Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(x^2+y^2-2x+10y+26=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+10y+25\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-5\end{cases}}\)
b,\(4x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+1=0\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)
c,\(5x^2+9y^2-12xy+4x+4=0\)
\(\Rightarrow\left(x^2+4x+4\right)+\left(4x^2-12xy+9y^2\right)=0\)
\(\Rightarrow\left(x+2\right)^2+\left(2x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\2x-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\2.\left(-2\right)-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}\)
d,\(5x^2+9y^2-6xy-4x+1=0\)
\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^x\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x+1=0\\x-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\-\frac{1}{2}-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{6}\end{cases}}\)
a, x(x-y)+2(x-y)=(x-y)(x+2)
b, \(x^2-6xy+9y^2=\left(x-3y\right)^2\)Thay x=16, y=2 có
\(x^2-6xy+9y^2=\left(x-3y\right)^2=\left(16-2\cdot3\right)^2=10^2=100\)
\(36x^2-9y^2-12x-6y\)
\(=\left(36x^2-12x+1\right)-\left(9y^2+6y+1\right)\)
\(=\left(6x-1\right)^2-\left(9y+1\right)\)
\(=\left(6x+9y\right)\left(6x-3y-2\right)\)
\(=3\left(2x+3y\right)\left(6x-3y-2\right)\)
1) \(x^2-16=\left(x-4\right)\left(x+4\right)\)
2)\(4a^{201}\)
3)\(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
4)\(25-9y^2=\left(5-3y\right)\left(5+3y\right)\)
5)\(\left(a+1\right)^2-16=\left(a+1-16\right)\left(a+1+16\right)=\left(a-15\right)\left(a+17\right)\)
6)\(x^2-\left(2+y\right)^2=\left(x-2-y\right)\left(x+2+y\right)\)
7) (a+b)2-(a-b)2
= \(\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=4ab\)
8 ) a^2 + 2ax + x^2
= ( a + x )2
9) x^2 - 4x + 4
= ( x-2)2
10) x^2-6xy+9y^2
= (x - 3y )2
11) x^3+8
= (x+2)( x2 - 2x + 4 )
12) a^3 + 27b^3
= (a + 3b ) ( a2 - 3ab + 9b2 )
13) 27x^3 - 1
= ( 3x -1 ) ( 9x2 + 3x +1)
14) 1/8 - b^3
= ( 1/2 - b ) ( 1/4 + 1/2b + b2)
15) a^3 - (a+b)3
= a3 - ( a3 + 3a2b + 3ab2 + b3)
= - 3a2b - 3ab2- b3= -b (3a2 + 3ab +b2)
16) 4x^2 + 4x + 1
= (2x +1 )2
\(4x^2+4x-9y^2+1\\ =\left(4x^2+4x+1\right)-9y^2\\ =\left(2x+1\right)^2-9y^2\\ =\left(2x+1\right)^2-\left(3y\right)^2\\ =\left[\left(2x+1\right)+3y\right]\left[\left(2x+1\right)-3y\right]\\ =\left(2x+1+3y\right)\left(2x+1-3y\right)\)
\(x^2-6xy+9y^2-25z^2\\ =\left(x^2-6xy+9y^2\right)-25z^2\\ =\left(x-3y\right)^2-25z^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left[\left(x-3y\right)+5z\right]\left[\left(x-3y\right)-5z\right]\\ =\left(x-3y+5z\right)\left(x-3y-5z\right)\)
\(x^2-xy+x-y\\ =\left(x^2-xy\right)+\left(x-y\right)\\ =x\left(x-y\right)+\left(x-y\right)\\ =\left(x-y\right)\left(x-1\right)\)
a: Sửa đề: \(-x^3-12x^2-48x-64\)
\(=-\left(x+4\right)^3\)
\(=-\left(-6+4\right)^3=-\left(-2\right)^3=-\left(-8\right)=8\)
b: \(=8x^3-y^3-8x^3+27y^3=26y^3=26\cdot\left(-3\right)^3=-702\)
c: \(=-\left(4x^4-12x^2y+9y^2\right)\)
\(=-\left(2x^2-3y\right)^2\)
\(=-\left(2x^2-2x-11\right)^2\)
4x2 - 9y2
= ( 2x )2 - ( 3y )2
= ( 2x - 3y ) ( 2x + 3y )
Với x = 26 ; y = 16 , ta có :
( 2 . 26 - 3 . 16 ) ( 2 . 26 + 3 . 16 )
= 4 . 100 = 400
\(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\) tại X=26 , y=16
\(\Rightarrow\left(2\cdot26-3\cdot16\right)\left(2\cdot26+3\cdot16\right)=400\)