* Đặt tên các biểu thức theo thứ tự là A,B,C,D,E.

Câu a)

Theo hằng đẳng thức đáng nhớ ta có:

\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)

\(=(a+b+c)^3-3[ab(a+b)+bc(b+c)+ca(c+a)+2abc]\)

\(=(a+b+c)^3-3[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)-abc]\)

\(=(a+b+c)^3-3[(a+b+c)(ab+bc+ac)]+3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)^3-3(ab+bc+ac)(a+b+c)\)

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

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

Do đó:

\(A=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)

Câu b)

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

Sử dụng kết quả (*) của câu a. Với \(a=x, b=-y, c=z\)

\(\Rightarrow x^3+(-y)^3+z^3-3x(-y)z=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)

Mặt khác xét mẫu số:

\((x+y)^2+(y+z)^2+(x-z)^2=x^2+2xy+y^2+y^2+2yz+z^2+x^2-2xz+z^2\)

\(=2(x^2+y^2+z^2+xy+yz-xz)\)

Do đó: \(B=\frac{(x-y+z)(x^2+y^2+z^2+xy+yz-xz)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{x-y+z}{2}\)

Câu c) Sử dụng kết quả (*) của phần a:

\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)

Và mẫu số:

\((x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2-xy-yz-xz)\)

Do đó: \(C=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=\frac{x+y+z}{2}\)

Câu d)

Xét tử số:

\(a^2(b-c)+b^2(c-a)+c^2(a-b)\)

\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)

\(=(b-c)(a^2-b^2)-(b^2-c^2)(a-b)\)

\(=(b-c)(a-b)(a+b)-(b-c)(b+c)(a-b)\)

\(=(a-b)(b-c)[a+b-(b+c)]=(a-b)(b-c)(a-c)\) (1)

Xét mẫu số:

\(a^4(b^2-c^2)+b^4(c^2-a^2)+c^4(a^2-b^2)\)

\(=a^4(b^2-c^2)-b^4[(b^2-c^2)+(a^2-b^2)]+c^4(a^2-b^2)\)

\(=(a^4-b^4)(b^2-c^2)-(b^4-c^4)(a^2-b^2)\)

\(=(a^2-b^2)(a^2+b^2)(b^2-c^2)-(b^2-c^2)(b^2+c^2)(a^2-b^2)\)

\(=(a^2-b^2)(b^2-c^2)[a^2+b^2-(b^2+c^2)]\)

\(=(a^2-b^2)(b^2-c^2)(a^2-c^2)\)

\(=(a-b)(b-c)(a-c)(a+b)(b+c)(c+a)\)(2)

Từ (1)(2) suy ra \(D=\frac{1}{(a+b)(b+c)(c+a)}\)

Câu e)

Theo phần d ta có:

\(TS=(a-b)(b-c)(a-c)\)

\(MS=ab^2-ac^2-b^3+bc^2\)

\(=b^2(a-b)-c^2(a-b)=(a-b)(b^2-c^2)=(a-b)(b-c)(b+c)\)

Do đó: \(E=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}=\frac{a-c}{b+c}\)

1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)

Từ (1), (2) và (3) suy ra:

\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)

<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm

5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) ⁽¹⁾

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) ⁽²⁾

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) ⁽³⁾

Từ (1),(2) và (3) suy ra:

\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)

<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)

mà x+y+z=3

=>\(x^2+y^2+z^2+3\ge2.3=6\)

<=> \(x^2+y^2+z^2\ge6-3=3\)

<=> \(A\ge3\)

Dấu "=" xảy ra khi x=y=z=1

Vậy GTNN của A=x²+y²+z² là 3 khi x=y=z=1

b, Ta có: x+y+z=3

=> \(\left(x+y+z\right)^2=9\)

<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)

<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)

mà \(x^2+y^2+z^2\ge3\) (theo a)

=> \(9-2xy-2yz-2xz\ge3\)

<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)

<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)

<=> \(B\le3\)

Dấu "=" xảy ra khi x=y=z=1

Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1

Câu 1:

Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)

\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)

\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)

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

\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)

Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)

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

\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)

\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)

Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)

5 , a³+b³+c³\(\ge\) 3abc

\(\Leftrightarrow\) a³+3a²b+3ab²+b³+c³-3a²b-3ab²-3abc\(\ge\) 0

\(\Leftrightarrow\) (a+b)³+c³-3ab(a+b+c) \(\ge0\)

\(\Leftrightarrow\) (a+b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c) \(\ge0\)

\(\Leftrightarrow\) (a+b+c)(a²+b²+c²-ab-bc-ca)\(\ge0\) (1)

ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)

(a-b)²+(b-c)²+(c-a)²\(\ge0\)

<=> 2a²+2b²+2c²-2ac-2cb-2ab\(\ge0\)

<=>a²+b²+c²-ab-bc-ac\(\ge\) 0 (3)

Từ (1)(2)(3)=> pt luôn đúng

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

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

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Bạn đăng từ từ thôi!

Dài quá

4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

=> (a+d)² - (b+c)²= (a-d)² - (c-b)²

=> a²+ d²+ 2ad - b²- c²- 2bc=a² + d² - 2ad - c²-b²+2bc

Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)

1) a²+b²+c²+3=2(a+b+c) =>(a-1)²+(b-1)²+(c-1)²=0

=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm

a.

\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)

\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)

\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(*)

Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)

Suy ra (*) đúng => đpcm

Dấu "=" xảy ra khi a = b

b.

\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)

\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)

\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)

\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)

Theo câu a. thì điều này đúng

Dấu "=" khi a=b=c