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Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
\(x^2\left(x^2+5\right)-4x^2-20=0\)
⇔ \(x^4+5x^2-4x^2-20=0\)
⇔\(x^4+x^2-20=0\)
thay x\(^2\) bằng t ( t ≥ 0 ) ta có:
pt⇔ \(t^2+t-20=0\)
⇔ \(t^2+5t-4t-20=0\)
⇔ \(\left(t-4\right)\left(t+5\right)\)
⇔\(\left[{}\begin{matrix}t-4=0\\t+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=4\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)
* \(t=4\) ⇔ \(x^2=4\) ⇔ x = \(\pm2\)
\( {x^2}\left( {{x^2} + 5} \right) - 4{x^2} - 20 = 0\\ \Leftrightarrow {x^4} + 5{x^2} - 4{x^2} - 20 = 0\\ \Leftrightarrow {x^4} + {x^2} - 20 = 0 \)
Đặt \(x^2=t(t\ge0)\)
PT trở thành: \(t^2+t-20=0\)
\(\Leftrightarrow t=4\)(thỏa điều kiện); \(t=-5\)(không thỏa điều kiện)
Với \(t=4 \Rightarrow x^2=4 \Rightarrow x = \pm2\)
Vậy \(S=\left\{2;-2\right\}\)
\(x^4+x^3+x+1=4x^2\)
⇔\(x^4+x^3-4x^2+x+1=0\)
⇔\(\left(x^3-2x^2+x\right)+\left(x^4-2x^2+1\right)=0\)
⇔\(x\left(x-1\right)^2+\left(x^2-1\right)^2=0\)
⇔\(x\left(x-1\right)^2+\left(x-1\right)^2\left(x+1\right)^2=0\)
⇔\(\left(x-1\right)^2\left[x\left(x+1\right)^2\right]=0\)
⇔\(\left(x-1\right)^2\left(x^2+3x+1\right)=0\)
⇔\(\left(x-1\right)^2=0\) hay \(x^2+3x+1=0\)
⇔\(x=1\) hay \(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{5}{4}=0\)
⇔\(x=1\) hay \(\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\).
⇔\(x=1\) hay \(\left(x+\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)\left(x+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=0\)
⇔\(x=1\) hay \(x=-\dfrac{3+\sqrt{5}}{2}\) hay \(x=-\dfrac{3-\sqrt{5}}{2}\).
-Vậy \(S=\left\{1;-\dfrac{3+\sqrt{5}}{2};-\dfrac{3-\sqrt{5}}{2}\right\}\).
a: \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{2\left(x-3\right)}{2-x}\)
\(=\dfrac{4+4x+x^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{2\left(x-3\right)}\)
\(=\dfrac{5x^2+4x+4-4+4x-x^2}{\left(2+x\right)}\cdot\dfrac{1}{2\left(x-3\right)}\)
\(=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{1}{2\left(x-3\right)}=\dfrac{4x\left(x+2\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-3}=\dfrac{2x}{x-3}\)
b: |x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=0(nhận) hoặc x=4(nhận)
Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)
Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)
c: A>0
=>x/x-3>0
=>x>3 hoặc x<0
=>x>3
\(x^3-5x^2+8x-4=0\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\Leftrightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\Leftrightarrow\left(x-2\right)^2\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right..Vậy:x\in\left\{1;2\right\}\)
a, \(x^8+x^7+1\)
= \(x^7\left(x+1\right)+1\)
= \(x^7\left(x+1\right)+1+x-x\)
= \(x^7\left(x+1\right)+\left(x+1\right)-x\)
= \(\left(x^7+1\right)\left(x+1\right)-x\)
a) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\) \(=x^6\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
b) \(x^4+64\)
\(=\left(x^2+8\right)^2-16x^2\)
\(=\left(x^2+8+4x\right)\left(x^2+8-4x\right)\)
nhì nhằng ai giải được