\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

83. He officially became a professional athlete in 1997 when he was 24 years old.

84. Lars stopped complaining about the hotel staff as he had understood the situation.

\(a,x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\\ b,2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\left(x+1\right)^2\\ c,3x^3y-12x^2y+12xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\left(x+y\right)^2\\ e,x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\\ f,9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(9x^2-4y^2\right)\left(x-2\right)=\left(3x-2y\right)\left(3x+2y\right)\left(x-2\right)\)

Tick plz

\(a,A=8x^3+12x^2+6x+1=\left(2x+1\right)^2\)

Thay \(x=99\) vào A, ta được:

\(A=\left(2\cdot99+1\right)^3=199^3=7880599\)

\(b,B=8x^3-60x^2+150x-125=\left(2x-5\right)^3\)

Thay \(x=105\) vào B, ta được:

\(B=\left(105\cdot2-5\right)^3=205^3=8615125\)

Tick plz

\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{11\sqrt{x}-5}{x-25}\left(x\ge0,x\ne25\right)\\ \Leftrightarrow P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-11\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{x-4\sqrt{x}-5+x+5\sqrt{x}-11\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{2x-10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{2\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+5}\)

\(b,\) Tại \(x=4\)

\(P=\dfrac{2\cdot2}{2+5}=\dfrac{4}{7}\)

\(c,x=51-10\sqrt{26}=\left(\sqrt{26}-5\right)^2\)

Thay vào P, ta được:

\(P=\dfrac{2\left(\sqrt{26}-5\right)}{\sqrt{26}-5+5}=\dfrac{2\sqrt{26}-10}{\sqrt{26}}=\dfrac{52-10\sqrt{26}}{26}=\dfrac{26-5\sqrt{26}}{13}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}-\dfrac{3\sqrt{x}}{x+\sqrt{x}-2}\left(x\ge0\right)\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

Lần sau ghi đề rõ ra nha bạn

\(3,\)

\(a,\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(x>0,x\ne9\right)\\ =\dfrac{x+3+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(b,x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ \Leftrightarrow x=5+\sqrt{2}-4-\sqrt{2}=1\)

Thay vào \(B\), ta được: \(B=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

\(c,\) Ta có \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)

Mà \(x>0\)

\(\Leftrightarrow\sqrt{x}>0\\ \Leftrightarrow\sqrt{x}+3>3\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}< \dfrac{2}{3}\\ \Leftrightarrow1-\dfrac{2}{\sqrt{x}+3}>1-\dfrac{2}{3}=\dfrac{1}{3}\)

Tick full hộ nha 🤩