xí câu 1:))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)

Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )

Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )

Vậy ta có đpcm

Đẳng thức xảy ra <=> a=2 => x=y=2

1.

1) a ( b + c )

2) a ( b-c+d)

3) x ( a-b-c+d)

4) ( b+c ) (a - d )

5) a (c-d) + b (c-d) =(c-d) (a + b )

6) a ( x+y) + b ( y+x) = (x+y) ( a+b)

2.

1) a - b + c - a - c = -b

2) a + b - b + a + c = 2a + c

3) - a - b + c + a - b - c = -2b

4) ab + ac - ab - ad = ac-ad = a (c-d)

5) ab - ac + ad +  ac = ab + ad = a (b+d)

a.

\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

(luôn đúng)

b. Áp dụng BĐT \(x^2+y^2\ge2xy\)

\(a^2+b^2\ge2ab,a^2+1\ge2a,b^2+1\ge2b\)\(\Rightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\Leftrightarrow a^2+b^2+1\ge ab+a+b\)

c. Tương tự câu b

Áp dụng BĐT Cô si ta có

i. \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},\frac{1}{b}+\frac{1}{c}\ge\frac{2}{\sqrt{bc}},\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ca}}\)

\(\Rightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

k. Tương tự câu i

1 a(b+c)

2 a(b-c+d)

3 x(a-b-c+d)

4 (b+c)(a-d)

5 a(c-d)+b(c-d)

(c-d)(a+b)

6 ax+by+bx+ay

ax+ay+bx+by

a(x+y)+b(x+y)

(x+y)(a+b)

làm được nhiu ây thui, mí bài kia tự làm nhak

hihhhi

bài 2 \

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

phá ngoặc

=a-b+c-a-c

=-b

2 làm giống bài 1 í. phá ngoặc hớt, mí bài còn lại cũng lm tương tự

phá ngoặc là được thui :)))))

Bài 2 :

1/  (a - b + c) - (a + c) 

= a - b + c - a - c

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

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

= -b

2/  (a + b) - (b - a) + c =

= a + b - b + a + c

= 2a + (b - b) + c

= 2a + c

3/  -(a + b - c) + (a - b - c) 

= -a - b + c + a - b - c

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

= -2b 

4/  a(b + c) - a(b + d) 

= ab + ac - (ab + ad)

= ab + ac - ab - ad

= (ab - ab) + ac - ad

= a.(c - d) 

5/  a(b - c) + a(d + c) 

= ab - ac + ad + ac

= ab + ad + (-ac + ac)

= a(b + d)

Bài 17 :

1) ab + ac = a ( b + c )

2) ab  - ac + ad = a ( b - c + d )

3) ax - bx - cx + dx = x ( a- b - c + d )

4) a(b + c) – d(b + c) = ( b + c ) ( a - d )

5) ac – ad + bc – bd = a( c - d ) + b ( c - d ) = ( c- d ) ( a + b )

6) ax + by + bx + ay = a( x+ y ) + b ( x + y ) = ( x + y ) (a +b )

Bài 18:

1/ (a – b + c) – (a + c) = a - b + c - a - c = -b

2/ (a + b) – (b – a) + c = a + b - b + a + c = 2a + 2

3/ - (a + b – c) + (a – b – c) = -a -b + c + a - b - c = -2b

4/ a(b + c) – a(b + d) = a ( b + c - b - d ) = a( c - d )

5/ a(b – c) + a(d + c) = a ( b - c + d + c ) = a ( b+ d ) 

Bài 1

1,ab+ac=a.(b+c)

2,ab-ac+ad=a.(b-c+d)

3,ax-bx-cx+dx=x.(a-b-c+d)

4,a.(b+c)-b.(b+c)=(b+c).(a-b)

5,ac-ad+bc-bd=a.(c-d)+b.(c-d)=(c-d).(a+b)

6,ax+by+bx+ay=a.(x+y)+b.(x+y)=(x+y).(a+b)

Bài 2

1,(a-b+c)-(a+c)=-b

 =a-b+c-a-c

 =(a-a)+(c-c)-b

 =0+0-b

 =-b

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

 =a+b-b+a+c

 =(a+a)+(b-b)+c

 =2a+0+c

 =2a+c

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

 =-a-b+c+a-b-c

 =(-a-a)+[-b+(-b)]+(c-c)

 =0+(-2b)+0

 =-2b

4,a.(b+c)-a.(d+c)=a.(c-d)

   =ab+ac-ab-ad

   =a.(b+c-b-d)

   =a.(c-d)

5,a.(b-c)+a.(d+c)=a.(b+d)

  =ab-ac+ac+ad

  =a.(b-c+c+d)

  =a.(b+d)

3)

1.

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)

Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)

\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)

\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)

\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)

\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)

2.

Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)

Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)

\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)

Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)

\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)

\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)

(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị