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dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)+64x^4\)
=\(\left[\left(x^2-2x+2\right)^4+2.10x^2\left(x^2-2x+2\right)^2+100x^4\right]\)-100x4+64x2
=\(\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^2\)
=\(\left[\left(x^2-2x+2\right)^2+4x^2\right].\left[\left(x^2-2x+2\right)^2+16x^2\right]\)
\(\left(x^2-2x+3\right)\left(\frac{1}{2x}-5\right)\)
\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)
\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)
\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)
\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)
học tốt
(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15
\(\left(4x-5\right)\left(2x+30\right)-4\left(x+2\right)\left(2x-1\right)+\left(10x+7\right)\)
\(=8x^2+110x-150-8x^2-12x+8+10x+7\)
\(=108x-135\)
(3x-1)2-5(2x+1)2+(6x-3)(2x+1)=(x-1)2
<=> (3x-1)2+2(3x-1)(2x+1)+(2x+1)2-6(2x+1)2=(x-1)2
<=> (5x)2-6(4x2+4x+1)-(x2-2x+1)=0
<=> -22x-7=0
=> x=-7/22
\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-5\left(2x+1\right)+6x-3\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-10x-5+6x-3\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-4x-8\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1-4x\left(2x+1\right)-8\left(2x+1\right)=x^2-1\)
\(\Leftrightarrow9x^2-6x+1-8x^2-4x-16x-8=x^2-1\)
\(\Leftrightarrow\left(9x^2-8x^2-x^2\right)-\left(4x+6x+16x\right)+\left(1-8\right)=-1\)
\(\Leftrightarrow0-26x-7=-1\)
\(\Leftrightarrow-26x=-1+7\)
\(\Leftrightarrow-26x=6\)
\(\Leftrightarrow x=\frac{-3}{13}\)
2x . ( x-2) - x+2 = 0
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)
\((2x-5)^2-64x^2\\=(2x-5)^2-(8x)^2\\=(2x-5-8x)(2x-5+8x)\\=(-6x-5)(10x-5)\)
Lời giải:
$(2x-5)^2-64x^2=(2x-5)^2-(8x)^2$
$=(2x-5-8x)(2x-5+8x)$
$=(-6x-5)(10x-5)$
$=-5(6x+5)(2x-1)$