3, A=(x-3)^2+(x-11)^2

\(\Rightarrow\)(X^2-3^2)+(x^2-11^2)

\(\Rightarrow\)(X^2-9)+(X^2-121)

Ta có :X^2 \(\ge\)0 và X^2 \(\ge\)0

\(\Rightarrow\)X^2 - 9 \(\le\)-9 và X^2- 121 \(\le\)-121

\(\Rightarrow\)(X^2-9)+(X^2-121)\(\le\)-130

Dấu = xảy ra khi : X=0

Vậy : Min A = -130 khi x=0

Mình mới lớp 7 sai thì thôi nhé

Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)

Vậy \(x=y=\frac{1}{2}\)

Ta có \(\Leftrightarrow x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow x^4+x^2+x^3+x+x^2+1=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)x\left(x^2+1\right)+\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+1=0\left(ktm\right)\\x^2+1=0\left(ktm\right)\end{cases}}\)

=> Pt vô nghiệm

đpcm.

\(x^4+x^3+x^2+x+1=0\)

\(\Rightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=0\)

\(\Rightarrow x^5-1=0\)

\(\Rightarrow x^5=1\)

\(\Rightarrow x=1\)

Nhưng thay vào PT ko đúng nên PT vô nghiệm

1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)

<=> 2x² - 8x + 3x - 12 + x² - 2x + 5x  - 10 - 3x² + 12x + 5x - 20 = 0

<=> 15x - 20 = 0

<=> 15x = 20

<=> x = 4/3

B1.a/ (x-2)(x^2+2x+2)

     b/ (x+1)(x+5)(x+2)

     c/ (x+1)(x^2+2x+4)

B2.

1a) x³ - 2x - 4 = 0

<=> (x³ - 4x) + (2x - 4) = 0

<=> x(x² - 4) + 2(x - 2) = 0

<=> x(x - 2)(x + 2) + 2(x - 2) = 0

<=> (x - 2)(x² + 2x + 2) = 0

<=> x - 2 = 0 (vì x² + 2x + 2 \(\ne\)0)

<=> x = 2

Vậy S = {2}

b) x³ + 8x² + 17x + 10 = 0

<=> (x³ + 5x²) + (3x² + 15x) + (2x + 10) = 0

<=> x²(x + 5) + 3x(x + 5) + 2(x + 5) = 0

<=> (x² + 3x + 2)(x + 5) = 0

<=> (x² + x + 2x + 2)(x + 5) = 0

<=> (x + 1)(x + 2)(x + 5) = 0

<=> x + 1 = 0 hoặc x + 2 = 0 hoặc x + 5 = 0

<=> x = -1 hoặc x = -2 hoặc x = -5

Vậy S = {-1; -2; -5}

c) x³ + 3x² + 6x + 4 = 0

<=> (x³ + x²) + (2x² + 2x) + (4x + 4) = 0

<=> x²(x + 1) + 2x(x + 1) + 4(x + 2) = 0

<=> (x² + 2x + 4)(x + 2) = 0

<=> x + 2 = 0

<=> x = -2

Vậy S = {-2}

Bài 1:

\(M=x^4-x^3-x^3+x^2+2x^2-2x+2\)

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

\(=3x^2-3x+6+2\)

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

\(=3\left(x^2-x\right)+8=3\cdot3+8=17\)