(6x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1
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1 tháng 3 2022
b: \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5-6x}{4-x^2}\)
\(=\dfrac{4x-8+2x+4+6x-5}{\left(x-2\right)\left(x+2\right)}=\dfrac{12x-9}{\left(x-2\right)\left(x+2\right)}\)
c: \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
e: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-x^2+7x+78}{x\left(x+6\right)}\)
\(=\dfrac{-x^2+13x-6x+78}{x\left(x+6\right)}\)
\(=\dfrac{-x\left(x-13\right)-6\left(x-13\right)}{x\left(x+6\right)}\)
\(=\dfrac{\left(13-x\right)\left(x+6\right)}{x\left(x+6\right)}=\dfrac{13-x}{x}\)
HV
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NT
1
18 tháng 2 2020
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CG
0
26 tháng 11 2016
1)\(2x^2+9y^2-6xy-6x-12y+2004\)
\(=x^2+x^2-6xy+9y^2-6x-12y+2004\)
\(=x^2+\left(x-3y\right)^2-10x+4x-12y+2004\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975\)
\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975\)
\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)
Dấu "=" khi \(\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)
Vậy Min=1975 khi \(\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)
2)\(x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt \(t=x^2+x\) ta có:
\(t\left(t-4\right)=t^2-4t+4-4\)
\(=\left(t-2\right)^2-4\ge-4\)
Dấu "=" khi \(t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)
Vậy Min=-4 khi \(\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)
3)\(\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]\)
\(=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]\)
Đặt \(t=x^2+5x+5\) ta có:
\(t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" khi \(t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}\)\(\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)
Vậy Min=\(-\frac{1}{4}\) khi \(x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)
4)\(\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)
\(=\left(x^2-4x+3\right)\left(x^2-4x+5\right)\)
Đặt \(t=x^2-4x+3\) ta có:
\(t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)
Dấu "=" khi \(t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2\)
Vậy Min=-1 khi x=2
11 tháng 8 2017
Câu a :
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)
Câu b :
\(2x^2+3=-5x\)
\(\Leftrightarrow2x^2+3+5x=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\2x+3=0\Rightarrow x=-\dfrac{3}{2}\end{matrix}\right.\)
Mấy câu sau khó quá ko bt làm :)
NM
9 tháng 9 2019
1. -6x .(x2-5x+4)-(x+1)2
=-6x3+30x2-24x-x2-2x-1
-6x3+29x2-26x-1
2. (X+3)2-4x(x-7)
=x2+6x+9-4x2+28x
=-3x2+34x+9
3.(5x-2)2-(3x-2). (X+1)
=25x2- 20x+4-3x2-3x+2x+2
=22x2-21x+6
\((6x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1\)
\(\Leftrightarrow36x^2+12x+1-2\left(x^3+3x^2+3x+1\right)+2\left(x^3-1\right)=1\)
\(\Leftrightarrow36x^2+12x+1-2x^3-6x^2-6x-2+2x^3-2=1\)
\(\Leftrightarrow30x^2+6x-4=0\)\(\Leftrightarrow2\left(15x^2+3x-2\right)=0\)
\(\Leftrightarrow15x^2+3x-2=0\)\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2-\dfrac{43}{20}=0\)
\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2=\dfrac{43}{20}\Leftrightarrow x=\pm\dfrac{\sqrt{129}}{30}-\dfrac{1}{10}\)