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a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
c) Đặt \(t=x^2+x+1\) thì
\(t\left(t+1\right)-12=t^2+t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
d) \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+11\) thì
\(\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Rồi nha bạn 
phân tích đa thức thành nhân tử
a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)
\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) \(x^2+2xy+y^2-x-y-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y-4\right)\left(x+y+3\right)=0\)
ý a pạn đưa về dạng ax+b=0 khi chuyển 16 sang và rút gọn 2 biểu thức còn lại đưa về dạng (a+b)2+(a-b)2-16=0. thế thôi. hai biểu thức (x+3)4+(x-2) 4 tự phân tích nhé
a: \(9x^2-6x+3\)
\(=\left(9x^2-6x+1\right)+2\)
\(=\left(3x-1\right)^2+2\ge2\)
b: \(6x-x^2+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10\le10\)
a)đặt x^2-5x=y
<=> y^2+10y+24=0
<=>(y^2+2.5y+25)=1
<=>(y+5)^2=1
\(\left[\begin{matrix}y+5=1\\y+5=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}y=-4\\y=-6\end{matrix}\right.\)
với y=-4 <=> x^2-5x=-4<=> x(x-4)-(x-4)=0
<=> (x-4)(x-1)=0=>\(\left[\begin{matrix}x=1\\x=4\end{matrix}\right.\)
với y=-6<=> x^2-5x=-6<=> x(x-2)-3(x-2)=(x-2)(x-3)=>\(\left[\begin{matrix}x=2\\x=3\end{matrix}\right.\)
d) trôi hết đề bạn đăng quá nhiều
(x+2)(x+3)(x+4)(x+5)-24=0
<=>[(x+2)(x+5)][(x+3)(x+4)]-24=0
<=>(x^2+7x+10)(x^2+7x+12)-24=0
đặt x^2+7x+11=t
<=> (t-1)(t+1)-24=0
<=>t^2-1-25=0
<=>t^2=25=> t=+-5
với t=5
x^2+7x+11=5<=> x^2+7x+6=0
{a-b+c=0}=> x=-1 hoặc -6
với t=-5
x^2+7x+11=-5<=> x^2+7x+17=0=> vô nghiệm