\(x^3+17x\)

\(x.x.x+x.17\)

Vì \(17+1\)mới chia hết cho 3.

Nếu x > 3  1 đơn vị thì có x³ = 68 là nhỏ nhất

Ta có x nhỏ nhất là 4,thỏa mãn điều kiện x . 17 : 3 dư 2.

Vậy ta có \(x\in N\left|x⋮4\right|x^3+17x⋮3|x\ge4\)

a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)

\(\Leftrightarrow-3x=7\)

\(\Leftrightarrow x=-\frac{7}{3}\)

b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)

\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)

\(\Leftrightarrow72x^3+55x^2-30x-6=0\)

Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)

a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6

=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6

=> x² + 3x + 2x + 6 - x² - 7x - x - 7 = 6

=> x² + 5x + 6 - x² - 7x - x - 7 = 6

=> (x² - x²) + (5x - 7x - x) + (6 - 7) = 6

=> -3x - 1 = 6

=> -3x = 7

=> x = -7/3

b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> (8x - 3)(9x² + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33

=> 8x(9x² + 12x + 4) - 3(9x² + 12x + 4) - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33

=> 72x³ + 96x² + 32x - 27x² - 36x - 12  - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0

=> 72x³ + (96x² - 27x² - 10x² - 4x²) + (32x - 36x  - 16x -  7x + 2x - 5x)  + (-12  - 28 + 1 +  33) = 0

=> 72x³ + 55x² - 30x - 6 = 0

=> x vô nghiệm

Ta có: \(M-N=77^2+75^2+....+1^2-\left(76^2+74^2+...+2^2\right)\)

\(=77^2+75^2+....+1^2-76^2-74^2-...-2^2\)

\(=\left(77^2-76^2\right)+\left(75^2-74^2\right)+...+\left(3^2-2^2\right)+1^2\)

\(=\left(77-76\right)\left(77+76\right)+\left(75-74\right)\left(75+74\right)+...+\left(3-2\right)\left(3+2\right)+1\)

\(=77+76+75+74+...+3+2+1\)

\(=\frac{\left[\left(77-1\right):1+1\right].\left(1+77\right)}{2}=\frac{77.78}{2}=3003\)

Thay vào S, ta có: \(S=\frac{M-N-3}{3000}=\frac{3003-3}{3000}=\frac{3000}{3000}=1\)

A = x² - 4xy + 5y² + 10x - 22y + 2044

= ( x² - 4xy + 4y² + 10x - 20y + 25 ) + ( y² - 2y + 1 ) + 2018

= [ ( x² - 4xy + 4y² ) + ( 10x - 20y ) + 25 ] + ( y - 1 )² + 2018

= [ ( x - 2y )² + 2( x - 2y ).5 + 5² ] + ( y - 1 )² + 2018

= ( x - 2y + 5 )² + ( y - 1 )² + 2018 ≥ 2018 ∀ x, y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

=> MinA = 2018 <=> x = -3 ; y = 1

Bài làm :

 Bình phương hai vế của a + b + c = 0 ta được :

\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)   ( 1 )

Bình phương hai vế của ( 1 ) ta được :

\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)  ( vì a + b + c = 0 nên 2abc . 0 = 0 )

=> đpcm 

Phần còn lại tương tự bạn tự làm nhé

Học tốt

Ta có :

\(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)( 1 )

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

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)( 2 )

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)( 3 )

Ta lại có : 

\(\left(ab+bc+ca\right)^2\)

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

\(=a^2b^2+b^2c^2+c^2a^2+2abc.0\)

\(=a^2b^2+b^2c^2+c^2a^2\)( 4 )

Thay ( 4 ) vào ( 2 ) ta được :

\(a^4+b^4+c^4+2\left(ab+bc+ca\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)( 5 )

Từ ( 1 ) => \(ab+bc+ca=\frac{-a^2-b^2-c^2}{2}\)

\(\Rightarrow2\left(ab+bc+ca\right)^2=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)( 6 )

Từ ( 3 ) ; ( 5 ) và ( 6 ) => Đpcm

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 )

= x² + 5x + 6 - ( x² + 3x - 10 )

= x² + 5x + 6 - x² - 3x + 10

= 2x + 16

b) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) + 10

= -5x² - 2x + 16 + 4( x² - x - 2 ) + 2( x² - 4 ) + 10

= -5x² - 2x + 16 + 4x² - 4x - 8 + 2x² - 8 + 10

= x² - 6x + 10

c) 4( x - 1 )( x + 5 ) - ( x + 2 )( x + 5 ) - 3( x - 1 )( x + 2 )

