\(25\left(x-3\right)^2-\left(2x-7\right)^2\)(*)

Đặt \(x-3=t\)và \(2x-7=z\)thay vào (*) ta được:

\(25t^2-z^2\)

\(=\left(5t-z\right)\left(5t+z\right)\)thay t=x-3 và y=2x-7 ta được:

\(=\left(5x-15-2x+7\right)\left(5x-15+2x-7\right)\)

\(=\left(3x-8\right)\left(7x-22\right)\)

C2 nhân ra rồi phân tích

\(25\left(x-3\right)^2-\left(2x-7\right)^2\)

\(=5^2.\left(x-3\right)^2-\left(2x-7\right)^2\)

\(=\left[5.\left(x-3\right)\right]^2-\left(2x-7\right)^2\)

\(=\left[5\left(x-3\right)-\left(2x-7\right)\right]\left[5\left(x-3\right)+\left(2x-7\right)\right]\)

\(=\left(5x-15-2x+7\right)\left(5x-15+2x-7\right)\)

\(=\left(3x-8\right)\left(7x-22\right)\)

Đề sai e nhé

\(E=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)+2017\)

\(=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy+2017\)

\(=\left(x^3-3x^2y+3xy^2-y^3\right)+\left(x^2-2xy+y^2\right)+2017\)

\(=\left(x-y\right)^3+\left(x-y\right)^2+2017\)

\(=\left(-3\right)^3+\left(-3\right)^2+2017\)

\(=-27+9+2017\)

\(=1999\)

SAI DE

 P=2x²+y²-2xy-6x+2y+2024

=>2P=4x²+2y²-4xy-12x+4y+4048

=(2x-y-3)²+y²-2y+1+4038

=(2x-y-3)²+(y-1)²+4038> hoặc = 4038

Dấu = xảy ra <=>2x-y-3=0 và y-1=0=>x=2;y=1=>2p=4038=>p=2019

Vậy Pmin=2019<=>x=2;y=1

Ta có: 

P = 2x² + y² - 2xy - 6x + 2y + 2024

P = (x² - 2xy + y²) - 2(x - y) + 1 + (x² - 4x + 4) + 2019

P = [(x - y)² - 2(x - y) + 1] + (x - 2)² + 2019

P = (x - y - 1)² + (x - 2)² + 2019 \(\ge\)2019 \(\forall\)x;y

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

Vậy MinP = 2019 <=> x = 2 và y = 1

Ta có: 4x² - y² + 4x + 4y - 3

= (4x² - 4x + 1) - (y² - 4y + 4)

= (2x - 1)² - (y - 2)²

= (2x - 1 -y + 2)(2x - 1 + y - 2)

= (2x - y + 1)(2x + y - 3)

\(4x^2-y^2+4x+4y-3\)

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

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

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

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

Bài làm

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

( x - 3 )( 2x - 1 + x + 4 ) + 3 = 0

( x - 3 )( 3x + 3 ) = 0

=> x - 3 = 0 hoặc 3x - 3 = 0

=> x      = 3 hoặc 3x      = 3

=> x      = 3 hoặc   x      = 3 : 3

=> x      = 3 hoặc   x      = 1

Vậy x = 3; x = 1

# Học tốt #

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

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

\(\Leftrightarrow\left(3-x\right)\left(2x-1-x-4\right)=-3\)

\(\Leftrightarrow\left(3-x\right)\left(x-5\right)=-3\)

\(\Leftrightarrow x^2-8x+15=-3\)

\(\Leftrightarrow x^2-8x+18=0\)

Ta có: \(\Delta=8^2-4.18=-8< 0\)

Vậy pt vô nghiệm