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

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

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

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

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

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

Bài 1: 

a) (x+y)²=9²=81

=> x²+2xy+y²=81

=> x²+2.14+y²=81

=> x²+y²=53

=> x²-2xy+y²=81-2.14=25

=> (x-y)²=25

=> x-y=5 hoặc x-y=-5

b) Câu a đã tính được x²+y²=53

c) Ta có: x³+y³=(x+y)(x²-xy+y²)=9(53-14)=9.39=351

Bài 2: 

Ta có: \(x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1\)

Mà x+y=1

\(\Rightarrow1^2-4.1+1=-2\)

Bài 3: 

Ta có: (x+y)³=x³+3x²y+3xy²+y³ 

= x³+y³+3xy(x+y)

Mà x+y=1 => (x+y)³=x³+y³+3xy=1³=1

Bài 4: 

Ta có: \(\left(x+y\right)^2=4^2=16\)

\(\Rightarrow x^2+2xy+y^2=16\Rightarrow10+2xy=16\)

\(\Rightarrow2xy=6\Rightarrow xy=3\)

Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4.\left(10-3\right)\)

\(=4.7=28\)

Bài 5: 

Ta có: \(x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

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

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

Mấy bài này đầu hè làm hết rồi:))

Bài 1:

a) \(xy=14\Rightarrow x=\frac{14}{y}\)

Thay vào: \(\frac{14}{y}+y=9\)

\(\Leftrightarrow y^2+14-9y=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x=7\\y=2\end{cases}}\Rightarrow x-y=5\)

+ Nếu: \(\hept{\begin{cases}x=2\\y=7\end{cases}}\Rightarrow x-y=-5\)

b) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^2=81\)

\(\Leftrightarrow x^2+2xy+y^2=81\)

\(\Rightarrow x^2+y^2=81-2xy=81-2.14=53\)

c) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^3=9^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)

\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)

B = x² - 4x + 2

B = ( x² - 4x + 4 ) - 2

B = ( x - 2 )² - 2

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

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

=> MinB = -2 <=> x = 2

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

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow\left(x-2\right)^2-2\ge-2\)

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

Vậy Bmin = - 2 <=> x = 2

Bài làm:

a) \(x^2-2xy+y^2-zx+yz\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(\left(x-y\right)\left(x-y-z\right)\)

a/ \(x^2-2xy+y^2-zx+yz.\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

c/ \(x^2-y^2-2x-2y.\)

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

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

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

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

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