Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)

b)\(4x^2+4x+5+y^2-4y\)

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

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

c) \(4x^2+5y^2+4xy-12y+9\)

\(=\left(4x^2+4xy+y^2\right)+\left(4y^2-12y+9\right)\)

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

a) \(x^2-6x+10+y^2+2y\)

\(=\left(x^2-6x+9\right)+\left(y^2+2y+1\right)\)

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

b) (x-5)² +9=0

=> (x-5)² = -9 (vô lí)

=> Pt vô nghiệm

a)64x³+48x²+12x+1=27

<=> (4x)³ +3.(4x)².1 +3.4x.1 +1=27

<=> (4x+1)³ =27

Mà 3³ = 27

=> (4x+1)³=3³

=>4x+1=3

=>4x=3-1

=>4x=2

=>x=2/4=1/2

a) 64x³ + 48x² + 12x + 1 = 27

\(\Rightarrow\) (4x)³ + 3 . (4x)² . 1 + 3 . 4x . 1² + 1³ = 27

\(\Rightarrow\) (4x + 1)³ = 27

\(\Rightarrow\) 4x + 1 = 3

\(\Rightarrow\) 4x = 2

\(\Rightarrow\) x = 0,5

a,Ta có

\(64x^3+48x^2+12x+1=27\)

<=>\(\left(4x\right)^3+3\cdot\left(4x\right)^2\cdot1+3\cdot4x\cdot1^2+1^3=27\)

<=>\(\left(4x+1\right)^3=27\)

<=>4x+1=3

<=>x=-1/2

Bài 1:

1) \(a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)\)

\(=ab-ac+bc-ba+ca-cb\)

\(=0\)

2) \(a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)\)

\(=abz-acy+bcx-baz+cay-cbx\)

\(=0\)

Bài 2:

Ta có:

\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)

\(=\dfrac{\left(x^2+bx\right)+\left(ax+ab\right)}{\left(3bx-ax\right)+\left(3ab-a^2\right)}\)

\(=\dfrac{x\left(x+b\right)+a\left(x+b\right)}{x\left(3b-a\right)+a\left(3b-a\right)}\)

\(=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}\)

\(=\dfrac{x+b}{3b-a}\)

Ta có : \(a^3+b^3=c\left(3ab-c^2\right)\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) ( Vì \(a+b+c=3\) )

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

Mà : \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow A=675\left(1^{2018}+1^{2018}+1^{2018}\right)+1=675.3+1=2026\)

+) ta có : \(D=x^2+y^2+2xy-4x-4y+100\)

\(=\left(x+y\right)^2-4\left(x+y\right)+100=3^2-4.3+100=97\)

+) ta có : \(2x^2+y^2=4y-4x-6\Leftrightarrow2x^2+4x+2+y^2-4y+4=0\)

\(\Leftrightarrow2\left(x+1\right)^2+\left(y-2\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

thế vào \(A\) ta có :

\(A=\dfrac{2x^{100}+5\left(y-3\right)^{2011}}{x+y}=\dfrac{2.\left(-1\right)^{100}+5\left(2-3\right)^{2011}}{-1+2}=-3\)

x - 4 = 2 => x = 2 + 4 => x = 6

x + y = 4 mà x = 6 => y = 4 - 6 => y = -2

=> xy = 6 \(\times\) (-2) = -12

x³ - y³ = 6³ - (-2)³ = 224

Ta có:\(x-4=2\Rightarrow x=6^{\left(1\right)}\)

Thay \(^{\left(1\right)}\) vào \(x+y=4\) ,ta được:

\(6+y=4\Rightarrow y=-2^{\left(2\right)}\)

Thay \(^{\left(1\right),\left(2\right)}\) vào xy ,ta được:

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

thay \(^{\left(1\right),\left(2\right)}\) vào \(x^3-y^3\), ta được:

\(x^3-y^3=6^3-\left(-2\right)^3=216-\left(-8\right)=216+8=224\)

Tổng của các bình phương:

\(x^2+2(x+1)^2+3(x+2)^2+4(x+3)^2\)

\(=x^2+(x+1)^2+(x+1)^2+(x+2)^2+(x+2)^2+(x+2)^2+(2x+6)^2\)

\(a.A=5x-x^2\)

\(=-\left(x^2-5x\right)=-\left[\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\right]=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

\(\Rightarrow Max_A=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

\(b.B=x-x^2=-\left(x^2-x\right)=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(\Rightarrow Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(c.C=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)=-\left(x-2\right)^2+7\le7\)

\(\Rightarrow Max_C=7\Leftrightarrow x=2\)

a) Ta có:

\(A=5x-x^2\)

\(=-\left(x^2-5x\right)\)

\(=-\left(x^2-5x\right)-6,25+6,25\)

\(=-\left(x^2-5x+6,25\right)+6,25\)

\(=-\left(x-2,5\right)^2+6,25\)

Ta lại có:

\(\left(x-2,5\right)^2\ge0\)

\(\Rightarrow-\left(x-2,5\right)^2\le0\)

\(\Rightarrow-\left(x-2,5\right)^2+6,25\le6,25\)

\(\Rightarrow A\le6,25\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2,5\right)^2=0\)

\(\Leftrightarrow x-2,5=0\)

\(\Leftrightarrow x=2,5\)

Vậy Max_A = 6,25 \(\Leftrightarrow x=2,5\)

\(d.D=-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x-3\right)^2-2\le-2\)

\(\Rightarrow Max_D=-2\Leftrightarrow x=3\)

\(e.E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left[\left(x+4\right)^2-21\right]=-\left(x+4\right)^2+21\le21\)

\(\Rightarrow Max_E=21\Leftrightarrow x=-4\)

\(f.F=4x-x^2+1=-\left(x-4x-1\right)=-\left[\left(x-2\right)^2-5\right]=-\left(x-2\right)^2+5\le5\)

\(\Rightarrow Max_F=5\Leftrightarrow x=2\)

Bài 1:

\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)

\(\Leftrightarrow-3x^4+3x^2=0\)

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

hay \(x\in\left\{0;1;-1\right\}\)

a²+b²+c²=ab+bc+ca

<=>2(a²+b²+c²)=2(ab+bc+ca)

<=>2a²+2b²+2c²=2ab+2bc+2ca

<=>2a²+2b²+2c²-2ab-2bc-2ca=0

<=>(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)=0

<=>(a-b)²+(b-c)²+(c-a)²=0

Mà \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\) (đpcm)

vì \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Với mọi a,b,c ta luôn có \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{matrix}\right.\)

mà \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

=> a-b=0 và b-c=0 và a-c=0

=> a=b=c

xét tam giác ABC có AB=AC=BC (a=b=c)

=> tam giác ABC đều