= 4( x² + 4x - 5 ) - ( x² + 7x + 10 ) - 3( x² + x - 2 )

= 4x² + 16x - 20 - x² - 7x - 10 - 3x² - 3x + 6

= 6x - 24

d) ( x - 1 )( x⁵ + x⁴ + x³ + x² + x + 1 )

= x⁶ + x⁵ + x⁴ + x³ + x² + x - x⁵ - x⁴ - x³ - x² - x - 1

= x⁶ - 1

a) \(P\left(a,b\right)=3a^2-2ab+b^2=3a^2-3ab+ab-b^2\)\(=3a\left(a-b\right)+b\left(a-b\right)=\left(a-b\right)\left(3a+b\right)\)

b) \(P\left(a,b\right)=0\Leftrightarrow\orbr{\begin{cases}a-b=0\\3a+b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=b\\a=\frac{-b}{3}\end{cases}}}\)

+) \(a=b\Leftrightarrow M=\frac{a^2+a.a+2a^2}{2a^2-a^2}=4\)

+) \(a=\frac{-b}{3}\Rightarrow M=\frac{\left(\frac{-b}{3}\right)^2+\left(\frac{-b}{3}\right).b+2b^2}{2.\left(\frac{-b}{3}\right)^2-b^2}=\frac{\frac{16}{9}b^2}{\frac{-7}{9}b^2}=\frac{-16}{7}\)

cảm ơn Đặng Ngọc Quỳnh nhé :>

a) 4x² - 5xy + y² = 4x² - 4xy - xy + y² = 4x( x - y ) - y( x - y ) = ( x - y )( 4x - y )

b) x² - 4xy + 3y² = x² - xy - 3xy + 3y² = x( x - y ) - 3y( x - y ) = ( x - y )( x - 3y )

c) 9x² + 6xy - 8y² = 9x² - 6xy + 12xy - 8y² = 9x( x - 2/3y ) + 12y( x - 2/3y ) = ( x - 2/3y )( 9x + 12y )

d) 2x² + 3xy - 5y² = 2x² - 2xy + 5xy - 5y² = 2x( x - y ) + 5y( x - y ) = ( x - y )( 2x + 5y )

e) x² - 35y² - 2xy = x² + 5xy - 7xy - 35y² = x( x + 5y ) - 7y( x + 5y ) = ( x + 5y )( x - 7y )

f) 2x² + 10xy + 8y² = 2( x² + 5xy + 4y² ) = 2( x² + xy + 4xy + 4y² ) = 2[ x( x + y ) + 4y( x + y ) ] = 2( x + y )( x + 4y )

g) x² - 10xy + 16y² = x² - 2xy - 8xy + 16y² = x( x - 2y ) - 8y( x - 2y ) = ( x - 2y )( x - 8y )

h) 4x² + 4xy - 15y² = 4x² - 6xy + 10xy - 15y² = 4x( x - 3/2y ) + 10y( x - 2/3y ) = ( x - 2/3y )( 4x + 10y )

i) -7xy + 3x² + 2y² = 3x² - xy - 6xy + 2y² = 3x( x - 1/3y ) - 6y( x - 1/3y ) = ( x - 1/3y )( 3x - 6y )

j) 56y² + 4x² - 36xy = 4( x² - 9xy + 14y² ) = 4( x² - 2xy - 7xy + 14y² ) = 4[ x( x - 2y ) - 7y( x - 2y ) ] = 4( x - 2y )( x - 7y )

Mình viết xuôi theo dạng ax² + bx + c nhé ;-; cho dễ làm

a) 2x² + 7x + 3 = 2x² + x + 6x + 3 = x( 2x + 1 ) + 3( 2x + 1 ) = ( 2x + 1 )( x + 3 )

b) 3x² - 8x + 4 = 3x² - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )

c) 3x² - 7x + 2 = 3x² - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )

d) -6x² + 7x - 2 = -6x² + 3x + 4x - 2 = -3x( 2x - 1 ) + 2( 2x - 1 ) = ( 2x - 1 )( 2 - 3x )

e) -3x² + 7x - 2 = -3x² + 6x + x - 2 = -3x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 1 - 3x )

f) 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 2x - 1 )

g) 3x² - 8x + 4 = 3x² - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )

h) 6x² - 11x + 3 = 6x² - 2x - 9x + 3 = 2x( 3x - 1 ) - 3( 3x - 1 ) = ( 3x - 1 )( 2x - 3 )

i) 2x² + 3x - 27 = 2x² - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 )

j) 4x² - 5x + 1 = 4x² - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )