\(A=x^2+9x+25\)

\(=x^2+2x\frac{9}{2}+\frac{81}{4}+\frac{19}{4}\)

\(=\left(x+\frac{9}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

Dấu"="xảy ra khi \(\left(x+\frac{9}{2}\right)^2=0\Rightarrow x=\frac{-9}{2}\)

Vậy \(Min_A=\frac{19}{4}\Leftrightarrow x=\frac{-9}{2}\)

b,\(B=4x^2-8x+\frac{21}{2}\)

\(=4\left(x^2-2x+1\right)+\frac{13}{2}\)

\(=4\left(x-1\right)^2+\frac{13}{2}\ge\frac{13}{2}\forall x\)

Dấu"="xảy ra khi \(4\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(Min_B=\frac{13}{2}\Leftrightarrow x=1\)

c,\(C=-x^2+2x+\frac{5}{2}\)

\(=-\left(x^2-2x-\frac{5}{2}\right)\)

\(=-\left(x^2-2x+1\right)+\frac{7}{2}\)

\(=-\left(x-1\right)^2+\frac{7}{2}\le\frac{7}{2}\forall x\)

Dấu"="xảy ra khi \(-\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy\(Max_C=\frac{7}{2}\Leftrightarrow x=1\)

Bài 1.

A = x² + 9x + 25

= ( x² + 9x + 81/4 ) + 19/4

= ( x + 9/2 )² + 19/4 ≥ 19/4 ∀ x

Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2

=> MinA = 19/4 <=> x = -9/2

B = 4x² - 8x + 21/2

= 4( x² - 2x + 1 ) + 13/2

= 4( x - 1 )² + 13/2 ≥ 13/2 ∀ x

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinB = 13/2 <=> x = 1

C = -x² + 2x + 5/2

= -( x² - 2x + 1 ) + 7/2

= -( x - 1 )² + 7/2 ≤ 7/2 ∀ x

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MaxC = 7/2 <=> x = 1

D = -9x² - 12x + 27/2

= -9( x² + 4/3x + 4/9 ) + 35/2

= -9( x + 2/3 )² + 35/2 ≤ 35/2 ∀ x

Đẳng thức xảy ra <=> x + 2/3 = 0 => x = -2/3

=> MaxD = 35/2 <=> x = -2/3

Bài 2.

a) 4x² + 9y² + 12x + 12y + 13 = 0

<=> ( 4x² + 12x + 9 ) + ( 9y² + 12y + 4 ) = 0

<=> ( 2x + 3 )² + ( 3y + 2 )² = 0 (*)

\(\hept{\begin{cases}\left(2x+3\right)^2\ge0\forall x\\\left(3y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(2x+3\right)^2+\left(3y+2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}2x+3=0\\3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{2}{3}\end{cases}}\)

=> x = -3/2 ; y = -2/3

b) 16x² + 4y² - 8x + 12y + 10 = 0

<=> ( 16x² - 8x + 1 ) + ( 4y² + 12y + 9 ) = 0

<=> ( 4x - 1 )² + ( 2y + 3 )² = 0 (*)

\(\hept{\begin{cases}\left(4x-1\right)^2\ge0\forall x\\\left(2y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(4x-1\right)^2+\left(2y+3\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}4x-1=0\\2y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{3}{2}\end{cases}}\)

=> x = 1/4 ; y = -3/2

Ta có: \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=2\left(4x^2+20x+25\right)+3\left(4x+1\right)\left(4x-1\right)\)

\(=8x^2+40x+50+48x^2-3\)

\(=56x^2+40x+47\)

2( 2x + 5 )² - 3( 4x + 1 )( 1 - 4x )

= 2( 4x² + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )

= 8x² + 40x + 50 + 3( 16x² - 1 )

= 8x² + 40x + 50 + 48x² - 3

= 56x² + 40x + 47

Ta có: \(\left(a+b+c\right)^2-\left(a+b\right)^2-c^2\)

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

\(=2bc+2ca\)

