4.

\(\lim\limits_{x\rightarrow8}f\left(x\right)=\lim\limits_{x\rightarrow8}\dfrac{\sqrt[3]{x}-2}{x-8}=\lim\limits_{x\rightarrow8}\dfrac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}=\lim\limits_{x\rightarrow8}\dfrac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\)

\(=\dfrac{1}{4+4+4}=\dfrac{1}{12}\)

\(f\left(8\right)=3.8-20=4\)

\(\Rightarrow\lim\limits_{x\rightarrow8}f\left(x\right)\ne f\left(8\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=8\)

5.

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

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{2}{\sqrt[]{1+2x}+1}-\dfrac{3}{1+\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)=\dfrac{2}{1+1}-\dfrac{3}{1+1+1}=0\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(3x^2-2x\right)=0\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=f\left(0\right)\)

\(\Rightarrow\) Hàm liên tục tại \(x=0\)

6.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{4x+1}-\sqrt[3]{6x+1}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{4x+1}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{6x+1}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{-x^2}{\sqrt[]{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{-1}{\sqrt[]{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-1}{1+1}+\dfrac{12}{1+1+1}=\dfrac{7}{2}\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(2-3x\right)=2\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)\ne\lim\limits_{x\rightarrow0^-}f\left(x\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=0\)

Nghề bác sĩ khiến tôi có thêm động lực học hành

câu trên phân tích như sau:

nghề bác sĩ là chủ ngữ , khiến tôi có thêm động lực học hành là vị ngữ

tôi là chủ ngữ , có thêm động lức học hành là vị ngữ

Mong bạn có thể tách ra để hỏi vì câu hỏi hơi nhiều :))

6:

a: \(\Leftrightarrow\dfrac{\sqrt{4}-\sqrt{1}}{4-1}+\dfrac{\sqrt{7}-\sqrt{4}}{7-4}+...+\dfrac{\sqrt{3n+4}-\sqrt{3n+1}}{3}=8\)

=>\(-\sqrt{1}+\sqrt{4}-\sqrt{4}+\sqrt{7}-...-\sqrt{3n+1}+\sqrt{3n+4}=24\)

=>\(\sqrt{3n+4}=24+1=25\)

=>3n+4=625

=>3n=621

=>n=207

b: \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\cdot\sqrt{n+1}}=\dfrac{4}{5}\)

=>\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

=>\(1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

=>n+1=25

=>n=24

Dcm giúp tớ với

Nhanh nhanh dùm đi mà

2A:

a: \(a^3+12a^2+48a+64\)

\(=a^3+3\cdot a^2\cdot4+3\cdot a\cdot4^2+4^3\)

\(=\left(a+4\right)^3\)

b: \(-b^3+6b^2-12b+8\)

\(=\left(-b\right)^3+3\cdot\left(-b\right)^2\cdot2+3\cdot\left(-b\right)\cdot2^2+2^3=\left(-b+2\right)^3\)

c: \(\left(m-n\right)^6-6\left(m-n\right)^4+12\left(m-n\right)^2-8\)

\(=\left\lbrack\left(m-n\right)^2\right\rbrack^3-3\cdot\left\lbrack\left(m-n\right)^2\right\rbrack^2\cdot2+3\cdot\left(m-n\right)^2\cdot2^2-2^3\)

\(=\left\lbrack\left(m-n\right)^2-2\right\rbrack^3\)

d: \(\frac{8}{27}a^3-\frac83a^2b+8b^2a-8b^3\)

\(=\left(\frac23a\right)^3-3\cdot\left(\frac23a\right)^2\cdot2b+3\cdot\left(\frac23a\right)\cdot\left(2b\right)^2-\left(2b\right)^3\)

\(=\left(\frac23a-2b\right)^3\)

Bài 2B:

a: \(\frac18x^3+\frac34x^2y^2+\frac32xy^4+y^6\)

\(=\left(\frac12x\right)^3+3\cdot\left(\frac12x\right)^2\cdot y^2+3\cdot\frac12x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)

\(=\left(\frac12x+y^2\right)^3\)

b: \(m^3+9m^2n+27mn^2+27n^3\)

\(=m^3+3\cdot m^2\cdot\left(3n\right)+3\cdot m\cdot\left(3n\right)^2+\left(3n\right)^3\)

\(=\left(m+3n\right)^3\)

c: \(8u^3-48u^2v+96uv^2-64v^3\)

\(=\left(2u\right)^3-3\cdot\left(2u\right)^2\cdot4v+3\cdot\left(2u\right)\cdot\left(4v\right)^2-\left(4v\right)^3\)

\(=\left(2u-3v\right)^3\)

d: \(\left(z-t\right)^3+15\left(z-t\right)^2+75\left(z-t\right)+125\)

\(=\left(z-t\right)^3+3\cdot\left(z-t\right)^2\cdot5+3\cdot\left(z-t\right)\cdot5^2+5^3=\left(z-t+5\right)^3\)