Tìm x,y,z biết: x2 + y2 - 4x = 6z - 2y - z2 - 14
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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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(10x^2y+25x^3+xy^2=x\left(5x+y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)=-2x^3+6x^2+4\)
\(A=\left(a^2+\left(a-1\right)\right)\left(a^2-\left(a-1\right)\right)=a^4-\left(a-1\right)^2=a^4-\left(a^2-2a+1\right)=a^4-a^2+2a-1\)
\(B=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)
\(C=9y^2+12y+4-\left(4x^2-12xy+9y^2\right)-12xy=12y+4-4x^2\)
\(D=x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x+1=-x^3+6x^2-x+4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left(a_1+a_2+...+a_{10}\right)⋮6\left(gt\right)\)
\(\Rightarrow\left(a_1^3+a_2^3+...+a_{10}^3\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x\in\left\{-1;-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(x^5+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(2x^2-x-1=\left(2x+1\right)\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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