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\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)
Suy ra x=0 hoặc x=-26/3
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5x2 + 5y2 + 8xy + 2y - 2x + 2 = 0
=> (4x2 + 4y2 + 8xy) + (x2 - 2x + 1) + (y2 + 2y + 1) = 0
=> 4(x + y)2 + (x - 1)2 + (y + 1)2 = 0
Mà 4(x + y)2 , (x - 1)2 , (y + 1)2 lớn hơn hoặc bằng 0.
=> 4(x + y)2 = (x - 1)2 = (y + 1)2 = 0
=> x + y = x - 1 = y + 1 = 0. => x - 2 = -1
M = ( x +y ) 2013 + ( x - 2 ) 2014 + ( y + 1 )2015 = 02013 + (-1)2014 + 02015 = 1
\(\left(x-1\right)^2+\left(y+1\right)^2+2\left(x+y\right)^2=0\)
Suy ra \(x=1,y=-1\). Tới đây bạn tự giải tiếp nha.
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Mình cũng mới hỏi câu này luôn ấy, mình có cách làm nhưng sợ không đúng thôi.
P = x4y4 + x4 + y4 + 1 + 12x2y2 – 16xy – 4
P = x4y4 + x4 + y4 + 1 + 16x2y2 – 16xy + 4 – 4x2y2 – 8
P = x4y4 + x4 + y4 + 1 + (4xy – 2)2 – 4x2y2 – 8
P = (x4 – 2x2y2 + y4) + (x4y4 – 2x2y2 + 1) – 8 + (4xy – 2)2
P = (x2 – y2)2 + (x2y2 – 1)2 – 8 + (4xy – 2)2
P = (x + y)2(x – y)2 + (xy + 1)2(xy – 1)2 + (4xy – 2)2 – 8
P = 4(x – y)2 + (xy + 1)2(xy – 1)2 + 4(2xy – 1)2 – 8
MinP = Min 4(x – y)2 + min (xy + 1)2(xy – 1)2 + min 4(2xy – 1)2 – 8
Min 4(x – y)2 = 0 => x – y = 0 => x = y = 1 => MinP = – 4
Min (xy + 1)2(xy – 1)2 = 0 =>
TH1: xy = -1 (không có x,y thỏa mãn)
TH2: xy = 1 => x = y = 1 => Min P = – 4
Min 4(2xy – 1)2 = 0 => xy = \(\frac{1}{2}\)(không có x,y thỏa mãn)
Vậy thì kết quả là -4, Violympic chưa mở nên mình chưa thử kết quả được, thân ái.
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a: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-x^2+2x+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow5x^2-20x-41=5x^2-6x+27\)
=>-14x=68
hay x=-34/7
b: \(\Leftrightarrow x^2-25-x^3+6x^2-12x+8-7x^2+x^3+1=\left(x+3\right)^3-x^3-9x^2\)
\(\Leftrightarrow-12x-16=x^3+9x^2+27x+27-x^3-9x^2=27x+27\)
=>-39x=43
hay x=-43/39
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\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
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\(a,\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
\(b,\left(x^2+1\right)^2-4x^2\)
\(=x^4+2x^2+1-4x^2\)
\(=x^4-2x^2+1\)
\(\left(x^2-1\right)^2\)
\(c\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-8y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-8y+4+y-2\right)\)
\(=\left(y+2\right)\left(y^2-7y+2\right)\)
a) ( a3 - b3) + ( a - b)2
= (a-b) (a2 + ab + b2 ) + (a-b)2
= (a-b) (a2 + ab + b2 +a -b )
hok tốt
( 5 - 2x )( x + 1 ) + 2( x - 1 )2 = ( 5x + 5 - 2x2 + 2x ) + 2x2 - 4x + 2
= 7x + 5 - 2x2 + 2x2 - 4x + 2
= 3x + 7 .
Đề bài là gì hả bạn!