Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(-537x^2+5054x=-541x^2+5092x\)
\(-537x^2+5054x+541x^2-5092x=0\)
\(4x^2-38x=0\)
\(x\left(2x-19\right)=0\)
\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)
Bài 3 :
Ta có : \(A=x^2+x+2012\)
=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)
=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)
- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)
<=> \(x=-\frac{1}{2}\)
Vậy MinA = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .
Bài 1 :
a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .
b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
=> \(x\ne\pm1\)
Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
=> \(\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}=\frac{x^2-3}{x^2-1}\)
=> \(\left(x+1\right)^2-4\left(x-1\right)=x^2-3\)
=> \(x^2+2x+1-4x+4=x^2-3\)
=> \(-2x=-3-5\)
=> \(x=4\left(TM\right)\)
Vậy phương trình có nghiệm là x = 4 .
c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)
=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)
=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)
=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)
=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)
=> \(10x+2012=0\)
=> \(x=-\frac{2012}{10}\)
Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .
Bài 3:
Giải:
Ta có : A = x2 + x + 2012
= x2 + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)
= (x + \(\frac{1}{2}\))2 + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)
⇒ Amin = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))2 = 0 ⇔ x = \(-\frac{1}{2}\)
Vậy Amin = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)
Chúc bạn học tốt@@
a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)
\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(8x^3-46x^2-60x-180=0\)
=> vô nghiệm
b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)
=> 3x-6=0
<=> x=2 (ktm)
Vậy pt vô nghiệm
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
Ta có
\(-x^2+6x-8=-\left(x^2-6x+8\right)=-\left(x^2-2x-4x+8\right)=-\left[x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=-\left(x-2\right)\left(x-4\right)\)
MTC:\(\left(x-2\right)\left(x-4\right)\)
\(\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{x^2-x-4x+4+x^2+3x-2x-6}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{2x^2-4x-2}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(2x^2-4x-2+2=0\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\Rightarrow x=0\\x-4=0\Rightarrow x=4\end{cases}}\)
a) trước nha
T i c k cho mình nha bạn cảm ơn sẽ làm típ câu b)
b) x3 - 8 - (x - 2) (x + 2) = 0
=> x3 - (2)3 - x2 - 22 = 0
mk làm bậy thui!!! mới lên lớp 8 mà!! 6756886787696969768658585685685685858978467