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\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
Bài 1
a) (x5 + 4x3 - 6x2) : 4x2
= 4x2(\(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)) : 4x2
= \(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)
b) (x3 - 8) : (x2 + 2x + 4)
= (x - 2)(x2 + 2x + 4) : (x2 + 2x + 4)
= x - 2
c) (3x2 - 6x) : (2 - x)
= -(6x - 3x2) : (2 - x)
= -3x(2 - x) : (2 - x)
= -3x
d) (x3 + 2x2 - 2x - 1) : (x2 + 3x + 1)
= [(x3 - 1) + (2x2 - 2x)] : (x2 + 3x + 1)
= [(x - 1)(x2 + x + 1) + 2x(x - 1)] : (x2 + 3x + 1)
= (x - 1)(x2 + x + 1 + 2x) : (x2 + 3x + 1)
= (x - 1)(x2 + 3x + 1) : (x2 + 3x + 1)
= x - 1
Bài 2
a) (x - 4)2 - (x - 2)(x + 2) = 6
x2 - 8x + 16 - (x2 - 4) = 6
x2 - 8x + 16 - x2 + 4 = 6
-8x + 20 = 6
\(\Rightarrow\) -8x = - 14
\(\Rightarrow\) x = \(\dfrac{7}{4}\)
b) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
9(x2 + 2x + 1) - (9x2 - 4) = 10
9x2 + 18x + 9 - 9x2 + 4 = 10
18x + 13 = 10
\(\Rightarrow\) 18x = -3
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Nhớ tik mik nha 
không lần sau mik ko giúp đâu 
AK... có j ko hiểu thì bn cứ bình luận bên dưới 
1) \(=9x^2-1\)
2) \(=9x^4-y^2\)
3)\(=25x^2-\dfrac{9}{4}\)
4) \(=x^3-1\)
5) \(=x^6-8\)
6) \(=x^3-64\)
7) \(=27x^3+8\)
8) \(=x^3-64\)
9) \(=x^3-\dfrac{1}{27}\)
10) \(x^3+\dfrac{1}{27}\)
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
a, \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\left(ĐKXĐ:x\ne\pm2;\pm5\right)\)
\(\frac{x+9}{\left(x-5\right)\left(x+2\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}=\frac{1}{x+2}\)
\(\frac{\left(x+9\right)\left(x+5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}=\frac{\left(x+5\right)\left(x-5\right)}{\left(x+2\right)\left(x+5\right)\left(x-5\right)}\)
Khử mẫu : \(\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)=\left(x+5\right)\left(x-5\right)\)
\(x^2+14x+45-x^2-17x-30=x^2-25\)
\(-3x+15-x^2+25=0\)
\(-3x-x^2+40=0\)( giải delta ta đc )
\(x_1=-5;x_2=8\)
b, \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1ĐKXĐ\left(x\ne1;\frac{1}{3}\right)\)
\(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=1\)
\(\frac{x-1}{\left(3x-1\right)\left(x-1\right)}+\frac{\left(2x+2\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=\frac{\left(3x-1\right)\left(x-1\right)}{\left(3x-1\right)\left(x-1\right)}\)
Khửi mẫu \(x-1+\left(2x+2\right)\left(3x-1\right)-3x^2-1=\left(3x-1\right)\left(x-1\right)\)( bn tự nốt nhé)
c, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)
\(x^2+6x+9-10\ge x^2+5x+6-4\)
\(x-3\ge0\Leftrightarrow x\ge3\)
a) \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\); ĐKXĐ: x # -2; x # +-5
<=> \(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+2}\)
<=> \(\frac{\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}\)
<=> (x + 9)(x + 5) - (x + 15)(x + 2) = (x - 5)(x + 5)
<=> -3x + 15 = x^2 - 25
<=> -3x + 15 - x^2 + 25 = 0
<=> -3x + 40 - x^2 = 0
<=> x^2 + 3x - 40 = 0
<=> (x - 5)(x + 8) = 0
<=> x - 5 = 0 hoặc x + 8 = 0
<=> x = 5 (ktm0 hoặc x = -8 (tm)
b) \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1\); ĐKXĐ: x # 1/3; x # 1
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{x\left(3x-1\right)-\left(3x-1\right)}=1\)
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=1\)
<=> \(\frac{x-1}{\left(x-1\right)\left(3x-1\right)}+\frac{2\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}\)
<=> x - 1 + 2(x + 1)(3x - 1) - 3x^2 + 1 = (x - 1)(3x - 1)
<=> 5x - 4 + 3x^2 = 3x^2 - 4x + 1
<=> 5x - 4 = -4x + 1
<=> 5x + 4x = 1 + 4
<=> 9x = 5
<=> x = 5/9 (tm)
c) (x + 3)^2 - 10 >= (x + 3)(x + 2) - 4
<=> x^2 + 3x + 3x + 9 - 10 >= x^2 + 2x + 3x + 6 - 4
<=> x^2 + 6x + 9 - 10 >= x^2 + 5x + 6 - 4
<=> x^2 + 6x - 1 >= x^2 + 5x + 2
<=> x^2 + 6x - 1 - x^2 - 5x - 2 >= 0
<=> x - 3 >= 0
<=> x >= 3