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\(A=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{4}{2}=2\)

Dấu ''='' xảy ra khi a = b = 1 

Vậy GTNN của A bằng 2 tại a = b = 1   

\(A=a^2+b^2\)

\(=\left(a+b\right)^2-2ab\)

\(=4-2ab\)

Giả sử \(a;b\ge0\)

Áp dụng bất đẳng thức Cô-si cho hai số a;b dương thì ta có:

\(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)

\(\Leftrightarrow1\ge ab\)

\(\Rightarrow-2ab\ge-2\)

\(\Leftrightarrow4-2ab\ge2\)

\(\Leftrightarrow A\ge2\)

Vậy \(MinA=2\)

Dấu '' = '' xảy ra khi: \(a=b=1\)

1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)

2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)

3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)

4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)

5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)

\(1,\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2,\)

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow x^2-x+9x-9=0\)

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

\(3,\)

\(6x\left(x-2\right)=x-2\)

\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)

\(4,\)

\(7\left(x-2020\right)^2-x+2020=0\)

\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)

\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)

\(5,\)

\(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

\(6,\)

\(x^2-2x-3=0\)

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

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

\(1,\)

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

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

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

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

\(2,\)

\(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

\(=\left(x+y-5\right)\left(x+y+5\right)\)

\(3,\)

\(x^2y-x^3-9y+9x\)

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

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(x^2-9\right)\left(y-x\right)\)

\(=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

\(4,\)

\(x^4+2x^3+x^2\)

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

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

\(5,\)

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

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

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

2, \(x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)

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

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

5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)

\(B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2+1\right)\)

\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

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

\(=-8\)

Vậy biểu thức không phụ thuộc vào biến

\(B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)

\(=-6x^2-2+6x^2-6=-8\)

Vậy biểu thức ko phụ thuộc vào giá trị biến x 

\(4x^2-12x-7\)

\(=4x^2+2x-14x-7\)

\(=2x\left(2x+1\right)-7\left(2x+1\right)\)

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

Cách 1:

\(4x^2-12x-7\)

\(=\left(4x^2-12x+9\right)-16\)

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

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

\(=\left(2x-7\right)\left(2x+1\right)\)

Cách 2:

\(4x^2-12x-7\)

\(=4x^2+2x-14x-7\)

\(=2x\left(2x+1\right)-7\left(2x+1\right)\)

\(=\left(2x+1\right)\left(2x-7\right)\)

Lời giải :

Thấy ảnh không ạ?

Hơi mờ bạn thông cảm!

a, Xét tam giác AMB và NMC :

MN=MA

CM=MB

CMN=AMB(2 góc đối đỉnh)

suy ra tam giác AMB=NMC (1)

từ (1) suy ra góc NCM=MBA suy ra CN//AB suy ra góc NCA=CAB mà CAB=90⁰(do vuông tại A) suy ra NCA =90⁰ suy ra IC vuông góc với CA

Đặt x + 4 = a ; 2a - 5 = b ; 1 - 3x = c

Nhận thấy a + b + c = 0 

=> a + b = -c

<=> (a + b)⁵ = (-c)⁵ 

<=> a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵ = -c⁵ 

<=> a⁵ + b⁵ + c⁵ = -5ab(a³ + 2a²b + 2ab² + b³) 

                           = -5ab[(a³ + b³) + 2ab(a + b)] 

                           = -5ab(a + b)(a² + b² + ab)

                           = 5abc(a² + b² + ab) = 0 

=> 5(x + 4)(2x - 5)(1 - 3x)[(x + 4)² + (2x - 5)² + (x + 4)(2x - 5)] = 0

<=> 5(x  + 4)(2x - 5)(1 - 3x) = 0 (vì [(x + 4)² + (2x - 5)² + (x + 4)(2x - 5) > 0 với mọi x) 

=> x = -4 hoặc x = 2,5 hoặc x = 1/3 

Vậy \(x\in\left\{-4;2,5;\frac{1}{3}\right\}\)là nghiệm phương trình 

Có đoạn này em không hiểu, tại sao (x+4)^2 + (2x-5)^2 + (x+4)(2x-5) > 0 với mọi x ạ?

d) \(25x^6-\frac{4y^2}{49}=\left(5x^3\right)^2-\left(\frac{2y}{7}\right)^2=\left(5x^3-\frac{2y}{7}\right)\left(5x^3+\frac{2x}{7}\right)\)

e) \(27x^3-\frac{1}{8}=\left(3x\right)^3-\left(\frac{1}{2}\right)^3=\left(3x-\frac{1}{2}\right)\left(9x^2+\frac{3}{2}x+\frac{1}{4}\right)\)

f ) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)

g) \(8x^3+125=\left(2x\right)^3+5^3=\left(2x+5\right)\left(4x^2-10x+25\right)\)

h) \(x^3+\frac{y^3}{8}=x^3+\left(\frac{y}{2}\right)^3=\left(x+\frac{y}{2}\right)\left(x^2-\frac{xy}{2}+\frac{y^2}{4}\right)\)

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