Bài 2 :
a, \(=\left(a^2+a-1\right)\left[a^2-\left(a-1\right)\right]=a^4-\left(a-1\right)^2=a^4-a^2+2a-1\)
b, \(=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)\)
\(=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)
c,
\(=\left(2+3y\right)^2-\left(2x-3y\right)^2-12xy=\left(2+3y-2x+3y\right)\left(2+3y+2x-3y\right)-12xy\)
\(=\left(2-2x+6y\right)\left(2+2x\right)-12xy=4+4x-4x-4x^2+12y+12xy-12xy=4-4x^2+12y\)
d, \(=\left(x+1-x+1\right)\left[\left(x+1\right)^2+x^2-1+\left(x-1\right)^2\right]-2\left(x^3-1\right)\)
\(=2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-2\left(x^3-1\right)\)
\(=2\left(3x^2+1\right)-2\left(x^3-1\right)=6x^2+2-2x^3+2=-2x^3+6x^2+4\)
Bài 10:
a) \(x^2+x+1\)=\(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)=\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\).
-GTNN của biểu thức là \(\dfrac{3}{4}\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}\).
b) \(2+x-x^2=-\left(x^2-x-2\right)=-\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{9}{4}\right)=-\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)-GTLN của biểu thức là \(\dfrac{9}{4}\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}\).
c) \(x^2-4x+1=\left(x^2-4x+4\right)-3=\left(x-2\right)^2-3\ge-3\).
-GTNN của biểu thức là \(-3\Leftrightarrow x-2=0\Leftrightarrow x=2\).
d) \(4x^2+4x+11=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\).
-GTNN của biểu thức là \(10\Leftrightarrow2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\).
e) \(3x^2-6x+1=\left(x\sqrt{3}\right)^2-2.x\sqrt{3}.3.\sqrt{3}+9.3-26=\left(x\sqrt{3}-3\sqrt{3}\right)^2-26\ge-26\)-GTNN của biểu thức là \(-26\Leftrightarrow x\sqrt{3}-3\sqrt{3}=0\Leftrightarrow\sqrt{3}\left(x-3\right)=0\Leftrightarrow x=3\).
f) \(x^2-2x+y^2-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\)-GTNN của biểu thức là \(1\Leftrightarrow x-1=0;y-2=0\Leftrightarrow x=1;y=2\).
g) \(b\left(b+1\right)\left(b+2\right)\left(b+3\right)=\left(b^2+3b\right)\left(b^2+3b+2\right)\).
- Đặt \(t=b^2+3b\) ta được:
\(\left(b^2+3b\right)\left(b^2+3b+2\right)=t\left(t+2\right)=t^2+2t=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)-GTNN của biểu thức là \(-1\Leftrightarrow t+1=0\Leftrightarrow b^2+3b+1=0\Leftrightarrow b^2+2.\dfrac{3}{2}b+\dfrac{9}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(b+\dfrac{3}{2}\right)^2-\dfrac{5}{4}=0\Leftrightarrow\left(b+\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)\left(b+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=0\Leftrightarrow b=-\dfrac{3+\sqrt{5}}{2}hayb=-\dfrac{3-\sqrt{5}}{2}\)
Bài 8:
a: =>x(x-4)(x+4)=0
hay \(x\in\left\{0;4;-4\right\}\)
b: =>2x(x-5)(x+5)=0
hay \(x\in\left\{0;5;-5\right\}\)
c: \(\Leftrightarrow x^2\left(x-4\right)-9\left(x-4\right)=0\)
=>(x-4)(x-3)(x+3)=0
hay \(x\in\left\{4;3;-3\right\}\)
d: \(\Rightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
=>(x+9)(x-1)=0
hay \(x\in\left\{-9;1\right\}\)
e: \(\Leftrightarrow\left(x-3\right)^2\cdot\left[\left(x+3\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+2\right)\left(x+4\right)=0\)
hay \(x\in\left\{3;-2;-4\right\}\)
f: \(\Leftrightarrow x^3-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=0\)
=>(x-1)(x+2)(x-1)=0
hay \(x\in\left\{1;-2\right\}\)
