Đại số lớp 8

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

\(\Rightarrow x^2+2y^2\ge\dfrac{1}{3}\)

dấu = xảy ra: \(\dfrac{x}{1}=\dfrac{\sqrt{2}y}{\sqrt{2}}\Leftrightarrow x=y=\dfrac{1}{3}\)

Theo đề: x +2y =1

<=> x = 1 - 2y

Ta có: A = x² + 2y²

= (1-2y)² +2y²

= 1-2y+4y²+2y²

= 1-2y + 6y²

= 6( y² - \(\dfrac{1}{3}\)y+\(\dfrac{1}{36}\)) + \(\dfrac{5}{6}\)

= 6(y-\(\dfrac{1}{6}\))² +\(\dfrac{5}{6}\)

mà 6 (y-\(\dfrac{1}{6}\))² \(\ge\)0 với mọi y

=> 6(y-\(\dfrac{1}{6}\))² +\(\dfrac{5}{6}\)\(\ge\)\(\dfrac{5}{6}\) với mọi y

=> A\(\ge\)\(\dfrac{5}{6}\)

dấu "=" xảy ra khi A nhận GTNN

<=> y = \(\dfrac{1}{6}\), x = \(\dfrac{2}{3}\)

vậy GTNN của A là \(\dfrac{5}{6}\) khi y=\(\dfrac{1}{6}\), x=\(\dfrac{2}{3}\)

Ta có: \(\left(x+3\right)\left(x-11\right)+2003\)

\(=x^2-8x+-33+2003\)

\(=x^2-8x+16+1954\)

\(=\left(x-4\right)^2+1954\)

Do \(\left(x-4\right)^2\ge0\) với mọi x

=> \(\left(x-4\right)^2+1954>0\) với mọi x

<=> \(\left(x+3\right)\left(x-11\right)+2003>0\) với mọi x

=> (x+3)(x-11)+2003 luôn dương với mọi giá trị của x

(x + 3)(x - 11)+ 2003

= x² + 3x - 11x - 33 + 2003

= x² - 8x - 33 + 2003

= x² - 4.2x + 16 - 49 + 2003

= (x - 4)² + 1954, luôn dương (đpcm)

Phương trình này nghiệm xấu lắm. Mà nhắm lớp 8 giải không nổi đâu

sửa đề: \(a^3+b^3+c^3=3abc\)

Giải:

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

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

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(a-c\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)(đpcm)

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

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

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

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

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

Vì \(\left(a-b\right)^2\ge0\) với mọi a, b ; \(\left(a-c\right)^2\ge0\) với mọi a, c ; \(\left(b-c\right)^2\ge0\) với mọi b, c.

Do đó \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\) khi \(a-b=a-c=b-c=0\), suy ra a = b = c

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

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)

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

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

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

Vì \(\left(a-b\right)^2\ge0\forall a,b\)

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

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

Do đó \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra \(\Leftrightarrow a=b,a=c,b=c\)

\(\Rightarrow a=b=c\)

c sai đề

1)\(8-12x+6x^2-x^3\)

\(=-\left(x^3-6x^2+12x-8\right)\)

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

2)\(15x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

4)

sao câu này cũng sai đề thế mắt m` lé à

\(\left\{{}\begin{matrix}x^2+2x+1=y^2+11\\\left(x+1\right)^2-y^2=11\\\Rightarrow\left(x+1\right)^2=36\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\end{matrix}\right.\) \(\)

Vậy: \(\left(x,y\right)=\left(5,5\right);\left(5,-5\right);\left(-7,5\right);,\left(-7,-5\right)\)