a, \(P\left(x\right)=3x^3+2x^3-2x+7-x^2-x=5x^3-3x+7-x^2\)

\(Q\left(x\right)=-3x^3+x-14-2x-x^2-1=-3x^3-x-x^2-15\)

b, \(M\left(x\right)=5x^3-3x+7-x^2-3x^3-x-x^2-15=2x^3-2x^2-4x-8\)

\(N\left(x\right)=5x^3-3x+7-x^2+3x^3+x+x^2+15=8x^3-2x+22\)

c, \(P\left(x\right)=-Q\left(x\right)\Leftrightarrow5x^3-3x+7-x^2=3x^3+x+x^2+15\)

\(\Leftrightarrow2x^3-2x^2-4x-8=0\)

a) \(P\left(x\right)=3x^3+2x^3-2x+7-x^2-x\\ =\left(3x^3+2x^3\right)-x^2+\left(-2x-x\right)+7\\ =5x^3-x^2-3x+7\)

\(Q\left(x\right)=-3x^3+x-14-2x-x^2-1\\ =-3x^3-x^2+\left(x-2x\right)+\left(-14-1\right)\\ =-3x^3-x^2-x-15\)

b) \(M\left(x\right)=5x^3-x^2-3x+7+\left(-3x^3-x^2-x-15\right)\\ =\left(5x^3-3x^3\right)+\left(-x^2-x^2\right)+\left(-3x-x\right)+\left(7-15\right)\\ =2x^3-2x^2-4x-8\)

 \(N\left(x\right)=5x^3-x^2-3x+7-\left(-3x^3-x^2-x-15\right)\\ =5x^3-x^2-3x+7+3x^3+x^2+x+15\\ =\left(5x^3+3x^3\right)+\left(x^2-x^2\right)+\left(x-3x\right)+\left(15+7\right)\\ =8x^3-2x+22\)

c) \(P\left(x\right)=-Q\left(x\right)\Rightarrow P\left(x\right)+Q\left(x\right)=0\\ \Rightarrow M\left(x\right)=0\Rightarrow2x^3-2x^2-4x-8=0\\ \Rightarrow x^3-x^2-2x-4=0\)

Bạn xem lại đề nhé

2x² - 18x + 6x -6 = 16 + 25

2x² - 12x -47 =0

\(x=\pm\dfrac{\sqrt{130}+6}{2}\)

Hằng đẳng thức: \(a^2+2ab+b^2=\left(a+b\right)^2\)

Cách chứng minh: \(VT=\left(a^2+ab\right)+\left(ab+b^2\right)=a\left(a+b\right)+b\left(a+b\right)\\ =\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=VP\)

Áp dụng:

Kiểu đề 1: \(2x\left(x-9\right)+3\left(2x\right)-6=4^2+5^2\\ \Rightarrow2x^2-18x+6x-6=16+25\\ \Rightarrow2x^2-12x-47=0\\ \Rightarrow x^2-6x-\dfrac{47}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{47}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{65}{2}=\left(\dfrac{\pm\sqrt{130}}{2}\right)^2\\\)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{130}}{2}\\x-3=\dfrac{-\sqrt{130}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{130}}{2}\\x=\dfrac{6-\sqrt{130}}{2}\end{matrix}\right.\)

Kiểu đề 2: \(2x\left(x-9\right)+3\left(2x-6\right)=4^2+5^2\\ \Rightarrow2x^2-18x+6x-18=16+25\\ \Rightarrow2x^2-12x-59=0\\ \Rightarrow x^2-6x-\dfrac{59}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{59}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{77}{2}=\left(\dfrac{\pm\sqrt{154}}{2}\right)^2\\ \)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{154}}{2}\\x-3=\dfrac{-\sqrt{154}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{154}}{2}\\x=\dfrac{6-\sqrt{154}}{2}\end{matrix}\right.\)

làm phần c thôi nhé

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

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0,\forall x,y\)

Dấu \("="\) xảy ra khi:

\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

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

\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Ta có:

\(92^3\equiv2\left(mod6\right)\)

\(\Rightarrow92^{30}\equiv\left(92^3\right)^{10}\left(mod6\right)\equiv2^{10}\left(mod6\right)\equiv4\left(mod6\right)\)

\(\Rightarrow92^{90}\equiv\left(92^{30}\right)^3\left(mod6\right)\equiv4^3\left(mod6\right)\equiv4\left(mod6\right)\)

\(\Rightarrow92^{93}\equiv92^{90}.92^3\left(mod6\right)\equiv4.2\left(mod6\right)\equiv2\left(mod6\right)\)

\(139^2\equiv1\left(mod6\right)\)

\(\Rightarrow139^{20}\equiv\left(139^2\right)^{10}\left(mod6\right)\equiv1^{10}\left(mod6\right)\equiv1\left(mod6\right)\)

\(\Rightarrow92^{93}+139^{20}+3\equiv2+1+3\left(mod6\right)\equiv6\left(mod6\right)\equiv0\left(mod6\right)\)

Vậy \(\left(92^{93}+139^{20}+3\right)⋮6\)

Tk:

Số vô tỉ là số viết được dưới dạng số thập phân vô hạn không tuần hoàn.

\(#SGK\)

tk

Trong toán học, các số vô tỉ là tất cả các số thực không phải là số hữu tỉ, mà là các số được xây dựng từ các tỷ số (hoặc phân số) của các số nguyên.

Tổng quát: \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

\(\Rightarrow C=\dfrac{9\left(9+2\right)}{\left(9+1\right)^2}=\dfrac{9.11}{10^2}=\dfrac{99}{100}\)

Vậy \(C=\dfrac{99}{100}\)

Ta có:

2! = 1.2

3! = 2.3

4! > 3.4

...

100! > 99.100

⇒ \(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(D< 1-\dfrac{1}{100}\)

\(D< \dfrac{99}{100}\)

Mà \(\dfrac{99}{100}< 1\) ⇒ \(D< 1\)

1 D

2 A

3 B

4 B

5 C

6 C

7 C

8 A

9 B

10 C

11 A

12 A

13 C

14 C

15 A

16 C

17 A

18 B

19 C

20 A