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8 tháng 12 2017

Q = \(\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)để Q đạt GTLN => \(x^2+1phảiNhỏnhất\)

\(x^2+1\ge1=>x^2+1\)đạt GTNN là 1 khi x=0

vậy Q đạt GTLN =3 khi x = 0

9 tháng 12 2017

\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}\\ Q=\dfrac{3}{x^2+1}\\ Do\text{ }x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\\ Q=\dfrac{3}{x^2+1}\le3\forall x\\ \text{Dấu “=” xảy ra khi : }\\ x^2=0\\ \Leftrightarrow x=0\\Vậy\text{ }Q_{\left(Max\right)}=3\text{ }khi\text{ }x=0\)

13 tháng 11 2017

\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)

10 tháng 3 2017

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)

ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)

\(\Leftrightarrow3x^2+15x-378=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)

10 tháng 3 2017

@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T

15 tháng 10 2018

\(\left(x-3\right)^2-\left(x^2-3x\right)=0\)

\(\left(x-3\right).\left(x-3\right)-x.\left(x-3\right)=0\)

\(\left(x-3\right).\left(x-3-x\right)=0\)

\(\left(x-3\right).3=0\)

\(x-3=0=>x=3\)

15 tháng 10 2018

sorry đoạn này

\(\left(x-3\right).\left(-3\right)=0=>....\)

bn sửa lại cái dòng thứ 3 nha

P/S: Mk mới lớp 7 làm sai sót thì sorry 

a: ĐKXĐ: x<>1; x<>2; x<>3

\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)

\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)

\(=\dfrac{2x^3-4x^2}{\left(x-2\right)}\cdot\dfrac{1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\dfrac{2x^2}{x^4+x^2+1}\)

b:

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11 tháng 3 2018

đkxđ với mọi x

đặt a=x2+x+1

\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)

<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)

=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)

<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14

<=> 12a2 -a2+24a-21a+6-14=0

<=> 11a2+3a-8=0

<=> 11a2 +11a-8a-8=0

<=> (11a2 +11a)-(8a+8)=0

<=> 11a(a+1)-8(a+1)=0

<=> (a+1)(11a-8)=0

=> a=-1 và a=\(\dfrac{8}{11}\)

thay a=x2+x+1 ta đc

x2+x+1=-1

<=> x2+x+2 =0 (vô nghiệm)

và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )

vậy pt trên vô nghiệm

12 tháng 3 2018

c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0

( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)

\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)

\(< =>16=\left(x+4\right)^2\)

<=> x2 + 8x = 0

<=> x( x + 8) = 0

<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )

Vậy,....

10 tháng 12 2017

Bài 1:

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)

10 tháng 12 2017

đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "

bn xem lại đề đi

23 tháng 7 2017

\(Q=\left(\dfrac{1}{x+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2}{x^2-x+1}\right):\left(x+2\right)\)\(=\left(\dfrac{x^2-x+1}{x^3+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2\left(x+1\right)}{x^3+1}\right).\dfrac{1}{x+2}\)\(=\left(\dfrac{x^2-x+1+6x+3-2x-2}{\left(x+1\right)\left(x-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\left(\dfrac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{1}{x+2}=\dfrac{1}{\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)Ta có:

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) \(\Rightarrow\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

Vậy Max Q = \(\dfrac{4}{3}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)