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Bài 1:
a) \(a^2-6a+9=\left(a-3\right)^2\)
b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)
Bài 2:
a) \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)
\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)
b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)
\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
hay P=x+3
\(\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)+\left(x-3\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x-1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1+x-3\right)^2\)
\(=\left(3x-2\right)^2\)
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\(a^3+3a^2-6a-8\)
\(=a^3+4a^2-a^2-4a-2a-8\)
\(=\left(a^3+4a^2\right)-\left(a^2+4a\right)-\left(2a+8\right)\)
\(=a^2\left(a+4\right)-a\left(a+4\right)-2\left(a+4\right)\)
\(=\left(a+4\right)\left(a^2-a-2\right)\)
\(=\left(a+4\right)\left(a^2-2a+a-2\right)\)
\(=\left(a+4\right)\left[\left(a^2-2a\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left[a\left(a-2\right)+\left(a-2\right)\right]\)
\(=\left(a+4\right)\left(a-2\right)\left(a+1\right)\)
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\(2x^2-5x+2\)
\(=2x^2-4x-x+2\)
\(=\left(2x^2-4x\right)-\left(x-2\right)\)
\(=2x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
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\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x-4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-2\right)\)
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\(a^2-1+4b-4b^2\)
\(=a^2-\left(1-4b+4b^2\right)\)
\(=a^2-\left(1-2b\right)^2\)
\(=\left(a-1+2b\right)\left(a+1-2b\right)\)
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\(a^4+6a^2b+9b^2-1\)
\(=\left(a^4+6a^2b+9b^2\right)-1\)
\(=\left(a^2+3b\right)^2-1\)
\(=\left(a^2+3b-1\right)\left(a^2+3b+1\right)\)
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\(2x^3+16y^3\)
\(=2\left(x^3+8y^3\right)\)
\(=2\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Lần sau ghi đề tách riêng từng câu ra nhé em. Ghi dính chùm vậy khó nhìn lắm. Sẽ ít ai giải cho em
a. (x + y)2 = x2 + 2xy + y2
b. (x - 2y)2 = x2 - 4xy - 4x2
c. (xy2 + 1)(xy2 - 1) = x2y4 - 1
d. (x + y)2(x - y)2 = (x2 + 2xy + y2)(x2 - 2xy + y2) = x4 - (2xy + y2)2 = x4 - (4x2y2 + y4) = x4 - 4x2y2 - y4
Chucs hocj toots
Câu 2:
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)
e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)
III.
a) \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\)\(25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow\)\(10x=20\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
b) \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow\)\(9x^2-6x+1+2x^2+12x+18+11-11x^2=6\)
\(\Leftrightarrow\)\(6x=-24\)
\(\Leftrightarrow\)\(x=-4\)
Vậy....
II.
a) mk chỉnh lại đề câu a
\(a^2-2a+1=\left(a-1\right)^2\)
b) \(1-4a+4a^2=\left(1-2a\right)^2\)
c) \(a^2+9-6a=\left(a-3\right)^2\)
d) \(25a^2-20ab+4b^2=\left(5a-2b\right)^2\)