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e) Sửa đề:
$2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2$
$=2x^2(x-2)-8x(x-2)+(x-2)=(x-2)(2x^2-8x+1)$
f)
$x^3-3x+2=(x^3-x)-(2x-2)=x(x^2-1)-2(x-1)=x(x-1)(x+1)-2(x-1)$
$=(x-1)(x^2+x-2)=(x-1)(x^2-x+2x-2)=(x-1)[x(x-1)+2(x-1)]$
$=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$
g)
$x^3+3x^2=x^2(x+3)$
h)
$x^3+9x^2+26x+24=(x^3+9x^2+27x+27)-x-3$
$=(x+3)^3-(x+3)=(x+3)[(x+3)^2-1]=(x+3)(x+3-1)(x+3+1)$
$=(x+3)(x+2)(x+4)$
a)
$4x^2-3x-1=4x^2-4x+x-1=4x(x-1)+(x-1)=(4x+1)(x-1)$
b)
$6x^2-11x^2=-5x^2$
c)
\(x^2-7xy+12y^2=x^2-4xy-3xy+12y^2\)
\(=x(x-4y)-3y(x-4y)=(x-3y)(x-4y)\)
d)
\(x^2-2xy+y^2+3x-3y=(x^2-2xy+y^2)+(3x-3y)\)
\(=(x-y)^2+3(x-y)=(x-y)(x-y+3)\)
a,\(-4x^2+4x-1\)
\(\Leftrightarrow\left(-2x-1\right)^2\)
b,\(\left(2x+1\right)^2-4\left(x-1\right)^2\)
\(\Rightarrow\left[2x+1-2\left(x-1\right)\right].\left[2x+1+2\left(x-1\right)\right]\)
\(\Rightarrow\left(2x+1-2x+2\right)\left(2x+1+2x-2\right)\)
\(\Rightarrow3\left(4x-1\right)\)
c,\(\left(2x-y\right)^2-4x^2+12x-9\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(4x^2-12x+9\right)\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(2x-3\right)^2\)
\(\Leftrightarrow\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)
\(\Rightarrow\left(y+3\right)\left(4x+y-3\right)\)
d,\(\left(x+1\right)^2-4\left(x+1\right)y^2+4y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+2^2y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+4\left(y^2\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)-2y^2+\left(2y^2\right)^2\)
\(\Leftrightarrow\left(x+1-2y^2\right)^2\)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
Đặt \(x^2+3x+1=t\)
\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=t\left(t-4\right)-5\)
\(=t^2-4t-5\)
tự làm nốt ý này nhé.
những ý kia lát nx mình làm.