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1. a) = 16
b) = 29x^2 + 29 - 29x^2 = 29
2. =x^2-2x+1 + y^2 - 2y + 1 = (x-1)^2 + (y-1)^2
b) = a^2+4a+4 + b^2 + 4b + 4 = (a+2)^2 + (b+2)^2
bạn giải chi tiết cho mình đc k ? pls xin đáy và cảm ơn bạn vô cùng
Bài 1:
a) \(x^2\left(5x^3-x-6\right)\)
\(=x^2.5x^3-x^2.x-x^2.6\)
\(=5x^5-x^3-6x^2\)
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2xy\left(x-y\right)+y^2\left(x-y\right)\)
\(=x^3-x^2y-2x^2y+2xy^2+y^2x-y^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
Bài 2:
a) \(y^2+2y+1\)
\(=\left(y+1\right)^2\)
b) \(9x^2+y^2-6xy\)
\(=\left(3x\right)^2-2.3x.y+y^2\)
\(=\left(3x-y\right)^2\)
c) \(25a^2+4b^2+20ab\)
\(=\left(5a\right)^2+2.5a.2b+\left(2b\right)^2\)
\(=\left(5a+2b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) x^2 - x + 1/4
= x^2 - 2.x + 1/4 + (1/2)^2
= ( x - 1/2)^2
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
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 \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
a) (a - 2b)2 = a2 - 2.a.2b + 4b2
= a2 - 4ab + 4b2
b) m2 - 4n2 = m2 - (2n)2 = (m - 2n)(m + 2n)
. Bài 1:
a; 9m^2 + n^2 - 6mn
= (3m)^2 - 2.3m.n + (n)^2
= ( 3m-n )^2
b; x^2-x+1/4
= x^2-2.(x).1/2+(1/2)^2
= (x-1/2)^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) \(25a^2-20ab+4b^2=\left(5a-2b\right)^2\)