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a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
Câu a và câu c bn kia làm rồi nên mk làm câu b thôi nhé....
b) y2 + 4x + 2y - 2x+1 + 2 = 0
\(\Leftrightarrow\) (y2 + 2y + 1) + 4x - 2x.2 + 1 = 0
\(\Leftrightarrow\) (y + 1)2 + [(2x)2 - 2.2x.1 + 1] = 0
\(\Leftrightarrow\) (y + 1)2 + (2x - 1)2 = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\2^x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
Vậy...................
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
d. ĐKXĐ: x khác 1, x khác 3
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+1+8=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\) (loại)
Vậy pt vô nghiệm
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100