`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)

`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`

`= -x^5 + 5x^4 + 2x^2 + 2x - 4`

`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)

`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`

`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

`= x^4 + 3x^3 + 2x - 4`

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`

a) Ta có:

A = (a - 1)x^3 + 4x^2 + 8x + 1

b) Ta có:

B = mx^4 - 3x^4 + 3

B = (m - 3)x^4 + 3

Vậy đáp án là:

a) A = (a - 1)x^3 + 4x^2 + 8x + 1

b) B = (m - 3)x^4 + 3

A B C D

\(\widehat{A}+\widehat{D}=70^o+110^o=180^o\) 

=> ABCD là tứ giác nội tiếp (tứ giác có tổng 2 góc đối =180 là tứ giác nt)

\(\widehat{ABD}=\widehat{ACD}\) (góc nt cùng chắn cung AD) (1)

\(\widehat{CBD}=\widehat{CAD}\) (góc nt cùng chắn cung CD) (2)

Tg ADC cân tại D \(\Rightarrow\widehat{ACD}=\widehat{CAD}\) (3)

Từ (1) (2) (3) \(\Rightarrow\widehat{ABD}=\widehat{CBD}\)

560 (tận cùng 0) => Chia hết cho 5

605 (tận cùng 5) => Chia hết cho 5

650 (tận cùng 0) => Chia hết cho 5

600 => Không có chữ số 5

655 => Không có chữ số 0

Các số thoả mãn: 560; 605; 650

Tổng, hiệu nào sau đây chia hết cho 2?

A B C D M N P Q

\(S_{BMN}=\dfrac{1}{2}xBMxBN=\dfrac{1}{2}x\dfrac{AB}{4}x\dfrac{BC}{2}=\dfrac{1}{16}xS_{ABCD}\)

\(S_{CPN}=\dfrac{1}{2}xCNxCP=\dfrac{1}{2}x\dfrac{BC}{2}x\dfrac{CD}{2}=\dfrac{1}{8}xS_{ABCD}\)

\(S_{DPQ}=\dfrac{1}{2}xPDxDQ=\dfrac{1}{2}x\dfrac{CD}{2}x\dfrac{AD}{3}=\dfrac{1}{12}xS_{ABCD}\)

\(S_{AMQ}=\dfrac{1}{2}xAMxAQ=\dfrac{1}{2}x\dfrac{3xAB}{4}x\dfrac{2xAD}{3}=\dfrac{1}{4}xS_{ABCD}\)

\(\Rightarrow S_{MNPQ}=S_{ABCD}-\left(S_{BMN}+S_{CPN}+S_{DPQ}+S_{AMQ}\right)\)

Bạn tự thay số rồi tính nốt nhé

Lời giải:

Xét thừa số tổng quát:
\(1-\frac{1}{1+2+...+n}=\frac{(1+2+...+n)-1}{1+2+...+n}=\frac{\frac{n(n+1)}{2}-1}{\frac{n(n+1)}{2}}=\frac{n(n+1)-2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}\)

Thay $n=2,3,....,$ ta được:

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{(n-1)(n+2)}{n(n+1)}\)

\(=\frac{[1.2.3....(n-1)][4.5.6..(n+2)]}{(2.3.4..n)[3.4.5...(n+1)]}\)

\(=\frac{1}{n}.\frac{n+2}{3}=\frac{n+2}{3n}\)

\(\frac{1}{P}=\frac{3n}{n+2}\in\mathbb{Z}\) khi mà $3n\vdots n+2$

$\Leftrightarrow 3(n+2)-6\vdots n+2$

$\Leftrightarrow 6\vdots n+2$

$\Rightarrow n+2\in\left\{6\right\}$ (do $n+2\geq 4$ với mọi $n\geq 2$)

$\Rightarrow n=4$

\(VT=\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}=\dfrac{a}{ab+a+b+a^2}+\dfrac{b}{ab+a+b+b^2}\)

\(=\dfrac{a}{\left(a+b\right).\left(a+1\right)}+\dfrac{b}{\left(a+b\right).\left(b+1\right)}\)

\(=\dfrac{\left(a+b\right).\left(ab+a+ab+b\right)}{\left(a+b\right)^2.\left(a+1\right).\left(b+1\right)}=\dfrac{ab+1}{\left(a+b\right).\left(ab+a+b+1\right)}\)

\(=\dfrac{ab+1}{2.\left(a+b\right)}\)(1)

\(VP=\dfrac{ab+1}{\sqrt{2\left(1+a^2\right)\left(1+b^2\right)}}=\dfrac{ab+1}{\sqrt{2\left(a+b\right)^2.\left(a+1\right).\left(b+1\right)}}\)

\(=\dfrac{ab+1}{2\left(a+b\right)}\) (2)

Từ (1) (2) => ĐPCM

Giải

Với a,b > 0, ta có:

\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}=\dfrac{1+ab}{\sqrt{2\left(1+a^2\right)\left(1+b^2\right)}}\)

Tương đương

\(\dfrac{a+ab^2+b+a^2b}{\left(1+a^2\right)\left(1+b^2\right)}=\dfrac{1+ab}{\sqrt{2\left(1+a^2\right)\left(1+b^2\right)}}\\ \Leftrightarrow\dfrac{a+b+ab\left(a+b\right)}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+a^2\right)\left(1+b^2\right)}}=\dfrac{1+ab}{\sqrt{2\left(1+a^2\right)\left(1+b^2\right)}}\\ \Leftrightarrow\dfrac{\left(a+b\right)\left(ab+1\right)}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)}}=\dfrac{1+ab}{\sqrt{2}}\\ \Leftrightarrow\dfrac{\left(a+b\right)}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)}}=\dfrac{1}{\sqrt{2}}\)

Mặt khác, \(\left(1+a^2\right)\left(1+b^2\right)=\left(a^2+a+b+ab\right)\left(b^2+a+b+ab\right)\\ =\left(a+b\right)\left(a+1\right)\left(a+b\right)\left(b+1\right)\\ =\left(a+b\right)^2\left[\left(a+1\right)\left(b+1\right)\right]\\ =\left(a+b\right)^2\left(a+b+ab+1\right)\\ =2\left(a+b\right)^2\)

Do đó phương trình đã cho tương đương:

\(\Leftrightarrow\dfrac{\left(a+b\right)}{\sqrt{2\left(a+b\right)^2}}=\dfrac{1}{\sqrt{2}}\\\Leftrightarrow\dfrac{\left(a+b\right)}{\sqrt{2}.\left(a+b\right)}=\dfrac{1}{\sqrt{2}}\left(a,b>0\right)\\ \Leftrightarrow\dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\left(1\right)\) 

Vì phương trình (1) đúng nên phương trình ban đầu cũng đúng

Suy ra điều phải chứng minh

Mong mn trl mình ạ

Ta có \(x^2+\dfrac{1}{x^2}=7\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=9\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2=9\)

\(\Leftrightarrow x+\dfrac{1}{x}=3\)  (Do x > 0) (1)  

Từ (1) \(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\Leftrightarrow x^3+\dfrac{1}{x^3}+3.\left(x+\dfrac{1}{x}\right)=27\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

Ta lại có \(\left(x+\dfrac{1}{x}\right)^5=x^5+5x^3+10x+\dfrac{10}{x}+\dfrac{5}{x^3}+\dfrac{1}{x^5}=243\)

\(\Leftrightarrow F=x^5+\dfrac{1}{x^5}=243-5.\left(\dfrac{1}{x^3}+x^3\right)-10.\left(x+\dfrac{1}{x}\right)=123\)