Tìm x: 3(x-1)^2+(x+5)(2-3x)=25
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\(xy^2-y^2+1-x^2=\left(xy^2-y^2\right)+\left(1-x^2\right)\)
\(=y^2\left(x-1\right)-\left(x^2-1\right)=y^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(y^2-x-1\right)\)
xy 2 − y 2 + 1 − x 2 = (xy 2 − y 2 )+ (1 − x 2 )
= y 2 (x − 1 )− (x 2 − 1)
= y 2 (x − 1) −( x − 1)(x + 1)
= (x − 1 ) (y 2 − x − 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+1-\frac{1}{4}\)
\(=\left(x^2+2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\frac{3}{4}>0\)
Nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)hay \(x^2+x+1>0\)
mà theo đề bài \(x^2+x+1=0\)
Vậy \(x\in\varnothing\)
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x2 - y2 - 4 + 4y = x2 - ( y2 - 4y + 4 ) = x2 - ( y - 2 )2 = ( x - y + 2 )( x + y - 2 )
x2 - y2 - 4 + 4y
= x2 - ( y 2 - 4y + 4 )
= x 2 - ( y - 2 )2
= ( x - y + 2 )( x + y - 2 )
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Ta có: \(\frac{a}{abc+ab+a+1}=\frac{acd}{\left(abc+ab+a+1\right)cd}=\frac{acd}{abc^2d+abcd+acd+cd}\)
\(=\frac{acd}{c+1+acd+cd}\left(abcd=1\right)\)
\(\frac{b}{bcd+bc+b+1}=\frac{b}{bcd+bc+b+abcd}=\frac{1}{acd+cd+c+1}\)
\(\frac{d}{dab+da+d+1}=\frac{dc}{\left(dab+da+d+1\right)c}=\frac{dc}{abcd+acd+cd+c}=\frac{cd}{1+acd+cd+c}\)
=> \(\frac{acd}{acd+cd+c+1}+\frac{1}{acd+cd+c+1}+\frac{c}{acd+cd+c+1}+\frac{cd}{acd+cd+c+1}\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-2xy-1+2y=2y\left(1-x\right)+x^2-1\)
\(=2y\left(1-x\right)+\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x+1-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=x\left(x-2\right)\left(x+2\right)\left(x+4\right)+16\)
\(\Rightarrow A=x\left(x+2\right)\left(x-2\right)\left(x+4\right)+16\)
\(=\left(x^2+2x\right)\left(x^2+2x-8\right)+16\)
Đặt \(x^2+2x-4=t\)
\(\Rightarrow A=\left(t+4\right)\left(t-4\right)+16=t^2-16+16\)
\(=t^2=\left(x^2+2x-4\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2=\frac{1}{2}.\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\)
\(=\frac{1}{2}.\left[\left(x^2+y^2\right)^2-\left(2xy\right)^2\right]\)
\(=\frac{1}{2}.\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)\)
\(=\frac{1}{2}\left(x-y\right)^2.\left(x+y\right)^2\)
\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=25\)
\(3x^2-6x+3-3x^2-13x+10=25\)
\(-19x+13=25\)
\(-19=25-13\)
\(-19x=12\)
\(x=\frac{-12}{19}\)
\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=25\)
\(3\left(x^2-2x+1\right)+2x-3x^2+10-15x=25\)
\(3x^2-6x+3+2x-3x^2+10-15x=25\)
\(-18x+13=25\)
\(-18x=12\)
\(x=-\frac{2}{3}\)