Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
\(A=3\left(4x^2-4xy+y^2\right)-10\left(2x-y\right)+8\)
\(A=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)
\(A=\left(6x-3y-4\right)\left(2x-y-2\right)\)
a: \(=\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)-144\)
\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)
\(=\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192\)
\(=\left(x^2-5x\right)^2-32\left(x^2-5x\right)-6\left(x^2-5x\right)+192\)
\(=\left(x^2-5x-32\right)\left(x^2-5x-6\right)\)
\(=\left(x^2-5x-32\right)\left(x-6\right)\left(x+1\right)\)
b: \(=\left(12x^2-12xy+3y^2\right)-20x+10y+8\)
\(=\left[3\left(2x-y\right)^2\right]-10\left(2x-y\right)+8\)
\(=3\left(2x-y\right)^2-4\left(2x-y\right)-6\left(2x-y\right)+8\)
\(=\left(2x-y\right)\left(6x-3y-4\right)-2\left(6x-3y-4\right)\)
\(=\left(6x-3y-4\right)\left(2x-y-2\right)\)
Ta có: 4x2 + 12xy + 10y2 + 4x + 4y + 2 = 0
<=> (4x2 + 12xy + 9y2) + 2(2x + 3y) + 1 + (y2 - 2y + 1) = 0
<=> (2x + 3y)2 + 2(2x + 3y) + 1 + (y - 1)2 = 0
<=> (2x + 3y + 1)2 + (y - 1)2 = 0
<=> \(\hept{\begin{cases}2x+3y+1=0\\y-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-\frac{1+3y}{2}\\y=1\end{cases}}\)
<=> \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)(tm)
Khi đó: P = \(\frac{x^2+y^2+xy}{3xy}=\frac{\left(-2\right)^2+1^2-2.1}{3.\left(-2\right).1}=-\frac{1}{2}\)
\(A=x^2+10y^2+2x-6xy-10y+25\)
=> \(A=x^2+2x\left(1-3y\right)+\left(1-3y\right)^2-\left(1-3y\right)^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-1+6y-9y^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-9y^2-4y+24\)
=> \(A=\left(x+1-3y\right)^2-\left(3y\right)^2-2.3y.\frac{2}{3}-\left(\frac{2}{3}\right)^2+\frac{220}{9}\)
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\)
Có \(\left(x+1-3y\right)^2\ge0\)với mọi x, y
\(\left(3y+\frac{2}{3}\right)^2\ge0\)với mọi y
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\ge\frac{220}{9}\)với mọi x, y
Dấu "=" xảy ra <=> \(\left(x+1-3y\right)^2=0\)<=> \(x+1-3y=0\)
và \(\left(3y+\frac{2}{3}\right)^2=0\)=> \(3y+\frac{2}{3}=0\)
=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
Bổ xung phần kết luận
KL: Amin = \(\frac{220}{9}\)<=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
\(a,x^2yz-x^3y^3z+xyz^2\)
\(=xyz\left(x-x^2y^2+z\right)\)
\(b,4x^3+24x^2-12xy^2\)
\(=4\left(x^3+6x^2-3xy^2\right)\)
\(c,15a^{m+2}b-45a^mb\)
\(=15a^m.a^2b-45a^mb\)
\(=15a^mb\left(a^2-3\right)\)
\(d,a^2-b^2+4bc-4c^2\)
\(=a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left(b-2c\right)^2\)
\(=\left(a-b+2c\right)\left(a+b-2c\right)\)
a) \(x^2yz-x^3y^3z+xyz^2\)
\(=xyz\left(x-x^2y^2+z\right)\)
b) \(4x^3+24x^2-12xy^2\)
\(=4x\left(x^2+6x-3y^2\right)\)
c) \(15a^{m+2}.b-45a^m.b\)
\(=15.\left(a^m.a^2-3a^m.b\right)\)
\(=15.a^m.\left(a^2-3b\right)\)
d) \(a^2-b^2+4bc-4c^2\)
\(=a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left[\left(b^2-2bc+c^2\right)-2bc+3c^2\right]\)
...... ;)))))))
a) ( a + b + c ) 2 + ( a + b - c ) 2 -2 x ( a+b) 2
2a+2b+2x+2a+2b-2c-2.(2a+2b)
2a+2b+2c+2a+2b-2c-4a-4b
4a+4b-4a-4b=0
b) 2x.( 2x -1 ) 2 -3x.( x+3 )( x-3) - 4x.(x+1).2
2x.(4x-2)-3x2-9x-3x2+9x-4x(2x+2)
8x2-4x-3x2-9x-3x2+9x-8x2-8x
-12x-3x2
c) ( a-b+c).2 -(b-c).2 + 2ab - 2ac
2a-2b+2c-2b+2c+2ab-2ac
2a-4b+4c+2ab-2ac
d) (3x+1).2 - 2(3x+1)( 3x+5 )+(3x+5).2
6x+2-6x-2-6x-10+6x+10=0
Quên chưa viết đề bài
Đề bài là:Tìm x,y
Tìm x, y thì phải có dấu "=" chứ