( a + b + c )² - ( a + b )² - c²

= [ ( a + b ) + c ]² - ( a + b )² - c²

= ( a + b )² + 2( a + b )c + c² - ( a + b )² - c²

= 2( a + b )c 

= 2ac + 2bc

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

\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)

\(\Leftrightarrow5x=-17\)

\(\Rightarrow x=-\frac{17}{5}\)

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

\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)

\(\Leftrightarrow10=1\)

=> vô nghiệm 

c) \(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\)

\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+8x+8-22x-27\)

\(\Leftrightarrow8x=-24\)

\(\Rightarrow x=-3\)

a) 2( x - 1 )² + ( x + 3 )² = 3( x - 2 )( x + 1 )

<=> 2( x² - 2x + 1 ) + x² + 6x + 9 = 3( x² - x - 2 )

<=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

<=> 2x² - 4x + x² + 6x - 3x² + 3x = -6 - 2 - 9

<=> 5x = -17

<=> x = -17/5

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

<=> x² + 4x + 4 - 2x + 6 = x² + 2x + 1

<=> x² + 4x - 2x - x² - 2x = 1 - 4 - 6

<=> 0x = -9 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x - 1 )² + ( x - 2 )² = 2( x + 4 )² - ( 22x + 27 )

<=> x² - 2x + 1 + x² - 4x + 4 = 2( x² + 8x + 16 ) - 22x - 27

<=> 2x² - 6x + 5 = 2x² + 16x + 32 - 22x - 27

<=> 2x² - 6x - 2x² - 16x + 22x = 32 - 27 - 5

<=> 0x = 0 ( đúng ∀ x ∈ R )

Vậy phương trình nghiệm đúng ∀ x ∈ R

1. Vì \(x=7\)\(\Rightarrow x+1=8\)

\(\Rightarrow A=x^{15}-8x^{14}+8x^{13}-8x^{12}+.......-8x^2+8x-5\)

\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-.......-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-......-x^3-x^2+x^2+x-5\)

\(=x-5=7-5=2\)

2. Gọi 3 số cần tìm lần lượt là \(a\), \(a+1\), \(a+2\)( \(a\inℕ\))

Tích của 2 số đầu là: \(a\left(a+1\right)\)

Tích của 2 số sau là: \(\left(a+1\right)\left(a+2\right)\)

Vì tích của 2 số đầu nhỏ hơn tích của 2 số sau là 50 nên ta có phương trình:

\(\left(a+1\right)\left(a+2\right)-a\left(a+1\right)=50\)

\(\Leftrightarrow\left(a+1\right).\left(a+2-a\right)=50\)

\(\Leftrightarrow2.\left(a+1\right)=50\)

\(\Leftrightarrow a+1=25\)

\(\Leftrightarrow a=24\)

Vậy 3 số cần tìm lần lượt là 24 , 25 , 26

1) Ta có: \(x=7\Rightarrow x+1=8\)

Thay vào:

\(A=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(A=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)

\(A=x-5=7-5=2\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)

b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)

c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)

d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

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

b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)

\(=3\left(x-2\right)\left(x+5\right)\)

c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

1. (x + 2)(x² - 2x + 4) - (x³ + 2x²) = 5

=> x(x² - 2x + 4) + 2(x² - 2x + 4) - x³ - 2x² - 5 = 0

=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ - 2x² - 5 = 0

=> (x³ - x³) + (-2x² + 2x² - 2x²) + (4x - 4x) + (8 - 5) = 0

=> -2x² + 3 = 0

=> -2x² = -3

=> x² = 3/2

=> x = \(\pm\sqrt{\frac{3}{2}}\)

2. \(\left(x+5\right)^2-6=0\)

=> x² + 10x + 25 - 6 = 0

=> x² + 10x + 19 = 0

=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)

3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)

