7x⁵( x - 1 ) - 3x²y⁶( 1 - x ) + 5xy³( x - 1 )

= 7x⁵( x - 1 ) + 3x²y⁶( x - 1 ) + 5xy³( x - 1 )

= x( x - 1 )[ 7x⁴ + 3xy⁶ + 5y³ )

Help me pls T^T

Dự đoán bđt xảy ra tại \(a=b=c\)

Đánh giá bđt trên theo bđt Bunhiacopxki dạng phân thức ta được

\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ca+2bc}\ge\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\)

Bài toán hoàn tất khi chỉ ra được \(\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\ge1\)Nhưng đánh giá này chính là\(\left(a+b+c\right)^2\ge3\left(ab+ca+bc\right)\)

Vậy bđt được chứng minh

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

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

\(\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{3\cdot\frac{\left(a+b+c\right)^2}{3}}=1\) (Bunhiacopxki dạng cộng mẫu)

Dấu "=" xảy ra khi: a = b = c

a) x² + 4y + 4y² + 26 - 10x = ( x² - 10x + 25 ) + ( 4y² + 4y + 1 ) = ( x - 5 )² + ( 2y + 1 )²

b) 4y² + 34 - 10x + 12y + x² = ( x² - 10x + 25 ) + ( 4y² + 12y + 9 ) = ( x - 5 )² + ( 2y + 3 )²

c) -10x + y² - 8y + x² + 41 = ( x² - 10x + 25 ) + ( y² - 8y + 16 ) = ( x - 5 )² + ( y - 4 )²

d) x² + 9y² - 12y + 29 - 10x = ( x² - 10x + 25 ) + ( 9y² - 12y + 4 ) = ( x - 5 )² + ( 3y - 2 )²

a) \(x^2+4y+4y^2+26-10x\)

\(=\left(x^2-10x+25\right)+\left(4y^2+4y+1\right)\)

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

b) \(4y^2+34-10x+12y+x^2\) đề ntn à?

\(=\left(4y^2+12y+9\right)+\left(x^2-10x+25\right)\)

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

c) \(-10x+y^2-8y+x^2+41\)

\(=\left(x^2-10x+25\right)+\left(y^2-8y+16\right)\)

\(=\left(x-5\right)^2+\left(y-4\right)^2\)

d) \(x^2+9y^2-12y+29-10x\)

\(=\left(x^2-10x+25\right)+\left(9y^2-12y+4\right)\)

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

\(\frac{1}{4}x^2\left(x-y\right)^2-4\left(x-3\right)^2\)

\(=\left(\frac{1}{2}x\right)^2\left(x-y\right)^2-2^2\left(x-3\right)^2\)

\(=\left[\frac{1}{2}x\left(x-y\right)\right]^2-\left[2\left(x-3\right)\right]^2\)

\(=\left(\frac{1}{2}x^2-\frac{1}{2}xy\right)^2-\left(2x-6\right)^2\)

\(=\left[\left(\frac{1}{2}x^2-\frac{1}{2}xy\right)-\left(2x-6\right)\right]\left[\left(\frac{1}{2}x^2-\frac{1}{2}xy\right)+\left(2x-6\right)\right]\)

\(=\left(\frac{1}{2}x^2-\frac{1}{2}xy-2x+6\right)\left(\frac{1}{2}x^2-\frac{1}{2}xy+2x-6\right)\)

Rút gọn thôi chứ phân tích sao được ._.

( x - 3 )² - ( 4x + 5 )² - 9( x + 1 )² - 6( x - 3 )( x + 1 )

= x² - 6x + 9 - ( 16x² + 40x + 25 ) - 9( x² + 2x + 1 ) - 6( x² - 2x - 3 )

= x² - 6x + 9 - 16x² - 40x - 25 - 9x² - 18x - 9 - 6x² + 12x + 18

= -30x² - 52x - 7

Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))

Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)

\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)

\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)

\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)

\(=\left(4x+7\right)\left(12x+17\right)\)

Bài 1:

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

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

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

\(=x^2\)

Bài 2:

đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)

Xét BT trái ta có:

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+5}\)

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

GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến

=> đpcm

Bài 1.

( x - y + z ) + ( z - y )² + ( x - y + z )( 2y - 2z )

= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )²

= [ ( x - y + z ) - ( z - y ) ]² 

= ( x - y + z - z + y )²

= x²

Bài 2. ĐKXĐ tự ghi nhé :))

\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)

\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)

\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)

\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)

\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)

\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)

\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)

=> đpcm

Câu 1:

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

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

b) \(x^4+2009x^2+2008x+2009\)

\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)

c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2-5=0\)

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

Câu 1.

a) 2x² + 5x - 3 = 2x² + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )

b) x⁴ + 2009x² + 2008x + 2009 

= x⁴ + 2009x² + 2009x - x + 2009 

= ( x⁴ - x ) + ( 2009x² + 2009x + 2009 )

= x( x³ - 1 ) + 2009( x² + x + 1 )

= x( x - 1 )( x² + x + 1 ) + 2009( x² + x + 1 )

= ( x² + x + 1 )[ x( x - 1 ) + 2009 ]

= ( x² + x + 1 )( x² - x + 2009 )

c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )

Câu 2. 

3x² + x - 6 - √2 = 0

<=> ( 3x² - 6 ) + ( x - √2 ) = 0

<=> 3( x² - 2 ) + ( x - √2 ) = 0

<=> 3( x - √2 )( x + √2 ) + ( x - √2 ) = 0

<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0

<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)

+) x - √2 = 0 => x = √2

+) 3( x + √2 ) + 1 = 0

<=> 3( x + √2 ) = -1

<=> x + √2 = -1/3

<=> x = -1/3 - √2

Vậy S = { √2 ; -1/3 - √2 }

Câu 3.

A = x( x + 1 )( x² + x - 4 )

= ( x² + x )( x² + x - 4 )

Đặt t = x² + x

A = t( t - 4 ) = t² - 4t = ( t² - 4t + 4 ) - 4 = ( t - 2 )² - 4 ≥ -4 ∀ t

Dấu "=" xảy ra khi t = 2

=> x² + x = 2

=> x² + x - 2 = 0

=> x² - x + 2x - 2 = 0

=> x( x - 1 ) + 2( x - 1 ) = 0

=> ( x - 1 )( x + 2 ) = 0

=> x = 1 hoặc x = -2

=> MinA = -4 <=> x = 1 hoặc x = -2

Áp dụng bđt : \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(1)

CM bđt đúng: Từ (1) => 3xy + 3yz + 3xz \(\le\)x² + y² + z² + 2xy + 2xz + 2yz

<=> 2x² + 2y² + 2z² - 2xy - 2yz - 2xz \(\ge\)0

<=> (x - y)² + (y - z)² + (x - z)² \(\ge\)0 (luôn đúng với mọi x;y;z)

Khi đó: P = \(ab+bc+ac\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

Dấu "=" xảy ra <=> a = b = c = 1

Vậy MaxP = 3 khi a = b = c = 1

Ta có đánh giá quen thuộc sau: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\Leftrightarrow\)\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)*đúng*

Áp dụng, ta được: \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

Đẳng thức xảy ra khi a = b = c = 1