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

<=> x² - 8x + 15 - ( 2x² + 3x - 2 ) + x² = 5

<=> 2x² - 8x + 15 - 2x² - 3x + 2 = 5

<=> -11x + 17 = 5

<=> -11x = -12

<=> x = 12/11

b) -2x( x - 1 ) + ( x - 1 )( 2x + 3 ) = x + 4

<=> -2x² + 2x + x² + x - 3 = x + 4

<=> 3x - 3 = x + 4

<=> 3x - x = 4 + 3

<=> 2x = 7

<=> x = 7/2

Bài làm :

\(a,\left(x-5\right)\left(x-3\right)-\left(x+2\right)\left(2x-1\right)+x^2=5\)

\(\Leftrightarrow x^2-3x-5x+15-\left(2x^2-x+4x-2\right)+x^2=5\)

\(\Leftrightarrow\left(x^2-2x^2+x^2\right)+\left(-3x-5x+x-4x\right)=5-2-15\)

\(\Leftrightarrow-11x=-12\)

\(\Leftrightarrow x=\frac{12}{11}\)

\(b,-2x\left(x-1\right)+\left(x-1\right)\left(2x+3\right)=x+4\)

\(\Leftrightarrow-2x^2+2x+2x^2+3x-2x-3-x=4\)

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

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{2}{7}\)

Học tốt nhé

\(\cdot\left(x+1\right)^2\ge0\)

\(\Rightarrow x^2+2x+1>0\)

\(\Rightarrow2x^2+4x+2\ge0\)

 \(\Rightarrow\left(3x^2+3x+3\right)-\left(x^2-x+1\right)\ge0\)

\(\Rightarrow3\left(x^2+x+1\right)\ge x^2-x+1\)

\(\Rightarrow\)\(\frac{x^2+x+1}{x^2-x+1}\ge\frac{1}{3}\) (1)

\(\cdot\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2x^2-4x+2\ge0\)

\(\Rightarrow\left(3x^2-3x+3\right)-\left(x^2+x+1\right)\ge0\)

\(\Rightarrow3\left(x^2-x+1\right)\ge x^2+x+1\)

\(\Rightarrow\frac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ(1),(2) => đpcm

a) \(A=x^2-2x+5\)

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

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

Vì \(\left(x-1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge0;\forall x\)

b) a sẽ làm tắt 1 vài bước nhé khi nào kiểm tra thì em làm theo mẫu a là được 

\(B=4x^2+4x+11\)

\(=4\left(x^2+x+\frac{11}{4}\right)\)

\(=4\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{11}{4}\right)\)

\(=4\left[\left(x+\frac{1}{2}\right)^2+\frac{10}{4}\right]\)

\(=4\left(x+\frac{1}{2}\right)^2+10\ge10;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(B_{min}=10\Leftrightarrow x=\frac{-1}{2}\)

c) Tìm GTLN nhé 

 \(C=5-8x-x^2\)

\(=-x^2-2.x.4-16+16+5\)

\(=-\left(x+4\right)^2+21\)

Vì \(-\left(x+4\right)^2\le0;\forall x\)

\(\Rightarrow-\left(x+4\right)^2+21\le21;\forall x\)

Dấu "="xảy ra\(\Leftrightarrow\left(x+4\right)^2=0\)

                     \(\Leftrightarrow x=-4\)

Vậy\(C_{max}=21\Leftrightarrow x=-4\)

A = x² - 2x + 5

= ( x² - 2x + 1 ) + 4

= ( x - 1 )² + 4 ≥ 4 > 0 ∀ x ( đpcm )

B = 4x² + 4x + 11

= ( 4x² + 4x + 1 ) + 10

= ( 2x + 1 )² + 10 ≥ 10 ∀ x

Đẳng thức xảy ra <=> 2x + 1 = 0 => x = -1/2

=> MinB = 10 <=> x = -1/2

C = 5 - 8x - x²

= -( x² + 8x + 16 ) + 21

= -( x + 4 )² + 21 ≤ 21 ∀ x

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

=> MaxC = 21 <=> x = -4