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Ta có: \(\left(a-b\right)^2\ge0,\forall ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\left(1\right)\)
Lại có: \(a^2+b^2\ge2ab\)
\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\left(2\right)\)
Từ (1) và (2) suy ra ĐPCM
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)
\(x^4+x^3+x+1=4x^2\)
⇔\(x^4+x^3-4x^2+x+1=0\)
⇔\(\left(x^3-2x^2+x\right)+\left(x^4-2x^2+1\right)=0\)
⇔\(x\left(x-1\right)^2+\left(x^2-1\right)^2=0\)
⇔\(x\left(x-1\right)^2+\left(x-1\right)^2\left(x+1\right)^2=0\)
⇔\(\left(x-1\right)^2\left[x\left(x+1\right)^2\right]=0\)
⇔\(\left(x-1\right)^2\left(x^2+3x+1\right)=0\)
⇔\(\left(x-1\right)^2=0\) 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 \(S=\left\{1;-\dfrac{3+\sqrt{5}}{2};-\dfrac{3-\sqrt{5}}{2}\right\}\).
Bài 1:
a) \(x^2-x+1\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0;\forall x\)
b) \(25x^2+10x+2\)
\(=25x^2+10x+1+1\)
\(=\left(5x+1\right)^2+1\ge1>0;\forall x\)
c) \(3x^2+2x+14\)
\(=3x^2+2x+\dfrac{1}{3}+\dfrac{41}{3}\)
\(=\left(\sqrt{3}x+\dfrac{\sqrt{3}}{3}\right)^2+\dfrac{41}{3}\ge\dfrac{41}{3}>0;\forall x\)
d) \(2x^2+y^2-2xy-2x+2\)
\(=x^2+y^2-2xy-2x+x^2+1+1\)
\(=\left(x-y\right)^2+\left(x-1\right)^2+1\ge1>0;\forall x\)
Vậy ...
\(x^2\left(x^2+5\right)-4x^2-20=0\)
⇔ \(x^4+5x^2-4x^2-20=0\)
⇔\(x^4+x^2-20=0\)
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 \(S=\left\{2;-2\right\}\)
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