xy - 4x = 29 - 5y

<=> x(y - 4) - 29 + 5y = 0

<=> x(y - 4) + 5(y - 4) - 9 = 0

<=> (x + 5)(y - 4) = 9 = 1.9 = 3.3

Lập bảng:

\(A=x^2-x\)

\(A=x^2-x+\frac{1}{4}-\frac{1}{4}\)

\(A=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)

Min \(A=\frac{-1}{4}\Leftrightarrow x=\frac{1}{2}\)

A=x²-x

Ta có: \(x^2\ge0\forall x\)

=> \(x^2-x\ge-x\forall x\)

Vậy Min_A= -x <=> x=0

Ơ, hình như não với bài của mình đang bị lag lag đâu đó '-'?

\(4x^2+2xy+4x+y+1\)

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

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

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

-3(x+4)(x-7)+7(x-5)(x-1)

=\(-3\left(x^2-3x-28\right)+7\left(x^2-6x+5\right)\)

= \(-3x^2+9x+84+7x^2-42x+35\)

= \(4x^2-33x+119\)

\(-3\left(x+4\right)\left(x+7\right)+7\left(x-5\right)\left(x-1\right)\)

\(=-3x^2+21x-12x+84+7x^2-7x-35x+35=4x^2-33x+119\)

1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )

2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)

\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)

\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )

Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))

1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

=> \(-4x^2+28x+4x^3-20x=28x^2-13\)

=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)

=> \(-4x^2+4x^3+8x-28x^2+13=0\)

=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)

=> \(-32x^2+4x^3+8x+13=0\)

=> vô nghiệm

2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)

=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)

=> \(-14x^2-56x+12=0\)

=> .... tự tìm

Câu c dấu bằng chỗ nào ?

\(A=2x^2+8x-20=2\left(x+2\right)^2-28\)

Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+2\right)^2-28\ge-28\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy Amin = - 28 <=> x = - 2

A = 2x² + 8x - 20

A = 2( x² + 4x + 4 ) - 28

A = 2( x + 2 )² - 28

2( x + 2 )² ≥ 0 ∀ x => 2( x + 2 )² - 28 ≥ -28

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

=> MinA = -28 <=> x = -2

a)\(\frac{x^2+3x+2}{3x+6}=\frac{x^2+2x+x+2}{3\cdot\left(x+2\right)}=\frac{\left(x^2+2x\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}=\frac{x\cdot\left(x+2\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}\)

\(=\frac{\left(x+2\right)\cdot\left(x+1\right)}{3\cdot\left(x+2\right)}=\frac{x+1}{3}\)

b) \(\frac{2x^2+x-1}{6x-3}=\frac{2x^2+2x-x-1}{3\cdot\left(2x-1\right)}=\frac{\left(2x^2+2x\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}\)

\(=\frac{2x\cdot\left(x+1\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{\left(2x-1\right)\cdot\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{x+1}{3}\)