đk : x khác 2; x khác 3; x khác 1

\(a.A=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right)\cdot\frac{x^2-4x+3}{x^4+x^2+1}\)

\(A=\left(\frac{x^2}{\left(x-2\right)\left(x-3\right)}+\frac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)

\(A=\left(\frac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)

\(A=\frac{x^2\left(x-1+x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)

\(A=\frac{x^2\left(2x-4\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\frac{2x^2}{x^4+x^2+1}\)

\(b.\frac{1}{A}=\frac{x^4+x^2+1}{2x^2}=\frac{x^2}{2}+\frac{1}{2}+\frac{1}{2x^2}\) (x khác 0)

\(\frac{1}{A}=\frac{2x^2}{4}+\frac{1}{2}+\frac{1}{2x^2}\)

có 2x^2/4 và 1/2x^2 > 0 áp dụng bđt cô si ta có 

\(\frac{2x^2}{4}+\frac{1}{2x^2}\ge2\sqrt{\frac{2x^2}{4}\cdot\frac{1}{2x^2}}=1\)

\(\Rightarrow\frac{1}{A}\ge\frac{3}{2}\)

\(\Rightarrow A\le\frac{2}{3}\)

DẤU = xảy ra khi 2x^2/4 = 1/2x^2 => 4x^4 = 4

=> x^4 = 1 

=> x = 1 (loại) hoặc x = -1  (thỏa mãn)

vậy max a = 2/3 khi x = -1

Ta có:

\(2x-x^{^2}-2\)

\(=-\left(x^{^2}-2x+2\right)\)

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

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

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

Do \(-\left(x-1\right)^2\le0\)nên \(-\left(x-1\right)^2-1=2x-x^{^2}-2< 0\)hay biểu thức đề cho luôn âm (đpcm)

\(2x-x^2-2=-\left(x-1\right)^2-1\le-1< 0\forall x\)

mình trả lời nè

(a+b+c)3−a3−b3−c3(a+b+c)3-a3-b3-c3

=[(a+b)+c)3−a3−b3−c3=[(a+b)+c)3-a3-b3-c3

=(a+b)3+3(a+b)2.c+3(a+b).c2+c3−a3−b3−c3=(a+b)3+3(a+b)2.c+3(a+b).c2+c3-a3-b3-c3

=a3+3a2b+3ab2+b3+3c(a+b)(a+b+c)+c3−a3+b3+c3=a3+3a2b+3ab2+b3+3c(a+b)(a+b+c)+c3-a3+b3+c3

=3ab(a+b)+3c(a+b)(a+b+c)=3ab(a+b)+3c(a+b)(a+b+c)

=3(a+b)(ab+ac+bc+c2)=3(a+b)(ab+ac+bc+c2)

=3(a+b)[a(b+c)+c(b+c)]=3(a+b)[a(b+c)+c(b+c)]

=3(a+b)(b+c)(a+c)

Ta có: $VT={\left(a+b+c\right)}^3-a^3-b^3-c^3$

$=\left[{\left(a+b+c\right)}^3-a^3\right]-\left(b^3+c^3\right)$

$=\left(b+c\right)\left[{\left(a+b+c\right)}^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)$

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

$=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]$

$=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP$ (Đpcm)

Thật ra mình làm theo đề thấy nó đáng ra phải là chứng minh chứ ko phải phân tích . chúc học tốt!

a) \(8x^3-y^3-6xy\left(2x-y\right)=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2-6xy\right)=\left(2x-y\right)\left(4x^2-4xy+y^2\right)\)

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

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

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

a) 8x³ - y³ - 6xy(2x - y)

= (2x)³ - y³ - 3.2x.y.(2x - y)

= (2x - y)³

b) (3x + 2)² - 2(x - 1)(3x + 2) + (x - 1)²

= (3x + 2 - x + 1)²

= (2x + 3)²

a)

Ta có: EF ≤ EI + IF

mà IF + EF = 1/2 AB + 1/2 CD

= 1/2 (AB + CD)

=> EF ≤ (AB+CD)/2 (đpcm)

Thay M , N bằng  E , F nha

HT :))

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