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

<=> x³ - 1 - x( x + 2 )( x - 2 ) = 5 

<=> x³ - 1 - x( x² - 4 ) = 5

<=> x³ - 1 - x³ + 4x = 5

<=> 4x - 1 = 5

<=> 4x = 6

<=> x = 6/4 = 3/2

b) 5x( x - 3 )² - 5( x - 1 )³ + 15( x + 4 )( x - 4 ) = 5

<=> 5x( x² - 6x + 9 ) - 5( x³ - 3x² + 3x - 1 ) + 15( x² - 16 ) = 5

<=> 5x³ - 30x² + 45x - 5x³ + 15x² - 15x + 5 + 15x² - 240 = 5

<=> 30x - 235 = 5

<=> 30x = 240

<=> x = 8

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

\(< =>x^3-1+x\left(4-x^2\right)=5\)

\(< =>x^3-1+4x-x^3=5\)

\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)

b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)

\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)

\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)

\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)

\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)

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 ?