\(f\left(-2\right)=-2m+1\)

\(\lim\limits_{x\rightarrow-2^+}f\left(x\right)=\lim\limits_{x\rightarrow-2^+}\dfrac{x^2-3x+2}{x^3+8}=\lim\limits_{x\rightarrow-2^+}\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\lim\limits_{x\rightarrow-2^+}\dfrac{x-1}{x^2-2x+4}=\dfrac{-2-1}{4-2.\left(-2\right)+4}=-\dfrac{1}{4}\)

\(f\left(-2\right)\ne\lim\limits_{x\rightarrow-2^-}f\left(x\right)\Leftrightarrow-2m+1\ne-\dfrac{1}{4}\Leftrightarrow m\ne\dfrac{5}{8}\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-1}+\sqrt[3]{\left(x-1\right)^3}}{\sqrt{x-1}}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^2-1\right)^{\dfrac{1}{2}}+x-1}{\left(x-1\right)^{\dfrac{1}{2}}}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^2-1\right)^{-\dfrac{1}{2}}.2+1}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}}\)

\(=\dfrac{1}{0}=+\infty\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt[3]{x}-1}{\sqrt{2}-\sqrt{x+1}}=\lim\limits_{x\rightarrow1^-}\dfrac{\left(x-1\right)\left(\sqrt{2}+\sqrt{x+1}\right)}{[\left(\sqrt[3]{x}\right)^2+\sqrt[3]{x}+1]\left(1-x\right)}=\lim\limits_{x\rightarrow1^-}\dfrac{-\left(\sqrt{2}+\sqrt{1+1}\right)}{1+1+1}=-\dfrac{2\sqrt{2}}{3}\)

\(f\left(1\right)=\sqrt{2}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)\ne\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne f\left(x\right)\)=> ham gian doan tai x=1

a.

\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{1}{2}-x^2\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow4x^3-3x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

- Với \(x=-1\) thế vào pt đầu: \(1+y^2=\dfrac{1}{2}\Rightarrow y^2=-\dfrac{1}{2}\) (vô nghiệm)

- Với \(x=\dfrac{1}{2}\) thế vào pt đầu: \(\dfrac{1}{4}+y^2=\dfrac{1}{2}\Rightarrow y=\pm\dfrac{1}{2}\)

\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)

Dễ thấy x = 0 không phải nghiệm ta nhân tử mẫu phương trình đầu cho 3x thì được

\(\Leftrightarrow\left\{{}\begin{matrix}3x^3+3xy^2=\dfrac{3x}{2}\left(1\right)\\x^3+3xy^2=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)

Lấy (1) - (2) thì đơn giản rồi ha

b.

Trừ vế cho vế:

\(\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)-\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=0\)

\(\Leftrightarrow\left(1+x^2\right)^2-\left(1+y^2\right)^2+\left(\dfrac{1+x^2}{y^2}\right)^2-\left(\dfrac{1+y^2}{x^2}\right)^2=0\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2\right)+\left(\dfrac{x^4+x^2-y^4-y^2}{x^2y^2}\right)\left(\dfrac{1+x^2}{y^2}+\dfrac{1+y^2}{x^2}\right)=0\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2\right)+\left(\dfrac{\left(x^2-y^2\right)\left(x^2+y^2+1\right)}{x^2y^2}\right)\left(\dfrac{x^2+1}{y^2}+\dfrac{y^2+1}{x^2}\right)=0\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2+\left(\dfrac{x^2+y^2+1}{x^2y^2}\right)\left(\dfrac{x^2+1}{y^2}+\dfrac{y^2+1}{x^2}\right)\right)=0\)

\(\Leftrightarrow x^2=y^2\) (ngoặc to hiển nhiên dương)

Thế vào pt đầu:

\(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\)

Ta có: \(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)\ge4x^2.2\sqrt{1.\dfrac{1}{x^4}}=8\)

Đẳng thức xảy ra khi và chỉ khi \(x^2=1\)

Vậy nghiệm của hệ là \(x^2=y^2=1\Rightarrow x;y\)

Chọn C

\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right)\)

\(2,HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{matrix}\right.\\ \Leftrightarrow\sqrt{2\left(x^2+y^2\right)}=x+y\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\\ \Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;4\right)\)

\(3,\text{Sửa: }\left\{{}\begin{matrix}\sqrt{x^2+3}+\left|y\right|=\sqrt{3}\left(1\right)\\\sqrt{y^2+5}+\left|x\right|=\sqrt{x^2+5}\left(2\right)\end{matrix}\right.\)

Ta thấy \(\sqrt{x^2+3}\ge\sqrt{3};\left|y\right|\ge0\Leftrightarrow VT\left(1\right)\ge\sqrt{3}=VP\left(1\right)\)

Dấu \("="\Leftrightarrow x=y=0\)

Thay vào \(\left(2\right)\Leftrightarrow\sqrt{5}+0=\sqrt{5}\left(tm\right)\)

Vậy \(\left(x;y\right)=\left(0;0\right)\)

a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{ax+1}-\sqrt[]{1-bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{bx}{1+\sqrt[]{1-bx}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{a}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{b}{1+\sqrt[]{1-bx}}\right)=\dfrac{a}{3}+\dfrac{b}{2}\)

Hàm liên tục tại \(x=0\) khi:

\(\dfrac{a}{3}+\dfrac{b}{2}=3a-5b-1\Leftrightarrow8a-11b=3\)