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a) 2x(x-5)=5(x-5)
<=> 2x(x-5)-5(x-5)=0
<=> (x-5) (2x-5)=0
<=> \(\orbr{\begin{cases}x-5=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{2}\end{cases}}}\)
b) x2-x-6=0
<=> x2-3x+2x-6=0
<=> x(x-3)+2(x-3)=0
<=> (x+2)(x-3)=0
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)
c) (x-1)(x2+5x-2)-x3+1=0
<=> (x-1)(x2+5x-2)-(x3-1)=0
<=> (x-1)(x2+5x-2)-(x-1)(x2+x+1)=0
<=> (x-1)(x2+5x-2-x2-x-1)=0
<=> (x-1)(4x-3)=0
<=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{4}\end{cases}}}\)
d) e) Bạn viết lại đề được không ạ?
a. 56x2y+712xy2+1118xy56x2y+712xy2+1118xy=30y36x2y2+21x36x2y2+22xy36x2y2=30y+21x+22xy36x2y2=30y36x2y2+21x36x2y2+22xy36x2y2=30y+21x+22xy36x2y2
b.4x+215x3y+5y−39x2y+x+15xy34x+215x3y+5y−39x2y+x+15xy3=3y2(4x+2)45x3y3+5xy2(5y−3)45x3y3+9x2(x+1)45x3y3=12xy2+6y2+25xy3−15xy2+9x3+9x245x3y3=6y2+25xy3−3xy2+9x3+9x245x3y3=3y2(4x+2)45x3y3+5xy2(5y−3)45x3y3+9x2(x+1)45x3y3=12xy2+6y2+25xy3−15xy2+9x3+9x245x3y3=6y2+25xy3−3xy2+9x3+9x245x3y3
c. 32x+3x−32x−1+2x2+14x2−2x32x+3x−32x−1+2x2+14x2−2x=32x+3x−32x−1+2x2+12x(2x−1)=32x+3x−32x−1+2x2+12x(2x−1)
=3(2x−1)2x(2x−1)+2x(3x−3)2x(2x−1)+2x2+12x(2x−1)=6x−3+6x2−6x+2x2+12x(2x−1)=8x2−22x(2x−1)=2(4x2−1)2x(2x−1)=(2x+1)(2x−1)x(2x−1)=2x+1x=3(2x−1)2x(2x−1)+2x(3x−3)2x(2x−1)+2x2+12x(2x−1)=6x−3+6x2−6x+2x2+12x(2x−1)=8x2−22x(2x−1)=2(4x2−1)2x(2x−1)=(2x+1)(2x−1)x(2x−1)=2x+1x
d. x3+2xx3+1+2xx2−x+1+1x+1x3+2xx3+1+2xx2−x+1+1x+1=x3+2x(x+1)(x2−x+1)+2xx2−x+1+1x+1=x3+2x(x+1)(x2−x+1)+2xx2−x+1+1x+1
=x3+2x(x+1)(x2−x+1)+2x(x+1)(x+1)(x2−x+1)+x2−x+1(x+1)(x2−x+1)=x3+2x+2x2+2x+x2−x+1(x+1)(x2−x+1)=x3+3x2+3x+1(x+1)(x2−x+1)=(x+1)3(x+1)(x2−x+1)=(x+1)2x2−x+1
a: Để B nguyên thì x^2+1+2 chia hết cho x^2+1
=>\(x^2+1\in\left\{1;2\right\}\)
hay \(x\in\left\{0;1;-1\right\}\)
b: \(B=\dfrac{x^2+3}{x^2+1}=1+\dfrac{2}{x^2+1}< =1+2=3\)
=>0<=B<=3
B=0 thì x^2+3=0(loại)
B=2 thì 2/x^2+1=1
=>x^2+1=2
=>\(x\in\left\{1;-1\right\}\)
B=3 thì 2/x^2+1=2
=>x^2+1=1
=>x=0