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5 tháng 1

làm hiệu trưởng hoặc bộ giáo dục

em lớp 6 tha em anh ơi

Đầu tư vào hđve là ngon luôn

ngu vãi HDPE mới đúng

5 tháng 1

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Học nhiều dậy sớm khi học lưu ý các ý chính và ghi nhớ để tránh bị dài dòng ko hiểu thì hỏi luôn và chúng ta sẽ gõ nhịp điệu khi học thuộc để nhớ nhanh, lâu

AH
Akai Haruma
Giáo viên
12 tháng 1 2018

Bài này sẽ có lời giải nếu sử dụng công thức toán cao cấp.

Bậc THPT không giải quyết những bài toán có nguyên hàm không sơ cấp.

20 tháng 2 2023

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AH
Akai Haruma
Giáo viên
14 tháng 12 2016

Giải như sau:

Ta biết rằng \(d\left(u\left(x\right)\right)=u\left(x\right)'d\left(x\right)\)

\(\Rightarrow\int\frac{x}{2-x^2}dx=\frac{1}{2}\int\frac{d\left(x^2\right)}{2-x^2}=-\frac{1}{2}\int\frac{d\left(2-x^2\right)}{2-x^2}=-\frac{1}{2}ln\left|2-x^2\right|+c\)

P/s: Muốn tính nguyên hàm mà tử nhỏ hơn mẫu thứ nhất bạn có thể phan tích mẫu ra thành các nhân tử có bậc nhỏ như bậc của tử số, rồi từ đó đặt ẩn phụ hoặc tách ghép hợp lý. Thứ 2 là bạn có thể sử dụng phương pháp $d(u(x))=u(x)'dx$ để đưa ẩn về cùng một mối ( như cách mình giải bài này). Nói chung mình diễn đạt có thể không rõ ràng một chút nhưng chủ yếu bạn làm nhiều tìm tòi nhiều sẽ quen thôi :)

 

31 tháng 5 2016

 

Câu này mình sót +3x ở sau cùng vế phải. Không hiểu vì sao đánh rồi mà lại bị mất

26 tháng 6 2016

1) bạn dùng dấu U 

điều kiện \(\begin{cases}m\ne0,m>-\frac{1}{4}\\m< 1\end{cases}\)

muons dễ nhìn thì vẽ trục số:  0 -1/4 1 x

=> điều kiện x \(\in\left(-\frac{1}{4};1\right)\backslash\left\{0\right\}\)

AH
Akai Haruma
Giáo viên
21 tháng 2 2017

Giải:

a) Xét \(y'=3x^2+2mx\)

Ta thấy \(y'=3x^2+2mx=0\)\(\Delta'=m^2>0\forall m\neq 0\) nên luôn có hai nghiệm phân biệt, đồng nghĩa với hàm số đã cho luôn có cực đại, cực tiểu với mọi \(m\neq 0\)

b) Đồ thị hàm số luôn cắt trục hoành tại điểm có hoành độ dương với mọi giá trị của $m$ nghĩa là phương trình \(x^3+mx^2-1=0\) luôn có nghiệm dương với mọi \(m\)

Xét hàm $y$ liên tục trên tập xác định.

Nếu \(m>0\)\(\left\{\begin{matrix} f(0)=-1<0\\ f(m+1)=(m+1)^3+m(m+1)^2-1>0\end{matrix}\right.\Rightarrow f(0).f(m+1)<0\)

Do đó phương trình luôn có nghiệm thuộc khoảng \((0;m+1)\), tức là nghiệm dương.

Nếu \(m<0\)\(\left\{\begin{matrix} f(0)=-1<0\\ f(1-m)=m^2-2m>0\forall m<0\end{matrix}\right.\Rightarrow f(0).f(1-m)<0\)

Do đó phương trình luôn có nghiệm thuộc khoảng \((0,1-m)\) , tức nghiệm dương

Từ hai TH ta có đpcm.

c) Để pt có $3$ nghiệm phân biệt thì \(y'=3x^2+2mx\) phải có hai nghiệm phân biệt \(x_1,x_2\) thỏa mãn \(f(x_1)f(x_2)<0\)

Kết hợp với định lý Viete:

\(\Leftrightarrow x_1^3+x_2^3+m(x_1^2+x_2^2)-1>0\)

\(\Leftrightarrow 4m^3-27>0\Leftrightarrow m>\frac{3}{\sqrt[3]{4}}\)