
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)
\(\Leftrightarrow2x^2+4x-4=0\)
\(\Leftrightarrow x^2+2x-2=0\)
\(\Leftrightarrow x^2+2x+1-3=0\)
\(\Leftrightarrow\left(x+1\right)^2=3\)
hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)
b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)
\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)
=>6=0(vô lý)
c: \(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
=>x=-2 hoặc x=2
đ: \(\Rightarrow2x^2-2x-5x+5=0\)
=>(x-1)(2x-5)=0
=>x=1 hoặc x=5/2

\(a.-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-5\right)^2\\ =-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-20x+25\right)\\ =-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+20x-25\\ =-3x^3+x^3-12x^2+3x^2-4x^2-12x-x+20x-3-25\\ =-2x^3-13x^2+7x-28\\ \)\(b.2\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\\ =2\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\\ =2\left(x^3+2x^2-9x-18\right)-x^3+3x+x^2-3-5x^3-40x^2-80x-x^2+10x-25\\ =2x^3+4x^2-18x-36-x^3+3x+x^2-3-5x^3-40x^2-80x-x^2+10x-25\\ =2x^3-x^3-5x^3+4x^2+x^2-40x^2-x^2-18x+3x-80x+10x-36-3-25\\ =-4x^3-36x^2-85x-64\)

\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

a) \(\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)
\(\Leftrightarrow3x^2+8x-3=3x^2-11x+10\)
\(\Leftrightarrow19x-13=0\)
\(\Leftrightarrow x=\frac{13}{19}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{13}{19}\right\}\)
b) \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2+9x-5=2x^2-x-3\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{5}\right\}\)
c) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là \(S=\left\{3\right\}\)
d) \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{33}\right\}\)
e) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)
\(\Leftrightarrow6x-4=-6x+8\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy tập nghiệm của phương trình là \(S=\left\{1\right\}\)
f) \(\left(x+1\right)\left(2x-3\right)-\left(3x-2\right)=2\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-x-3-3x+2=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2-4x-1=2x^2-4x+2\)
\(\Leftrightarrow-1=2\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\varnothing\)

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)

a) 2x( x - 7 ) - ( x + 3 )( x - 2 ) - ( x + 4 )( x - 4 )
= 2x2 - 14x - ( x2 + x - 6 ) - ( x2 - 16 )
= 2x2 - 14x - x2 - x + 6 - x2 + 16
= 22 - 15x
b) ( 2x + 5 )( x - 2 ) - 3( x + 2 )2 + ( x + 1 )2
= 2x2 + x - 10 - 3( x2 + 4x + 4 ) + x2 + 2x + 1
= 3x2 + 3x - 9 - 3x2 - 12x - 12
= -9x - 21
c) ( x + 3 )( x - 3 ) - ( x + 5 )( x - 1 ) - ( x - 4 )2
= x2 - 9 - ( x2 + 4x - 5 ) - ( x2 - 8x + 16 )
= x2 - 9 - x2 - 4x + 5 - x2 + 8x - 16
= -x2 + 4x - 20
d) 2x( x + 1 )2 - ( x - 1 )3 - ( x - 2 )( x2 + 2x + 4 )
= 2x( x2 + 2x + 1 ) - ( x3 - 3x2 + 3x - 1 ) - ( x3 - 8 )
= 2x3 + 4x2 + 2x - x3 + 3x2 - 3x + 1 - x3 + 8
= 7x2 - x + 9
e) ( x + 5 )( x - 5 )( x + 2 ) - ( x + 2 )3
= ( x2 - 25 )( x + 2 ) - ( x3 + 6x2 + 12x + 8 )
= x3 + 2x2 - 25x - 50 - x3 - 6x2 - 12x - 8
= -4x2 - 37x - 58

1,\(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)
\(=-6x+5\)
2,\(\left(x+1\right)^2-3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)
\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)
\(=-6x^2+6x+75\)
3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-1\right)^3-\left(x^3-27\right)\)
\(=x^3-3x^2+3x-1-x^3+27\)
\(=-3x^2+3x+26\)
4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)
\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)
\(=x^3+125-x^3-6x^2-12x-8\)
\(=-6x^2-12x+117\)
5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)
\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)
=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)
\(=-x^3+4x^2-4x+1\)
6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)
\(=3x-26\)
7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)
=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)
\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)
\(=-4x^2-27x-58\)
Nếu đúng thì tick cho mk nha ^_^
ố bn trùng tên nà
kết bn thui