=> x(x² - 3x + 9) + 3(x² - 3x + 9) - x³ - 2x = 0

=> x³ - 3x² + 9x + 3x² - 9x + 27 - x³ - 2x = 0

=> (x³ - x³) + (-3x² + 3x²) + (9x - 9x - 2x) + 27 = 0

=> -2x + 27 = 0

=> -2x = -27

=> x = 27/2

4. \(\left(x-2\right)^3-x^3+6x^2=7\)

=> x³ - 6x²  + 12x - 8 - x³ + 6x² = 7

=> (x³ - x³) + (-6x² + 6x²) + 12x - 8 = 7

=> 12x - 8 = 7

=> 12x = 15

=> x = 5/4

5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)

=> 3x² - 12x + 12 + 9x - 9 - 3x² - 3x + 9 = 12

=> (3x² - 3x²) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12

=> -6x + 12 = 12

=> -6x = 0

=> x = 0

6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)

=> 48x - 5x - 2 = 0

=> 43x - 2 = 0

=> 43x = 2

=> x = 2/43

Còn bài cuối tự làm :>

Anh Sang làm cầu kì quá ;-;

1. ( x + 2 )( x² - 2x + 4 ) - ( x³ + 2x² ) = 5

<=> x³ + 8 - x³ - 2x² = 5

<=> 8 - 2x² = 5

<=> 2x² = 3

<=> x² = 3/2

<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)

<=> \(x=\pm\sqrt{\frac{3}{2}}\)

2. ( x + 5 )² - 6 = 0

<=> ( x + 5 )² - ( √6 )² = 0

<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0

<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)

3. ( x + 3 )( x² - 3x + 9 ) - x³ = 2x

<=> x³ + 27 - x³ = 2x

<=> 27 = 2x

<=> x = 27/2

4. ( x - 2 )³ - x³ + 6x² = 7

<=> x³ - 6x² + 12x - 8 - x³ + 6x² = 7

<=> 12x - 8 = 7

<=> 12x = 15

<=> x = 15/12 = 5/4

5. 3( x - 2 )² + 9( x - 1 ) - 3( x² + x - 3 ) = 12

<=> 3( x² - 4x + 4 ) + 9x - 9 - 3x² - 3x + 9 = 12

<=> 3x² - 12x + 12 + 6x - 3x² = 12

<=> -6x + 12 = 12

<=> -6x = 0

<=> x = 0

6. ( 4x + 3 )² - ( 4x - 3 )² - 5x - 2 = 0

<=> 16x² + 24x + 9 - ( 16x² - 24x + 9 ) - 5x - 2 = 0

<=> 16x² + 24x + 9 - 16x² + 24x - 9 - 5x - 2 = 0

<=> 43x - 2 = 0

<=> 43x = 2

<=> x = 2/43

7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0

<=> -12x² - 13x + 14 - ( -12x² + 26x + 10 ) = 0

<=> -12x² - 13x + 14 + 12x² - 26x - 10 = 0

<=> -39x + 4 = 0

<=> -39x = -4

<=> x = 4/39

a) \(\left(x+y\right)^3-x^3-y^3\)

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

\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)

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

\(=3xy\left(x+y\right)\)

b) \(x^2+y^2+2xy+yz+xz\)

\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)

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

\(=\left(x+y\right)\left(x+y+z\right)\)

c) \(x^2-10xy-1+25y^2\)

\(=\left(x^2-10xy+25y^2\right)-1\)

\(=\left(x-5y\right)^2-1\)

\(=\left(x-5y-1\right)\left(x-5y+1\right)\)

d) \(ax^2-ax+bx^2-bx+a+b\)

\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)

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

\(=(a+b)(x^2-x+1)\)

e)\(x^2-2y+3xz+x-2y+3z\)

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

\(=x(x+1)-2y(x-1)+3z(x+1)\)

\(=(x+1)(x-2y+3z)\)

f) \(xyz-xy-yz-xz+x+y+z-1\)

\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)

\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)

\(=(z-1)(xy-y-x+1)\)

\(=(z-1)[y(x-1)-(x-1)]\)

\(=(z-1)(x-1)(y-1)\)

_Học tốt_