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12 tháng 5 2018

Đặt \(t=x^2+x-5\)

\(\Leftrightarrow\frac{t}{x}+\frac{3x}{t}+4=0\)

\(\Leftrightarrow\frac{t^2+4xt+3x^2}{xt}=0\)

\(\Leftrightarrow t^2+4xt+3x^2=0\)

\(\Leftrightarrow t^2+xt+3xt+3x^2=0\)

\(\Leftrightarrow t\left(t+x\right)+3x\left(t+x\right)=0\)

\(\Leftrightarrow\left(t+3x\right)\left(t+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+3x=0\\t+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2+x-5+3x=0\\x^2+x-5+x=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+4x-5=0\\x^2+2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+5x-x-5=0\\x^2+2x+1-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)-\left(x+5\right)=0\\\left(x+1\right)^2=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)\left(x+5\right)=0\\x+1=\sqrt{6};x+1=-\sqrt{6}\end{cases}}\)

\(\Leftrightarrow x=1,x=5,x=-1+\sqrt{6},x=-1-\sqrt{6}\)

Vậy \(S=\left\{1;5;-1+\sqrt{6};-1-\sqrt{6}\right\}\)

31 tháng 1 2020

Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
\(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)

31 tháng 1 2020

Câu b : \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)

<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm

8 tháng 2 2020

Bài 2 :

a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)

=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)

c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)

d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)

e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)

=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)

f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)

=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)

8 tháng 2 2020

Bài 1 :

a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)

=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)

=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)

=> \(12-3x-9-2x+4=0\)

=> \(-5x=-7\)

=> \(x=\frac{7}{5}\)

1 tháng 4 2019

a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

1 tháng 4 2019

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

31 tháng 1 2020

a) \(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)

 \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+8x}=\frac{9}{52}\)

\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+10\right)\left(x+8\right)}-\frac{9}{52}=0\)

\(\Leftrightarrow\frac{104\left(x+10\right)\left(x+8\right)+260\left(x+1\right)\left(x+10\right)+104\left(x+1\right)\left(x+3\right)-9\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

Đoạn này cậu tự phân tích tử rồi rút gọn nhé :D Vì hơi dài nên viết ra đây sẽ rối, k đẹp mắt cho lắm :>

\(\Leftrightarrow\frac{-927x^2+1782x+9072-9x^4-198x^3}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x^4+22x^3+103x^2-198x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x^4-3x^3+25x^3-75x^{^2}+178x^2-534x+336x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left[x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+25x^2+178x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+14x^2+11x^2+154x+24x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left[x^2\left(x+14\right)+11x\left(x+14\right)+24\left(x+14\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x+14\right)\left(x^2+11x+24\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)=0}\)

\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)\left(x+3\right)\left(x+8\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)}{52\left(x+1\right)\left(x+10\right)}=0\)

\(\Leftrightarrow-9x^2-99x+378=0\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Leftrightarrow\left(x+14\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+14=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-14\\x=3\end{cases}}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-14;3\right\}\)

b) \(ĐKXĐ:x\ne-1\)

 \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow x^2+\frac{x^2}{\left(x+1\right)^2}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4x^2\left(x^2+2x+1\right)+4x^2-5\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow4x^4+8x^3+4x^2+4x^2-5x^2-10x-5=0\)

\(\Leftrightarrow4x^2+8x^3+3x^2-10x-5=0\)

\(\Leftrightarrow4x^4+2x^3+6x^3+3x^2-10x-5=0\)

\(\Leftrightarrow2x^3\left(2x+1\right)+3x^2\left(2x+1\right)-5\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x^3+3x^2-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x^3-2x^2+5x^2-5x+5x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x^2\left(x-1\right)+5x\left(x-1\right)+5\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(2x^2+5x+5\right)=0\)

\(\Leftrightarrow2x+1=0\)                                 

hoặc \(x-1=0\)                                    

hoặc \(2x^2+5x+5=0\)                   

\(\Leftrightarrow\) \(x=-\frac{1}{2}\left(tm\right)\)

hoặc \(x=1\left(tm\right)\)

hoặc \(\left(x+\frac{5}{4}\right)^2+\frac{55}{16}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2};1\right\}\)

c) \(x^4-3x^3+2x^2-9x+9=0\)

\(\Leftrightarrow x^4-x^3-2x^3+2x^2-9x+9=0\)

\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)+\left(x^2-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2+x+3\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)

hoặc \(x-3=0\)

hoặc \(x^2+x+3=0\)

\(\Leftrightarrow\)\(x=1\left(tm\right)\)

hoặc \(x=3\left(tm\right)\)

hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là :\(S=\left\{1;3\right\}\)

3 tháng 2 2020

\(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+8\right)-\left(x+3\right)}{\left(x+3\right)\left(x+8\right)}+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+10\right)}\)

\(=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)

\(\Leftrightarrow x^2+11x+10=52\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Delta=11^2+4.42=289,\sqrt{289}=17\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+17}{2}=3\\x=\frac{-11-17}{2}=-14\end{cases}}\)

Vậy pt có 2 nghiệm là 3 và -14

5 tháng 3 2020

a) \(2\left(x-1\right)-a\left(x-1\right)=2a+3\)

\(\Leftrightarrow2a-2-ax+a=2a+3\)

\(\Leftrightarrow-2-ax+a=3\)

\(\Leftrightarrow-a\left(x-1\right)=5\)

\(\Leftrightarrow\left(x-1\right)=\frac{-5}{a}\Leftrightarrow x=\frac{a-5}{a}\)

5 tháng 3 2020

b) \(\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=3\)

\(\Leftrightarrow\frac{12x+12+8x+16+6x+18}{24}=3\)

\(\Leftrightarrow12x+12+8x+16+6x+18=72\)

\(\Leftrightarrow26x+46=72\)

\(\Leftrightarrow26x=26\Leftrightarrow x=1\)

4 tháng 4 2020

a) 5 - (x - 6) = 4(3 - 2x)

<=> 5 - x + 6 = 12 - 8x

<=> -x + 8x = 12 - 11

<=> 7x = 1

<=> x = 1/7

Vậy S = {1/7}

b) 2x(x - 3) + 5(x - 3) = 0

<=> (2x + 5)(x - 3) = 0

<=> \(\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)

Vậy S = {-5/2; 3}

c)ĐK: x \(\ne\)1; x \(\ne\)2

 \(\frac{3x-5}{x-2}-\frac{2x-5}{x-1}=1\)

<=> \(\frac{\left(3x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)

<=> 3x2 - 8x + 5 - 2x2 + 9x - 10 = x2 - 3x + 2

<=> x2 + x - 5 = x2 - 3x + 2

<=> x+ x  - x2 + 3x = 2 + 5

<=> 4x = 7

<=> x = 7/4 

Vậy S = {7/4}

28 tháng 4 2020

1,(3x-2)(4x+5)=0

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-5}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là ...

2,\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=15-28+19-22\)

\(\Leftrightarrow-8x=-16\)

=> x= 2

vậy..

3,\(\left(x^2-2x+1\right)-4=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1\right)-4=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}-4=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}=0\) ( vô nghiệm )

(vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\ge0\) )

từ đó suy ra phương trình vô nghiệm

5,\(\frac{4x+3}{2}-2+3x=\frac{2x-1}{10}+\frac{19x+2}{5}-1\)

\(\Leftrightarrow\frac{5\left(4x+3\right)}{10}-\frac{10\left(2-3x\right)}{10}=\frac{2x-1}{10}+\frac{2\left(19x+2\right)}{10}-\frac{10}{10}\)

\(\Leftrightarrow\frac{20x+15}{10}-\frac{20-30x}{10}=\frac{2x-1}{10}+\frac{38x+4}{10}-\frac{10}{10}\)

\(\Rightarrow20x+15-20+30x=2x-1+38x+4-10\)

\(\Leftrightarrow20x+30x-2x-38x=-15+20-1+4-10\)

\(\Leftrightarrow10x=-2\)

\(\Leftrightarrow x=-5\)

Vậy ....

p/s : thực ra mk cx chỉ ms học th nên giải bài tập về phương trình vẫn còn nhiều chỗ sai nữa,có gì mong mn giúp đỡ :)