![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(28\left(x-1\right)^2\)chẵn mà 37 lẻ nên \(y^2\)lẻ
Mà \(y^2\)là số chính phương và \(y^2\le37\)nên \(y^2\in\left\{1;9;25\right\}\)
\(TH1:y^2=1\Rightarrow\left(x-1\right)^2=\frac{36}{28}\left(L\right)\)
\(TH2:y^2=9\Rightarrow\left(x-1\right)^2=1\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
\(TH3:y^2=25\Rightarrow\left(x-1\right)^2=12\left(L\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) P = x(x + 2) + y(y - 2) - 2xy + 37
⇒ P = x2 + 2x + y2 - 2y - 2xy + 37
⇒ P = (x2 - 2xy + y2) + 2(x - y) + 37
⇒ P = (x - y)2 + 2.7 + 37
⇒ P = 72 +14 + 37
⇒ P = 49 + 51
⇒ P = 100
b) Q = x2(x + 1) - y2(y - 1) + xy - 3xy(x - y + 1) - 95
⇒ Q = x3 + x2 - y3 + y2 + xy - 3x2y + 3xy2 - 3xy - 95
⇒ Q = (x3 - 3x2y + 3xy2 - y3) + (x2 + xy - 3xy + y2) - 95
⇒ Q = (x - y)3 + (x - y)2 - 95
⇒ Q = 73 + 72 - 95
⇒ Q = 343 + 49 - 95
⇒ Q = 297
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Sửa đề: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=x^2+y^2+1+2x-2y-2xy+36\)
\(=\left(x-y+1\right)^2+36\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=\left(7+1\right)^2+36=8^2+36=100\)
Vậy: 100 là giá trị của biểu thức \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\) tại x-7=7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)
Vậy: Khi x-y=7 thì A=100
b) Ta có: \(x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy+10=4\)
\(\Leftrightarrow2xy=-6\)
\(\Leftrightarrow xy=-3\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)
Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:
\(A=2\cdot\left(10+3\right)=2\cdot13=26\)
Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26
\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)
\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(A=x^2+2x+y^2-2y-2xy+37\)
\(A=\left(x^2+y^2-2xy+1+2x-2y\right)+36\)
\(A=\left(x-y+1\right)^2+36\)
\(A=\left(7+1\right)^2+36\)
\(A=8^2+36\)
\(A=100\)
\(B=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)-95\) \((9^5\) \(sai\)\()\)
\(B=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy-95\)
\(B=\left(x^3-3x^2y+3xy^2-y^3\right)+\left(x^2+xy-3xy+y^2\right)-95\)
\(B=\left(x-y\right)^3+\left(x^2-2xy+y^2\right)-95\)
\(B=\left(x-y\right)^3+\left(x-y\right)^2-95\)
\(B=7^3+7^2-95\)
\(B=297\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\dfrac{y^4-1}{y^3+y^2+y+1}\)
\(=\dfrac{\left(y^2+1\right)\left(y^2-1\right)}{y^2\left(y+1\right)+\left(y+1\right)}\)
\(=\dfrac{\left(y^2+1\right)\left(y+1\right)\left(y-1\right)}{\left(y+1\right)\left(y^2+1\right)}\)
\(=y-1\)
d) \(\dfrac{2x^2-9x+7}{-2x^2-x+28}\)
\(=\dfrac{2x^2-2x-7x+7}{-\left(2x^2+8x-7x-28\right)}\)
\(=\dfrac{2x\left(x-1\right)-7\left(x-1\right)}{-\left(2x-7\right)\left(x+4\right)}\)
\(=-\dfrac{\left(2x-7\right)\left(x-1\right)}{\left(2x-7\right)\left(x+4\right)}\)
\(=\dfrac{1-x}{x+4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 1)(3x + 1) + (3x + 4)(3 - 2x)
= 6x2 + 2x - 3x - 1 + 9x - 6x2 + 12 - 8x
= 11
b) x(2x2 - 3) - x2(5x + 1) + x2
= 2x3 - 3x - 5x3 - x2 + x2
= -3x2 - 3x
c) x(x2 + x + 1) - x2(x + 1) - x + 5
= x3 + x2 + x - x3 - x2 - x + 5
= 5
d) (x - 2)(x + 1) - (x + 2)(x - 3)
= x2 + x - 2x - 2 - x2 + 3x - 2x + 6
= 4
e) (2x - y)(2x + y) + y2
= 4x2 - y2 + y2
= 4x2
Thay x = 5 vào biểu thức trên, ta có:
4x2 = 4.52= 100