Bài 3: Những hằng đẳng thức đáng nhớ

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

`= (x - y)(x^2 + xy + y^2) - 3xy`

`= 1 * (x^2 + xy + y^2) - 3xy`

`= x^2+ xy + y^2 - 3xy`

`= x^2 - 2xy + y^2`

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

`= (x + y)(x^2 - xy + y^2)`

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

`= (x - y)(x^2 + xy + y^2)`

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

\(9^n-2\cdot3^n+x^2+5+4x=0\)

\(\Leftrightarrow\left(9^n-2\cdot3^n+1\right)+\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left[\left(3^n\right)^2-2\cdot3^n\cdot1+1^2\right]+\left(x^2+2\cdot x\cdot2+2^2\right)=0\)

\(\Leftrightarrow\left(3^n-1\right)^2+\left(x+2\right)^2=0\)

Ta thấy: \(\left(3^n-1\right)^2\ge0\forall n\)

              \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(3^n-1\right)^2+\left(x+2\right)^2\ge0\forall x;n\)

Mặt khác: \(\left(3^n-1\right)^2+\left(x+2\right)^2=0\)

nên ta có: \(\left\{{}\begin{matrix}3^n-1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3^n=1\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n=0\\x=-2\end{matrix}\right.\left(tm\right)\)

Vậy \(n=0;x=-2\).

#\(Toru\)

\(a^2+b^2+c^2-ab-ac-bc=0\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left(a-b\right)^2\ge0\forall a;b\)

              \(\left(b-c\right)^2\ge0\forall b;c\)

              \(\left(a-c\right)^2\ge0\forall a;c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a;b;c\)

Mặt khác: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)

\(\Leftrightarrow a=b=c\left(dpcm\right)\)

#\(Toru\)

\(a)M=5-4x-x^2\\=-(x^2+4x-5)\\=-(x^2+4x+4)+9\\=-(x+2)^2+9\)

Ta thấy: \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+2\right)^2+9\le9\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy \(Max_{M}=9\) khi \(x=-2\)

\(---\)

\(b)N=4x-2x^2+3\\=-2(x^2-2x)+3\\=-2(x^2-2x+1)+5\\=-2(x-1)^2+5\)

Ta thấy: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-2\left(x-1\right)^2\le0\forall x\)

\(\Rightarrow-2\left(x-1\right)^2+5\le5\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy \(Max_{N}=5\) khi \(x=1\)

\(---\)

\(c)P=x(4-x)\\=4x-x^2\\=-(x^2-4x)\\=-(x^2-4x+4)+4\\=-(x-2)^2+4\)

Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+4\le4\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(Max_{P}=4\) khi \(x=2\)

\(---\)

\(d)Q=2021+4y-2x-x^2-y^2\\=-(x^2+2x)-(y^2-4y)+2021\\=-(x^2+2x+1)-(y^2-4y+4)+2026\\=-(x+1)^2-(y-2)^2+2026\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\left(x+1\right)^2\le0\forall x\\-\left(y-2\right)^2\le0\forall y\end{matrix}\right.\)

\(\Rightarrow-\left(x+1\right)^2-\left(y-2\right)^2\le0\forall x,y\)

\(\Rightarrow-\left(x+1\right)^2-\left(y-2\right)^2+2026\le2026\forall x,y\)

Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(Max_{Q}=2026\) khi \(x=-1;y=2\)

#\(Toru\)

\(a)x^2-81\\=x^2-9^2\\=(x-9)(x+9)\\b)64-9y^2\\=8^2-(3y)^2\\=(8-3y)(8+3y)\\c)(x+y)^2-z^2\\=(x+y-z)(x+y+z)\)

\(d)\)\(\dfrac{4u^2}{25}-\dfrac{v^2}{16}\)

\(=\left(\dfrac{2u}{5}\right)^2-\left(\dfrac{v}{4}\right)^2\)

\(=\left(\dfrac{2u}{5}-\dfrac{v}{4}\right)\left(\dfrac{2u}{5}+\dfrac{v}{4}\right)\)

#\(Toru\)

Ta có:

\(x^3+x^2z-xyz+y^2z+y^3\)

