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Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Bài 2:
a)A= \(6x^2\)\(-11x+3\)
<=>A=\(6x^2\)\(-2x-9x+3\)
<=>A=(\(6x^2\)\(-2x\))-\(\left(9x-3\right)\)
=>A=\(2x\left(3x-1\right)\)\(-3\left(3x+1\right)\)
<=>A=\(2x\left(3x-1\right)+3\left(3x-1\right)\)
=>A=(3x-1)(2x+3)
1a, 2x2+3(x2-1)=5x(x+1)
=> 2x2 +3 x2-3= 5x2+5x
=> 2x2 +3 x2-3- 5x2-5x=0
=>-3-5x=0 =>5x=-3 =>x=-3/5
b,2x(5-3x) +2x(3x-5)-3(x-7)=0
=>2x(5-3x) -2x(5-3x)-3(x-7)=0
=>-3(x-7)=0 =>x-7=0 =>x=7
c,3x(x+1)-2x(x+2)=-1-x
=> 3x2+3x-2x2-4x=-1-x
=>3x2+3x-2x2-4x+x=-1
=>x2=-1(vô lí)
2, A= 5x(x-4y) -4y(y-5x)-11/20
=> A=5x2-20xy-4y2+20yx -11/20
=>A=5x2-4y2-11/20
Thay x=-0,6 y=-0,75 vào ta có
A= 5. (-0,6)2-4(-0,75)2-11/20=-1
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
Bài 1.
a) 2x - x2
= x(2 - x)
b) 16x2 - y2
= (4x + y)(4x - y)
c) xy + y2 - x - y
= (xy + y2) - (x + y)
= y(x + y) - (x + y)
= (y - 1)(x + y)
d) x2 - x - 12
= x2 + 3x - 4x - 12
= (x2 + 3x) - (4x + 12)
= x(x + 3) - 4(x + 3)
= (x - 4)(x + 3)
Bài 2.
(2x + 3y)(2x - 3y) - (2x - 1)2 + (3y - 1)2
= (2x + 3y)(2x - 3y) + [(3y - 1)2 - (2x - 1)2]
= (2x + 3y)(2x - 3y) + (3y - 1 + 2x - 1)(3y - 1 - 2x + 1)
= (2x + 3y)(2x - 3y) + (3y + 2x - 2)(3y - 2x)
= (2x + 3y)(2x - 3y) - (2x + 3y - 2)(2x - 3y)
= (2x - 3y)(2x + 3y - 2x - 3y + 2)
= 2.(2x + 3y)
Thay x = 1; y = -1 và biểu thức đại số, ta có:
2[2.1 + 3.(-1)]
= 2(2 - 3)
= 2.(-1) = -2
Bài 3
a) 9x2 - 3x = 0
3x(3x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}3x=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\3x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) x2 - 25 - (x + 5) = 0
(x2 - 25) - (x + 5) = 0
(x - 5)(x + 5) - (x + 5) = 0
(x - 5 - 1)(x + 5) = 0
(x - 6)(x + 5) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
c) x2 + 4x + 3 = 0
x2 + x + 3x + 3 = 0
(x2 + x) + (3x + 3) = 0
x(x + 1) + 3(x + 1) = 0
(x + 3)(x + 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d) (3x - 1)(2x - 7) - (x + 1)(6x - 5) = 16
6x2 - 21x - 2x + 7 - 6x2 + 5x - 6x + 5 - 16 = 0
-24x - 4 = 0
\(\Rightarrow\)-24x = 4
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Bài 1:Phân tích đa thức thành nhân tử
a,2x−x2
=x(2-x)
b,
16x2−y2
=(4x-y)(4x+y)
c,xy+y2−x−y
=(xy+y2)-(x+y)
=y(x+y)-(x+y)
=(x+y)(y-1)
d,
x2−x−12
=x2-4x+3x-12
=(x2-4x)+(3x-12)
=x(x-4)+3(x-4)
=(x-4)(x+3)