Gợi ý: u – uv + v – v 2  = (1 – v)(u + z).

x² - 2x + 3 = ( x² - 2x + 1 ) + 2 = ( x - 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

x² - x + 1 = ( x² - x + 1/4 ) + 3/4 = ( x - 1/2 )² + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )

x² + 4x + 7 = ( x² + 4x + 4 ) + 3 = ( x + 2 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

-x² + 4x - 5 = -( x² - 4x + 4 ) - 1 = -( x - 2 )² - 1 ≤ -1 < 0 ∀ x ( đpcm )

-x² - x - 1 = -( x² + x + 1/4 ) - 3/4 = -( x + 1/2 )² - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )

-4x² - 4x - 2 = -4( x² + x + 1/4 ) - 1 = -4( x + 1/2 )² - 1 ≤ -1 < 0 ∀ x ( đpcm )

a, Ta co : A = 1999 * 2001

= ( 2000 - 1 ) *( 2000 + 1 )

= \(2000^2-1\)

Vây A < B

cậu ơi tối mình về mình làm tiếp cho bây giờ mình phải đi hok .

a) A = 1999.2001 và B = 2000²
Ta có :
A = 1999.2001
= ( 2000 - 1 )( 2000 + 1 )
= 2000² - 1²
= 2000² - 1
Mà : 2000² - 1 < 2000²
=> A < B

Bài 1 : Chứng minh phương trình vô nghiệm :

a, Ta có : \(x+2=x+5\)

=> \(x+2-x-5=0\)

=> \(-3=0\left(VL\right)\)

Vậy phương trình vô nghiệm .

b, Ta có : \(x^2-x+1=0\)

=> \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\) ( Vô lý )

Vậy phương trình vô nghiệm .

2) \(x^4-x^2+1=0\)(1)

Đặt: t=x², khi đó:

(1)\(\Leftrightarrow t^2-t+1=0\)

\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(2\right)\)

\(\Rightarrow\left(2\right)\) vô nghiệm => (1) vô nghiệm

Bài 1:

a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{2-u}{u+2}\)(1)

Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)

\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)

\(=\frac{-\left(u-2\right)}{u+2}\)

\(=\frac{2-u}{u+2}\)(2)

Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)

b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)

\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)

\(=v+3=VP\)(đpcm)

Bài 2:

a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)

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

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

hay \(M=2x^2-x-3\)

Vậy: \(M=2x^2-x-3\)

b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)

\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)

\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)

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

\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)

\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)

\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)

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

hay \(M=2x^2-5x+2\)

Vậy: \(M=2x^2-5x+2\)

Bài 3:

a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)

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

\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)

\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)

hay \(N=x^2+3x+2\)

Vậy: \(N=x^2+3x+2\)

n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)

\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)

\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)

hay \(N=\frac{2x-6}{x+3}\)

Vậy: \(N=\frac{2x-6}{x+3}\)

a) b 3 + 3 b 2 + 2 b 3 + 1 .          b) 0.

Mình camon nha ❤