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\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)
\(=\left(2x+5\right).\left(2x-5\right)-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x+5-2x-7\right).\left(2x-5\right)\)
\(=-2.\left(2x-5\right)\)
\(a^2x^2-a^2x^2-b^2x^2+b^2y^2\)
\(=a^2.\left(x^2-y^2\right)-b^2.\left(x^2-y^2\right)\)
\(=\left(a^2-b^2\right).\left(x^2-y^2\right)\)
\(=\left(a-b\right).\left(a+b\right).\left(x-y\right).\left(x+y\right)\)
\(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right).\left(x+y-6\right)\)
\(\left(x+2\right)^2-x^2+2x-1\)
\(=\left(x+2\right)^2-\left(x^2-2x+1\right)\)
\(=\left(x+2\right)^2-\left(x-1\right)^2\)
\(=[x+2-\left(x-1\right)].[x+2+\left(x-1\right)]\)
\(=\left(x+2-x+1\right).\left(x+2+x-1\right)\)
\(=3.\left(2x+1\right)\)
\(16x^2-y^2=\left(4x\right)^2-y^2=\left(4x-y\right).\left(4x+y\right)\)
\(1+27x^3=1^3+\left(3x\right)^3=\left(1+3x\right).\left(1-3x+9x^2\right)\)
Bài \(4a!\)
Ta có:
\(2x^2+y^2+2xy-2x+2y+5=0\)
\(\Leftrightarrow\) \(x^2+2xy+y^2+2x+2y+x^2-4x+5=0\)
\(\Leftrightarrow\) \(\left(x+y\right)^2+2\left(x+y\right)+1+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\) \(\left(x+y+1\right)^2+\left(x-2\right)^2=0\) \(\left(\text{*}\right)\)
Vì \(\left(x+y+1\right)^2\ge0\) và \(\left(x-2\right)^2\ge0\) với mọi \(x,y\)
nên từ \(\left(\text{*}\right)\) \(\Rightarrow\) \(\left(x+y+1\right)^2=0\) \(V\) \(\left(x-2\right)^2=0\)
\(\Leftrightarrow\) \(x+y+1=0\) \(V\) \(x-2=0\)
\(\Leftrightarrow\) \(x+y=-1\) \(V\) \(x=2\)
\(\Leftrightarrow\) \(x=2\) và \(y=-3\)
Vậy, cặp số cần tìm là \(\left(x;y\right)=\left(2;-3\right)\)
Bài \(3a.\)
Vì \(xy=13\) nên \(xy+1=14\)
Từ giả thiết suy ra \(xy\left(x+y\right)+x+y=2016\)
\(\Leftrightarrow\) \(\left(x+y\right)\left(xy+1\right)=2016\)
\(\Leftrightarrow\) \(x+y=144\)
Khi đó, \(\left(x+y\right)^2=144^2=20736\)
\(\Leftrightarrow\) \(x^2+2xy+y^2=20736\)
\(\Leftrightarrow\) \(x^2+y^2=20736-2xy=20736-26=20710\)
\(b,c\) tối giải cho
Bài \(4a.\) tối giải!
g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
ĐÂY NÀY:
( x +y) ^2 = a^2 => x^2 + 2xy + y^2 = a^2
=> 2xy = a^2 - ( x^2 + y^2) = a^2 -b
=> xy = a^2-b/2
Ta có E = x^3 + y^3 = ( x+ y)( x^2 - xy + y^2)
E = a ( b - a^2-b/2)
Answer:
\(5x^2-10xy+5y^2-20z^2\)
\(=5.\left(x^2-2xy+y^2-4z^2\right)\)
\(=5.[\left(x+y\right)^2-\left(2z\right)^2]\)
\(=5.\left(x+y-2z\right).\left(x+y+2z\right)\)
\(16x-5x^2-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(1-5x\right).\left(x-3\right)\)
\(x^2-5x+5y-y^2\)
\(=(x-y).(x+y)-5.(x-y)\)
\(=(x-y).(x+y-5)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3.(x^2-2xy+y^2-4z^2)\)
\(=3[\left(x-y\right)^2-\left(2z\right)^2]\)
\(=3.(x-y-2z).(x-y+2z)\)
\(x^2+4x+3\)
\(=(x^2+x)+(3x+3)\)
\(=x.(x+1)+3.(x+1)\)
\(=(x+1).(x+3)\)
\((x^2+1)^2-4x^2\)
\(=(x^2-2x+1).(x^2+2x+1)\)
\(=(x-1)^2.(x+1)^2\)
\(x^2-4x-5\)
\(=(x^2+x)-(5x+5)\)
\(=x.(x+1)-5.(x+1)\)
\(=(x-5).(x+1)\)
\(2P=2x^2+2y^2-2xy-2x+2y+2\)
= (x2 - 2xy + y2) + \(\frac{4}{3}\)(y - x) + \(\frac{4}{9}\)+ (x2 - \(\frac{2}{3}\)x + \(\frac{1}{9}\)) + (y2 + \(\frac{2}{3}\)y + \(\frac{1}{9}\)) + \(\frac{4}{3}\)
= (y - x + \(\frac{2}{3}\))2 + (x - \(\frac{1}{3}\))2 + (y + \(\frac{1}{3}\))2 + \(\frac{4}{3}\)\(\ge\frac{4}{3}\)
\(\Rightarrow P\ge\frac{2}{3}\)
Vậy GTNN là \(\frac{2}{3}\)đạt được khi x = \(\frac{1}{3}\); y = - \(\frac{1}{3}\)
Nhiều quá không muốn giải. Bạn chọn đi. Mình giúp bạn giải 1 câu (bạn thích câu nào mình giải câu đó cho ) :D