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\(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)+4\left(3-x\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}}\)
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
Bài làm
~ Bài này không phải lớp 8, mà là lớp 6 ~
a) 2(x+7)-2=18
2(x+7) = 12+2
2(x+7) = 20
x+7 = 20:2
x+7 = 10
x = 10 - 7
x = 3
Vậy x = 3
2( x + 7) - 2 = 18
2( x + 7) = 20
x + 7 =10
x = 3
2x( x - 5 ) = 4( x - 5 )
2x( x - 5 ) - 4( x - 5 ) = 0
( 2x - 4 )( x - 5 ) = 0
th1 : 2x - 4 = 0
2x = 4
x = 2
th2 : x - 5 = 0
x = 5
Vậy x = 2 hoặc x = 5
hok tốt
a) x(4x2 - 1) = 0
=> x(2x-1)(2x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.......\)
b) \(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)
\(\Rightarrow3x^2-6x+3-3x^2+13=0\\ \Rightarrow13-6x=0\\ \Rightarrow x=\dfrac{13}{6}\)
\(d.2x^2-5x-7=0\\ \Rightarrow2x^2+2x-\left(7x+7\right)=0\\ \Rightarrow2x\left(x+1\right)-7\left(x+1\right)=0\\ \Rightarrow\left(2x-7\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-7=0\Rightarrow x=\dfrac{7}{2}\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
Bài 2:
a)x3+2x2+x
=x(x2+2x+12)
=x(x+1)2
b)xy+y2-x-y
=(xy-x)+(y2-y)
=x(y-1)+y(y-1)
=(y-1)(x+y)
Bai 1:
a) = 2x^3 + 14x^2 - 2x^3 - x^2 + 9x - 12
= 13x^2 + 9x - 12
b) = x^2 - 2x + 1 - x^2 + 4x - 4x + 16
= -2x + 17
b, A=[(a+1)(a+7)][(a+3)(a+5)]+15
=>A=(a2+8a+7)(a2+8a+15)+15
Đặt a2+8a+11= t
=>a2+8a+7= t-4 và a2+8a+15= t+4
=>A=(t-4)(t+4)+15
=>A=t2-16+15
=t2-1=(t-1)(t+1)
Thay t = a2+8a+11
=>A=(a2+8a+11-1)(a2+8a+11+1)
=>A=(a2+8a+10)(a2+8a+12)
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y-2\right)\left(x+y+5\right)\)
\(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x^2-5x=x-1\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-\frac{1}{5}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
\(2\left(x-7\right)-x^2+7x=0\)
\(2\left(x-7\right)-x\left(x-7\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)