Áp dụng HĐT a² - b² = ( a + b )( a - b ) ta có :

50² - 49² + 48² - 47² + ... + 2² - 1²

= ( 50² - 49² ) + ( 48² - 47² ) + ... + ( 2² - 1² )

= ( 50 + 49 )( 50 - 49 ) + ( 48 + 47 )( 48 - 47 ) + ... + ( 2 + 1 )( 2 - 1 )

= 99.1 + 95.1 + ... 3.1

= 99 + 95 + ... + 3

= \(\frac{\left(99+3\right)\left[\left(99-3\right):4+1\right]}{2}\)

= 1275

Ta có: \(x^2+y^2-4x=6z-2y-z^2-14\)

\(x^2+y^2-4x-6z+2y+z^2+14=0\)

\(\left(x^2-4x+2^2\right)+\left(y^2+2y+1\right)+\left(z^2-6z+3^2\right)=0\)

\(\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)

\(\cdot\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

\(\cdot\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\)

\(\left(z-3\right)^2=0\Rightarrow z-3=0\Rightarrow z=3\)

hok tốt!

Ta có x² + y² - 4x = 6z - 2y - z² - 14

=> x² + y² - 4x - 6z + 2y + z² + 14 = 0

=> (x² - 4x + 4) + (y² + 2y + 1) + (z² - 6z + 9) = 0

=> (x - 2)² + (y + 1)² + (z - 3)² = 0

Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y+1=0\\z-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)

Vậy x = 2 ; y = - 1 ; z = 3

Ta có: x + y = a + b

\(\Rightarrow\left(x+y\right)^2=\left(a+b\right)^2\)

\(\Rightarrow x^2+y^2=a^2+b^2\)(đpcm)

đề hơi sai!!:))

hok tốt!

Ta có : x + y = a + b (1)

=> (x + y)³ = (a + b)³

=> x³ + y³ + 3x²y + 3y²x = a³ + b³ + 3ab² + 3a²b

=> 3x²y + 3y²x = 3ab² + 3a²b

=> 3xy(x + y) = 3ab(a + b)

=> 3xy = 3ab

=> xy = ab

Từ (1) => (x + y)² = (a + b)²

=> x² + y² + 2xy = a² + b² + 2ab

=> x² + y² = a² + b² (Vì xy = ab => 2xy = 2ab) (đpcm)

Giúp mk!! 

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

b. \(7x^2y^2+14xy^2-21^2y=7y\left(x^2y+2xy-63\right)\)

c. \(10x^2y+25x^3+xy^2=x\left(5x+y\right)^2\)

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

b. \(\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)\left(a^2-2x+4\right)=a^6-64\)

c. \(\left(2+3y\right)^2-\left(2x-3y\right)^2-12xy=4+12y-4x^2\)

d. \(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)=-2x^3+6x^2+4\)

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

\(B=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)

\(C=9y^2+12y+4-\left(4x^2-12xy+9y^2\right)-12xy=12y+4-4x^2\)

\(D=x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x+1=-x^3+6x^2-x+4\)

Xét hiệu:

\(\left(a_1^3+a_2^3+...+a_{10}^3\right)-\left(a_1+a_2+...+a_{10}\right)\)

\(=\left(a_1^3-a_1\right)+\left(a_2^3-a_2\right)+...+\left(a_{10}^3-a_{10}\right)\)

Ta dễ dàng chứng minh các biểu thức trong ngoặc đều chia hết cho 6

Lại có: \(\left(a_1+a_2+...+a_{10}\right)⋮6\left(gt\right)\)

\(\Rightarrow\left(a_1^3+a_2^3+...+a_{10}^3\right)⋮6\)

a) Ta có : x³ - x = 0

=> x(x² - 1) = 0

=> \(\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Vậy \(x\in\left\{0;1;-1\right\}\)

b) x² + 4x = 0

=> x(x + 4) = 0

=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

Vậy \(x\in\left\{0;-4\right\}\)

c) 9x² - 1 = 0

=> 9x² = 1

=> x² = \(\frac{1}{9}\)

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

Vậy \(x\in\left\{\frac{1}{3};-\frac{1}{3}\right\}\)

d) 5x² - 10x + 5 = 0

=> 5x² - 5x - 5x + 5 = 0

=> 5x(x - 1) - 5(x - 1) = 0

=> 5(x - 1)² = 0

=> (x - 1)² = 0

=> x - 1 = 0

=> x = 1

e) x² + 6x + 5 = 0

=> x² + 6x + 9 - 4 = 0

=> (x + 3)² = 4

=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

Vậy \(x\in\left\{-1;-5\right\}\)

a, \(x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b, \(5x^2-10x+5=0\)

\(\Leftrightarrow5x\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

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

2) \(=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

3) 

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

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

1.\(x^2-2021+2020=x^2-1=\left(x+1\right)\left(x-1\right)\)

2. \(x^4+64=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

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

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

1. \(11x-7x+6=4x+6=2\left(2x+3\right)\)

2. \(7x^2+xy-2x^2=5x^2+xy=x\left(5x+y\right)\)

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