\(=\left(x^3+y^3\right)+\left(x^2z-xyz+y^2z\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)

\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)

\(=0\cdot\left(x^2-xy+y^2\right)\)

\(=0\left(dpcm\right)\)

a)

\(A=3x^2-6x-5\\ =3\left(x^2-2x-\dfrac{5}{3}\right)\\ =3\left(x^2-2x+1\right)-8\\ =3\left(x-2\right)^2-8\ge-8\)

GTNN của A đạt `-8` khi và chỉ khi `x=2`

b)

\(B=x^2+y^2-4x-2y+11\\ =x^2-4x+4+y^2-2y+1+6\\ =\left(x-2\right)^2+\left(y-1\right)^2+6\ge6\)

GTNN của B đạt `6` khi và chỉ khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

c)

\(C=21-8y+4x+x^2+y^2\\ =x^2+4x+4+y^2-8y+16+1\\ =\left(x+2\right)^2+\left(y-4\right)^2+1\ge1\)

GTNN của C đạt `1` khi và chỉ khi \(\left\{{}\begin{matrix}x=-2\\y=4\end{matrix}\right.\)

d)

\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+72\\ =\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+72\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)+72\\ =\left(x^2+5x\right)^2-36+72\\ =\left(x^2+5x\right)^2+36\ge36\)

GTNN của D đạt `36` khi và chỉ khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(D=2023-8x+2y+4xy-y^2-5x^2\)

\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)

\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)

\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)

\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)

\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)

Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)

\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)

a) \(\left(3x-y\right)^2-\left(y+3x\right)^2=\left(3x-y+y+3x\right)\left(3x-y-y-3x\right)\)

\(=6x.\left(-2y\right)=-12xy\)

b) \(\left(3y-1\right)^2+\left(3x+2\right)^2-2\left(3x+2\right)\left(3y-1\right)\)

\(=\left(3y-1-3x-2\right)^2=\left(3y-3x-3\right)^2=9\left(x-y-1\right)^2\)

c) \(\left(x+y\right)^2-2\left(x^2-y^2\right)+\left(x-y\right)^2\)

\(=\left(x+y-x+y\right)^2=4y^2\)

d) \(\left(x+y-z\right)^2+\left(y-z\right)^2+2\left(x+y-z\right)\left(y-z\right)\)

\(=\left(x+y-z+y-z\right)^2=\left(x+2y-2z\right)^2\)

e) \(\left(x^2+2xy\right)^2-\left(x^2-2xy\right)^2+4x^2\left(x^2-2xy\right)\)

\(=\left(x^2+2xy-x^2+2xy\right)\left(x^2+2xy+x^2-2xy\right)+4x^2\left(x^2-2xy\right)\)

\(=8x^3y+4x^2\left(x^2-2xy\right)=4x^2\left(2xy+x^2-2xy\right)=4x^4\)

f) \(\left(x-y-z\right)^2+\left(x+y+z\right)^2+\left(x-y-z\right)\left(2x+2y+2z\right)\)

\(=\left(x-y-z+x+y+z\right)^2=4x^2\)

Lời giải:

a.

$=(3x-y-y-3x)(3x-y+y+3x)=-2y.6x=-12xy$

b.

$=[(3y-1)-(3x+2)]^2=(3y-3x-3)^2=9(y-x-1)^2$

c.

$=(x+y)^2-2(x-y)(x+y)+(x-y)^2=[(x+y)-(x-y)]^2=(2y)^2=4y^2$
d.

$=[(x+y-z)+(y-z)]^2=(x+2y-2z)^2$

e.

$=(x^2+2xy-x^2+2xy)(x^2+2xy+x^2-2xy)+4x^2(x^2-2xy)$

$=4xy.2x^2+4x^2(x^2-2xy)=8x^3y+4x^2(x^2-2xy)$

$=4x^2(2xy+x^2-2xy)=4x^2.x^2=4x^4$

f.

$=(x-y-z)^2+(x+y+z)^2+2(x-y-z)(x+y+z)$

$=[(x-y-z)+(x+y+z)]^2=(2x)^2=4x^2$