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

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

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

\(\Leftrightarrow-x=-1\Leftrightarrow x=1\)

Bài làm:

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

\(\Leftrightarrow x^3-1-2x=x^3-x\)

\(\Leftrightarrow x=-1\)

Mình làm câu đầu tượng trưng thui nhé, 2 câu sau tương tự vậy !!!!!!

a) pt <=> \(x^2-2xy+2y^2-2x-2y+5=0\)

<=> \(\left(x-y-1\right)^2+y^2-4y+4=0\)

<=> \(\left(x-y-1\right)^2+\left(y-2\right)^2=0\)    (1) 

TA LUÔN CÓ: \(\left(x-y-1\right)^2;\left(y-2\right)^2\ge0\forall x;y\)

=> \(\left(x-y-1\right)^2+\left(y-2\right)^2\ge0\)      (2)

TỪ (1) VÀ (2) => DẤU "=" SẼ PHẢI XẢY RA <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

VẬY \(\left(x;y\right)=\left(3;2\right)\)

Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath

Đây nha! Vô tcn xem ảnh!

a) \(\left(x^2+2x+2\right)\left(x^2+2x+3\right)=0\)

<=> \(\orbr{\begin{cases}x^2+2x+2=0\\x^2+2x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+1\right)^2+1=0\left(vl\right)\\\left(x+1\right)^2+2=0\left(vl\right)\end{cases}}\)

=> pt vô nghiệm

b) \(\left(x+3\right)\left(x-3\right)\left(x^2-11\right)+3=2\)

<=> \(\left(x^2-9\right)\left(x^2-11\right)+1=0\)

<=> \(\left(x^2-9\right)^2-2\left(x^2-9\right)+1=0\)

<=> \(\left(x^2-9-1\right)^2=0\)

<=> \(x^2-10=0\)

<=> \(x=\pm\sqrt{10}\)

c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

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

Đặt x + 4 = a

<=> \(\left(a-1\right)^4+\left(a+1\right)^4=2\)

<=> \(a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)

<=> \(a^4+12a^2=0\)

<=> \(a^2\left(a^2+12\right)=0\)

<=> a = 0 (vì a² + 12 > 0)

Vậy S = {0}

\(M=x^2-8x+5\)

\(\Leftrightarrow M=x^2-8x+16-11\)

\(\Leftrightarrow M=\left(x-4\right)^2-11\ge-11\)

Min M = -11 

\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)

\(N=-3x-6x-9\)

\(\Leftrightarrow N=-9x-9\le-9\)

Max N = -9

\(\Leftrightarrow x=0\)

1) x² - 6x + 9 = (x² - 3x) -(3x - 9) = x(x - 3) - 3(x - 3) = (x - 3)(x - 3)

2) 25 + 10x + x² = x² + 5x + 5x + 25 = x(x + 5) + 5(x + 5) = (x + 5)(x + 5)

5) x² - 10x + 25 = x² - 5x - 5x + 25 = x(x - 5) - 5(x - 5) = (x - 5)(x - 5)

6) x² + 4xy + 4y² = x² + 2xy + 2xy + 4y² = x(x + 2y) + 2y(x + 2y) = (x + 2y)(x + 2y)

7) (3x + 2)² - 4 = (3x + 2)² - 2² = (3x + 2 - 2)(3x + 2 + 2) = 3x(3x +  4)

8) 4x² - 49 = (2x)² - 7² = (2x - 7)(2x + 7)

9) \(\frac{9}{25}x^4-\frac{1}{4}=\left(\frac{2}{3}\right)^2.\left(x^2\right)^2-\left(\frac{1}{2}\right)^2=\left(\frac{2}{3}x^2\right)^2-\left(\frac{1}{2}\right)^2=\left(\frac{2}{3}x^2-\frac{1}{2}\right)\left(\frac{2}{3}x^2+\frac{1}{2}\right)\)

10) x³² - 1 = (x¹⁶)² - 1² = (x¹⁶ - 1)(x¹⁶ + 1)

11 4x² + 4x + 1 = 4x² + 2x + 2x + 1 = 2x(2x + 1) + (2x + 1) = (2x + 1)(2x + 1)

12 x² - 20x - 100 = x² - 10x - 10x + 100 = x(x - 10) - 10(x - 10) = (x - 10)(x - 10)

13 y⁴ - 14y² + 49 = y⁴ - 7y² - 7y² + 49 = y²(y² - 7) - 7(y² - 7) = (y² - 7)²

Mk chỉ làm được đến đó thôi

 Định làm hết nhưng bạn kia làm đúng rồi ! Còn 2 câu làm nốt !!!!

                                   Bài giải

\(3,\text{ }\frac{1}{4}a^2+2ab+4b^2=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)

\(4,\text{ }\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2\cdot\frac{1}{3}y^4+\left(y^4\right)^2=\left(y^4+\frac{1}{3}\right)^2\)

\(A=\left(x-2\right)^2+\left(x+3\right)^2-2\left(x+1\right)\left(x-1\right).\)

\(A=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(A=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(A=2x+15\)

\(B=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(B=\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(B=\left(x^8-y^8\right)\left(x^8+y^8\right)\)

\(B=x^{16}-y^{16}\)

VẬY \(B=x^{16}-y^{16}\)