Đại số lớp 8

M = (x + 2)(x + 4)(x + 6)(x + 8) + 16

M = [(x + 2)(x + 8)][(x + 4)(x + 6)] + 16

M = (x^2 + 2x + 8x + 16)(x^2 + 4x + 6x + 24) + 16

M = (x^2 + 10x + 16)(x^2 + 10x + 24) + 16

Đặt t = x^2 + 10x + 20

M = (t - 4)(t + 4) + 16

M = t^2 - 16 + 16 = t^2

Vậy ta có đpcm

\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)\left(x-3\right)\left(x+3\right)=\left(x-3\right)^2\left(x+3\right)\)

\(\left(3x-2\right)\left(3x+2\right)-\left(4-3x\right)^2=9x^2-4-16+24x-9x^2=24x-20=4\left(6x-5\right)\)

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

\(4x^2+20x+25-4x^2+1=-3\)

\(20x=-3-25-1\)

\(20x=-29\)

\(x=-\frac{29}{20}\)

a) (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

b) (a + b - c)² = a² + b² + c² + 2ab - 2ac - 2bc

c) (a - b - c)² = a² + b² + c² - 2ab - 2ac + 2bc

a)

( a + b + c)² = [(a + b) + c ]²

= (a + b)² + 2(a + b)c + c²

= a² + 2ab + b² + 2ac +2bc + c²

= a² + b² + c² + 2ab + 2ac + 2bc

b)

(a + b - c)² = [(a + b) - c ]²

= (a + b)² - 2(a + b)c + c²

= a² + 2ab + b² - 2ac - 2bc + c²

= a² + b² + c² + 2ab - 2ac - 2bc

c)

(a - b - c)² = [(a - b) - c ]²

= (a - b)² - 2(a - b)c + c²

= a² - 2ab + b² - 2ac + 2bc + c²

= a² + b² +c² - 2ab - 2ac + 2bc

\(x^6+3x^2y^2+y^6=\left(x^6+y^6\right)+3x^2y^2=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)+3x^2y^2\)

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

\(P=x^2-2xy+6y^2-12x+3y+45\)

\(=x^2-2x\left(y+6\right)+\left(y+6\right)^2-\left(y+6\right)^2+6y^2+3y+45\)

\(=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+\left(5y^2-9y+9\right)\)

\(=\left(x-y-6\right)^2+5\left(y-\frac{9}{10}\right)^2+\frac{99}{20}\)

\(\ge\frac{99}{20}\) . Đẳng thức xảy ra khi y = 9/10, x = 69/10

Vậy min P = 99/20 tại x = 69/10, y = 9/10

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Rightarrow\left[\left(a+d\right)+\left(b+c\right)\right]\left[\left(a+d\right)-\left(b+c\right)\right]-\left[\left(a-d\right)-\left(b-c\right)\right]\left[\left(a-d\right)+\left(b-c\right)\right]=0\)

\(\Rightarrow\left(a+d\right)^2-\left(b+c\right)^2-\left(a-d\right)^2+\left(b-c\right)^2=0\)

\(\Rightarrow a^2+d^2+2ad-b^2-c^2-2bc-a^2-d^2+2ad+b^2+c^2-2bc\)

\(\Rightarrow4ad-4bc\)

\(\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(t=x^2+x\) ta có pt trở thành:

\(t^2+4t=12\Leftrightarrow t^2+4t-12=0\)

\(\Leftrightarrow t^2-2t+6t-12=0\)\(\Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)+4=8\)

\(\Leftrightarrow\left(x^2+x+2\right)^2=8\)

\(\left[{}\begin{matrix}x^2+x+2=8\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\x^2+x+2=-8\left(loai\right)\end{matrix}\right.\)

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

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Leftrightarrow x=2\)

A = 100² - 99² + 98² - 97² + . . . + 2² - 1²

= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)

= 199 + 195 + . . . + 3

= 5050

B = 3(2² + 1)(2⁴ + 1) . . . (2⁶⁴ + 1) + 1

= (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1)(2⁶⁴ + 1) + 1

= (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2¹⁶ - 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2³² - 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2⁶⁴ - 1)(2⁶⁴ + 1) + 1

= 2¹²⁸ - 1 + 1

= 2¹²⁸

Câu C mk chép nhầm đề đó