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

Dài rứa :v

Bài 1.

1. ( x + 2y )² = x² + 4xy + 4y²

2. ( 2x + 3y )² = 4x² + 12xy + 9y²

3. ( 3x - 2y )² = 9x² - 12xy + 4y²

4. ( 5x - y )² = 25x² - 20xy + y²

5. ( x + 1/4 )² = x² + 1/2x + 1/16

6. ( 2x - 1/2 )² = 4x² - 2x + 1/4

7. ( 1/3x - 1/2y )² = 1/9x² - 1/3xy + 1/4y²

8. ( 3x + 1 )( 3x - 1 ) = ( 3x )² - 1² = 9x² - 1

9. ( x² + 2/5y )( x² - 2/5y ) = ( x² )² - ( 2/5y )² = x⁴ - 4/25y²

10. ( x/2 + y )( x/2 - y ) = ( x/2 )² - y² = x²/4 - y²

11. Chưa rõ đề

12. \(\left(\sqrt{2x}-y\right)^2=2x-2\sqrt{2x}\cdot y+y^2\)

13. ( 3/2x + 3y )² = 9/4x² + 9xy + 9y²

14. \(\left(\sqrt{2x}+\sqrt{8y}\right)^2=2x+\left(2\sqrt{2x}\sqrt{8y}\right)+8y\)

15. ( x + 1/6y + 3 )² = [( x + 1/6y ) + 3 ]² = x² + 1/3xy + 1/36y² + 6x + y + 9

16. ( 1/2x - 4y )² = 1/4x² - 4xy + 16y²

17. ( x/2 + 2y² )( x/2 - 2y² ) = ( x/2 )² - ( 2y² )² = x²/4 - 4y⁴

18. ( x² - 4 )( x² + 4 ) = ( x² )² - 4² = x⁴ - 16

19. ( x + y )² + ( x - y )² = x² + 2xy + y² + x² - 2xy - y² = 2x² + 2y² = 2( x² + y² )

20. ( 2x + 3 )² - ( x + 1 )² = [ 2x + 3 + x + 1 ][ 2x + 3 - ( x + 1 )] = [ 3x + 4 ][ x + 2 ]

1. (x + 2y)² = (x + 2y)(x + 2y) = x(x + 2y) + 2y(x + 2y) = x² + 2xy + 2xy + 4y² = x² + 4xy + 4y²

2. (2x + 3y)² = (2x + 3y)(2x + 3y) = 2x(2x + 3y) + 3y(2x + 3y) = 4x² + 6xy + 6xy + 9y² = 4x² + 12xy + 9y²

3. (3x - 2y)² = (3x - 2y)(3x - 2y) = 3x(3x - 2y) - 2y(3x - 2y) = 9x² - 6xy - 6xy + 4y² = 9x² - 12xy + 4y²

4. (5x - y)²  = (5x - y)(5x - y) = 5x(5x - y) - y(5x - y) = 25x² - 5xy - 5xy - y² = 25x² - 10xy - y²

5. (x + 1/4)² = (x + 1/4)(x + 1/4) 

= x(x + 1/4) + 1/4(x + 1/4)

= x² + 1/4x + 1/4x + 1/16

= x² + 1/2x + 1/16

6. (2x - 1/2)² = (2x - 1/2)(2x - 1/2)

= 2x(2x - 1/2) - 1/2(2x - 1/2)

= 4x² - x - x + 1/4

= 4x² - 2x + 1/4

7. (1/3x - 1/2y)² = (1/3x - 1/2y)(1/3x - 1/2y)

= 1/3x(1/3x - 1/2y) - 1/2y(1/3x - 1/2y)

= 1/9x² - 1/6xy - 1/6xy + 1/4y²

= 1/9x² - 1/3xy + 1/4y²

8. (3x + 1)(3x - 1)

= 3x(3x - 1) + 1(3x - 1)

= 9x² - 3x + 3x - 1

= 9x² - 1

9. \(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)

= x²(x² - 2/5y) + 2/5y(x² - 2/5y)

= x⁴ - 2/5x²y + 2/5x²y - 4/25y²

= x⁴ - 4/25y²

10. \(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)\)

= \(\frac{x}{2}\left(\frac{x}{2}+y\right)-y\left(\frac{x}{2}+y\right)\)

= \(\frac{x^2+2xy}{4}-\frac{2y^2+xy}{2}\)

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

= \(\frac{x^2-4y^2}{4}\)

11. Giữ nguyên

Làm nốt đi nhé chứ mỏi tay lắm rồi

Bài 4 : \(\left(x-10\right)^2-x\left(x+80\right)\)

= (x - 10)(x - 10) - x² - 80x

= x(x-  10) - 10(x - 10) - x² - 80x

= x² - 10x - 10x + 100 - x² - 80x

= (x² - x²) + (-10x - 10x - 80x) + 100

= -100x + 100 = -100(x - 1)

Thay x = 0,98 vào biểu thức trên ta có :

-100(x - 1) = -100(0,98 - 1) = -100.(-0,02) = 2

2. (2x + 9)² - x(4x + 31)

= (2x + 9)(2x + 9) - 4x² - 31x

= 2x(2x + 9) + 9(2x + 9) - 4x² - 31x

= 4x² + 18x + 18x + 18 - 4x² - 31x

= (4x² - 4x²) + (18x + 18x - 31x) + 18

= 5x + 18

Thay x = -16,2 vào biểu thức trên ta có : 5.(-16,2) + 18 = -63

3. Thay x = 4 vào biểu thức trên ta có :

4.4² - 28.4 + 49 = 4.16 - 112 + 49 = 64 - 112 + 49 = 1

4. Thay x = 1 vào biểu thức trên ta có :

9.1² + 42.1 + 49 = 9 + 42 + 49 = 100

5. Thay x = -1/5 ; y = -5 vào biểu thức trên ta có :

25. (-1/5)² - 2.(-1/5) . (-5) + 1/25 . (-5)²

= 25 . 1/25 - 2 + 1

= 1 - 2 + 1 = 0

Lần sau đăng ít bài thôi