\left(-x\right)=\dfrac{\left(-x\right)^2+1}{\left|2\left(-x\right)+1\right|+\left|2\left(-x\right)-1\right|}\)

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

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

\(=\dfrac{x^2+1}{\left|2x-1\right|+\left|2x+1\right|}\) \(=f\left(x\right)\)

Vậy hàm số \(y=f\left(x\right)=\dfrac{x^2+1}{\left|2x+1\right|+\left|2x-1\right|}\) là hàm số chẵn.

\(a,A=x^2+7x+7y-y^2\\ =x^2-y^2+7x+7y\\ =\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\\ =\left(x+y\right)\left(x-y+7\right)\)

\(b,B=x^2+2xy+y^2-3x-3y\\ =\left(x+y\right)^2-3\left(x+y\right)\\ =\left(x+y\right)\left(x+y-3\right)\)

Áp dụng đ/l Pytago vào tam giác vuông ABC, có :

\(BC^2=AB^2+AC^2\\ \Rightarrow AC^2=BC^2-AB^2\\ \Rightarrow AC=\sqrt{4^2-3^2}\\ =\sqrt{7}\left(m\right)\)

Chiều cao của cây lúc chưa gãy là :

\(4+\sqrt{7}\approx6,6\left(m\right)\)

\(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}+2}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}.\dfrac{\sqrt{x}+2}{2}\\ =\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{2}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)