E=(-a-b+c+d)-(d+c-b-2a)

E=-a-b+c+d-d-c+b+2a

E=-a+(-)b+c+d+(-d)+(-c)+b+2a

E=-a+(-b)+c+d+(-d)+(-c)+b+2a

E=(2a-a)+(-b+b)+(-d+d)+(-c+c)=a+0+0+0=a

thanks nhiều nha ĐỨC THỊNH

F=(a-2b-c-2d)-(3d-2c-3b+a)+15

F=a-2b-c-2d-3d+2c+3b-a+15

F=a+(-2b)+(-c)+(-2d)+(-3d)+2c+3b+(-a)+15

F=(-2b+3b)+(-c+2c)+[-2d+(-3d)]+(-a+a)+15

F=b+c+(-5d)+0+15=b+c+(-5d)+15

Ta có a+b+c-(a+b-2c)=-2-(-8)

<=>3c=6

=>c=2

=>a+b=-4; a-2b=-1

=>a+b-(a-2b)=-4-(-1)

<=>3b=-3

=>b=-1

=>a=-3

`@` `\text {Ans}`

`\downarrow`

`a,`

`(2x - 3)^2`

`= 4x^2 - 12x + 9`

`b,`

`(x + 1)^2`

`= x^2 + 2x + 1`

`c,`

`(2x + 5)(2x - 5)`

`= 4x^2 - 25`

`d,`

`(a + b - c)(a - b + c)`

`= a^2 - b^2 + bc - c^2 + cb`

`e,`

\((x + 1)^2 - 10(x + 1) + 25\)

`= x^2 + 2x + 1 - 10x - 10 + 25`

`= x^2 - 8x +16`

`@` `\text {Kaizuu lv uuu}`

`@` CT:

Bình phương của `1` tổng: `(A + B)^2 = A^2 + 2AB + B^2`

Bình phương của `1` hiệu: `(A - B)^2 = A^2 - 2AB + B^2`

`A^2 - B^2 = (A-B)(A+B)`

2, - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b -c - b + c + a + 1 - a - b - c + 3b

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

= 0 - 3b + 0 + (-a) + (-c) + 3b

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

= 0 + (-a) + (-c)

= (-a) + (-c)

3, ( b - c - 6 ) - ( 7 - a + b ) + c

= b - c - 6 - 7 + a - b + c

= (b-b) + (c-c) - (6+7) + a

= 0 + 0 + 13 + a

= 13 + a

6, 2a - { a - b [ a - b - ( a + b + c ) + 2b ] - c - b }

= 2a - { a - b [ a - b - a - b - c  + 2b ] - c - b }

= 2a - { a - b [ ( a - a ) - (b+b) - c + 2b ] - c - b }

= 2a - { a - b [ 0 - 0 - 2b - c + 2b ] - c - b }

= 2a - { a- b [ (2b - 2b) - c ] - c - b }

= 2a - { a - b [ 0 - c ] - c - b }

= 2a - { a - b.(-c) - c - b}

= 2a - a - b.(-c) - c - b

= 1a - (-b).c - c - b

= a - (-b).c - c.1 - b

= a - [(-b) - 1].c - b

ko chắc lắm

a) Mình sửa lại 1 chút ở VP=-3b

Ta có: VT=-2(a+b-2c)+(2a-b-4c)

=-2a-2b+4c+2a-b-4c=-3b

=> VT=VP (đpcm)

b) Ta có VT=(a-b-c)-(a-b+c)=a-b-c-a+b-c=-2c

=> VT=VP (đpcm)

Ta có: \(\frac{2a+b+c}{a}=\frac{a+2b+c}{b}=\frac{a+b+2c}{c}\)

\(\Rightarrow\frac{2a+b+c}{a}-1=\frac{a+2b+c}{b}-1=\frac{a+b+2c}{c}-1\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

Mà \(a,b,c\ne0\)

=> a = b= c

\(A=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

      \(=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}\)

        \(=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}\)

          \(=2+2+2=6\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)

=> a = b = c = d

=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)

D = 1 + 1 + 1 + 1 = 4

Ta có BĐT \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

Nên BĐT cần chứng minh là 

\(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\ge\frac{3}{2}\)

Đặt \(\hept{\begin{cases}a^2=x\\b^2=y\\c^2=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x,y,z>0\end{cases}}\)

Áp dụng BĐT AM-GM and Cauchy-Schwarz ta có:

\(Σ\frac{a^2}{a+b^2}=Σ\frac{x}{\sqrt{x}+y}=Σ\frac{x}{\sqrt{\frac{x\left(x+y+z\right)}{3}+y}}\)

\(=Σ\frac{6x}{2\sqrt{3x\left(x+y+z\right)}+6y}\geΣ\frac{6x}{3x+x+y+z+6y}=Σ\frac{6x}{4x+7y+z}\)

\(=Σ\frac{6x^2}{4x^2+7xy+xz}\ge\frac{6\left(x+y+z\right)^2}{Σ\left(4x^2+7xy+xz\right)}=\frac{3}{2}\)

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