`f(x)=1+x^3+x^5+.....+x^101`

`=1+(-1-1-.....-1)`

`=1+50.(-1)`

`=-49`

=>f(1)=1+1³+1⁵+...+1¹⁰¹

=1+1+...+1

=1+1*50 (tính theo số mũ)

=51

=>f(-1)=1+(-1)³+...+(-1)¹⁰¹

=1+(-1)+(-1)+...+(-1)

=1+(-1)*50

=-49

Chọn F(x)=5x-23

\(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5x-23-2}{x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{5x-25}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5\left(x-5\right)}{x-5}=5\)

=>f(x)=5x-23 thỏa mãn yêu cầu đề bài

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\cdot f\left(x\right)+10}+\sqrt{f^3\left(x\right)+1}-7}{x^2-25}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\left(5x-23\right)+10}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}-4+\sqrt{\left(5x-23\right)^3+1}-3}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15x-59-16}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3+1-9}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3-8}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23-2\right)\left[\left(5x-23\right)^2+2\left(5x-23\right)+4\right]}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15}{\sqrt{15x-59}+4}+\dfrac{5\cdot\left(25x^2-230x+529+10x-46+4\right)}{\sqrt{\left(5x-23\right)^3+1}+3}}{x+5}\)

\(=\dfrac{\dfrac{15}{\sqrt{15\cdot5-59}+4}+\dfrac{5\left(25\cdot5^2-220\cdot5+487\right)}{\sqrt{\left(5\cdot5-23\right)^3+1}+3}}{5+5}\)

\(=\dfrac{\dfrac{15}{8}+\dfrac{5\cdot12}{6}}{10}=\dfrac{19}{16}\)

Do \(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}\) hữu hạn nên \(f\left(x\right)-2=0\) có nghiệm \(x=5\)

\(\Rightarrow f\left(5\right)=2\)

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}-4+\sqrt{f^3\left(x\right)+1}-3}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{3\left[f\left(x\right)-2\right]}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{\left[f\left(x\right)-2\right]\left[f^2\left(x\right)+2f\left(x\right)+4\right]}{\sqrt{f^3\left(x\right)+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{f\left(x\right)-2}{x-5}.\dfrac{3}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{f\left(x\right)-2}{x-5}.\dfrac{f^2\left(x\right)+2f\left(x\right)+4}{\sqrt{f^3\left(x\right)+1}+3}}{x+5}\)

\(=\dfrac{5.\dfrac{3}{\sqrt{3.2+10}+4}+5.\dfrac{2^2+2.2+4}{\sqrt{2^3+1}+3}}{5+5}=\)

Ta có: f(1) = 1 + 1^3 + 1^5 + 1^7 +...+ 1^101

               =  1 + 50.1

               = 1 + 50           

               = 51

Vậy f(1) = 51

Có:  f(-1) = 1 + (-1)^3 + (-1)^5 + (-1)^7 + ... + (-1)^101

              = 1 + 50.(-1)

              = 1 - 50 

              = -49

Vậy f(-1) = -49

Chúc bạn học tốt nha

Ta có:

Với x = -2 

=> \(0.f\left(-2+1\right)=\left(-2+1\right).f\left(-2+3\right)\)

=> \(f\left(1\right)=0\) (1)

Với x = 0

=> \(2.f\left(1\right)=1.f\left(3\right)\) (2)

(1) ; (2) => \(f\left(3\right)=0\)(3)

Với  x = 2

=> \(4.f\left(3\right)=3.f\left(5\right)\)(4)

Từ (3) ; (4) => f ( 5 ) = 0.

Link ảnh: https://imgur.com/7VWoyPW

Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?

Em cần kiểu tự luận ạ

Làm tự luận thì hơi tốn thời gian đấy (đi thi sẽ không bao giờ đủ thời gian đâu)

Câu 1:

Kiểm tra lại đề, \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}\) hay một trong 2 giới hạn sau: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x}-1}{g\left(x\right)}\) hoặc \(\lim\limits_{x\rightarrow1}\dfrac{g\left(x\right)}{\sqrt[]{x}-1}\)

Vì đúng như đề của bạn thì \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}=\dfrac{1}{0}=\infty\), cả \(g\left(x\right)\) lẫn \(\sqrt{x}-1\) đều tiến tới 0 khi x dần tới 1

\(\text{1)}\)

\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)

\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)

\(f\left(3\right)=3^2-5\)

\(\text{2)}\)

\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)

\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)

\(\text{3)}\)

\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)

\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)

\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)

\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)

sai 1 chút chỗ cÂU 3

nhân vs 3 thì phải là 1/12

thay x bằng ? mik cũng ko bit làm lên vào đây tham khảo hihihihi