Xí trước phần b

Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

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

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

\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Cách làm khác của phần b ngắn gọn hơn:)

Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)

\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

8a³ - 36a²b + 54ab² - 27b³ - 8 

= ( 8a³ - 36a²b + 54ab² - 27b³ ) - 8 

= ( 2a - 3b )³ - 2³

= ( 2a - 3b - 2 )[ ( 2a - 3b )² + 2( 2a - 3b ) + 4 ]

= ( 2a - 3b - 2 )( 4a² - 12ab + 9b² + 4a - 6b + 4 )

Giúp tui vs

\(x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)

mà \(x\ge7\)\(\Rightarrow x\in\left\{10;15;30\right\}\)

Vậy \(x\in\left\{10;15;30\right\}\)

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

= [(x - 5)² + 2(2x - 1)(x - 5) + (2x - 1)²) - [2(x - 3)]²

= (x - 5 + 2x - 1)² - (2x - 6)²

= (3x - 6)² - (2x - 6)²

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

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

= [ ( x - 5 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )² ] - 2²( x - 3 )²

= ( x - 5 + 2x - 1 )² - ( 2x - 6 )²

= ( 3x - 6 )² - ( 2x - 6 )²

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

Ta có \(\sqrt{15-x}-\sqrt{x-2}=1\)

=> \(\left(\sqrt{15-x}-\sqrt{x-2}\right)^2=1\)

=> \(15-x-2\sqrt{15-x}.\sqrt{x-2}+x-2=1\)

=> \(13-2\sqrt{15-x}.\sqrt{x-2}=1\)

=> \(2\sqrt{15-x}.\sqrt{x-2}=12\)

=> \(\sqrt{15-x}.\sqrt{x-2}=6\)

=> \(\left(\sqrt{15-x}.\sqrt{x-2}\right)^2=6^2\)

=> (15 - x)(x - 2) = 36

=> 15x - 30 - x² + 2x = 36

=> -x² + 17x - 30 = 36

=> - x² + 17x - 66 = 0

=> - x² + 11x + 6x - 66 = 0

=> -x(x - 11) + 6(x - 11) = 0

=> (6- x)(x - 11) = 0

=> \(\orbr{\begin{cases}6-x=0\\x-11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=11\end{cases}}\)

Vậy \(x\in\left\{6;11\right\}\)là giá trị cần tìm

Từ dòng gần cuối sửa cho mk

Thay x = 6 vào đề bài 

(Tự tính)

=> Tm với đề

Thay x = 11 vào đề bài 

(Tự tính)

=> Ktm với đề

Vậy x = 6

a) 2x - x³ + 4y - 8y³

= ( 2x + 4y ) - ( x³ + 8y³ )

= 2( x + 2y ) - ( x + 2y )( x² - 2xy + 4y² )

= ( x + 2y )( 2 - x² + 2xy - 4y² )

b) -3x² + 11x + 14

= -3x² - 3x + 14x + 14

= -3x( x + 1 ) + 14( x + 1 )

= ( x + 1 )( 14 - 3x )

a) 2x - x³ + 4y - 8y³ 

= (2x + 4y) - (x³ + 8y³) 

= 2 (x + y) - [x³ + (2y)³] 

= 2 (x + y) - (x + y)(x² - 2xy + 4y²) 

= (x + y)( 2 - x² + 2xy - 4y²)      (Thật sự là câu này mình vẫn chưa chắc chắn lắm =)))

b) -3x² + 11x + 14 

= -3x² - 3x + 14x + 14 

= (-3x² - 3x) + (14x + 14) 

= -3x(x + 1) + 14(x + 1) 

= (-3x + 14)(x + 1)

=))