\(x^2-x+\frac{1}{4}=\left[x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(x-\frac{1}{2}\right)^2\) ( mình nghĩ phải là \(\frac{1}{4}\) chứ bạn ) 

\(4x^2-4x+1=\left[\left(2x\right)^2-2.2x.1+1^2\right]=\left(2x-1\right)^2\)

Chúc bạn học tốt ~ 

9 viên

thì số kẹo của P=25-9=16

                       L=55+9=64 VÀ 16.4=64

5 viên nha!

\(\frac{x-5}{x-1}+\frac{2}{x-3}=1\)(ĐKXĐ: x khác 1;3)

\(\Leftrightarrow\frac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=1\)

\(\Leftrightarrow\frac{x^2-8x+15+2x-2}{x^2-4x+3}=1\)

\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)\(\Rightarrow x^2-4x+3=x^2-6x+13\)

\(\Leftrightarrow x^2-4x+3-x^2+6x-13=0\)

\(\Leftrightarrow2x-10=0\Leftrightarrow2x=10\Leftrightarrow x=5\)(t/m ĐKXĐ)

Vậy nghiệm của pt là x=5.

ĐKXĐ: x khác 1, 3

\(\frac{x-5}{x-1}+\frac{2}{x-3}-1=0\Leftrightarrow\frac{\left(x-5\right).\left(x-3\right)}{\left(x-1\right).\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-1\right).\left(x-3\right)}-\frac{\left(x-1\right).\left(x-3\right)}{\left(x-1\right).\left(x-3\right)}=0\\ \)

\(\Leftrightarrow\frac{\left(x^2-8x+15\right)+\left(2x-2\right)-\left(x^2-4x+3\right)}{\left(x-1\right).\left(x-3\right)}=0\)

\(\Leftrightarrow\left(x^2-8x+15\right)+2x-2-\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow x^2-8x+15+2x-2-x^2+4x-3=0\)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)

\(y+2⋮x;x+2⋮y\Rightarrow\left(x+2\right)\left(y+2\right)⋮xy\Rightarrow xy+2x+2y+4⋮xy\Rightarrow2x+2y+4⋮xy\)

\(\Rightarrow2\left(x+y+2\right)⋮xy\Rightarrow2⋮xy\Rightarrow xy\inƯ\left(2\right)=1;2\)

\(xy=1\Rightarrow x=1,y=1\Rightarrow y+2=1+2=3⋮x=1\Rightarrow y+2⋮x\)

                                             \(x+2=1+2=3⋮y=1\Rightarrow x+2⋮y\)

\(\Rightarrow x=1,y=1\left(tm\right)\)

\(xy=2\Rightarrow x=1,y=2;x=2,y=1\Rightarrow x+2=1+2=3\)ko chia hết cho \(y=2\Rightarrow x+2\)ko chia hết cho y

\(\Rightarrow x=1,y=2\left(ktm\right)\Rightarrow x=2,y=1\left(ktm\right)\)

vậy x=1,y=1 

ko chắc lắm

\(M=\frac{20}{x^2+y^2}+\frac{11}{xy}=\frac{20}{x^2+y^2}+\frac{22}{2xy}=\frac{20}{x^2+y^2}+\frac{20}{2xy}+\frac{2}{2xy}\)

\(=20\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}>=20\cdot\frac{4}{x^2+2xy+y^2}+\frac{4}{\left(x+y\right)^2}\)

\(=\frac{80}{\left(x+y\right)^2}+\frac{4}{\left(x+y\right)^2}=\frac{84}{\left(x+y\right)^2}>=\frac{84}{2^2}=\frac{84}{4}=21\)

dấu = xảy ra khi \(\hept{\begin{cases}x+y=2\\x=y\end{cases}\Rightarrow x=y=1}\)

vậy min M là 21 khi x=y=1

\(16x-5x^2-3=\left(-5x^2+15x\right)+\left(x-3\right)=-5x.\left(x-3\right)+\left(x-3\right)\\ \)

                                 \(=\left(x-3\right).\left(1-5x\right)\)

\(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x+2\right)\)

\(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-\left(5x^2-16x+3\right)\)
\(=-\left(5x^2-15x-x+3\right)\)

\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)

\(=-\left[\left(5x-1\right)\left(x-3\right)\right]\)

Câu 2:

\(x^2-x-6\)

\(=x^2+2x-3x-6\)

\(=x\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x-3\right)\left(x+2\right)\)

a) \(\frac{1}{4}x^2-5xy+25y^2=\left(\frac{1}{2}x\right)^2-5xy+\left(5y\right)^2\)

\(=\left(\frac{1}{2}x-5y\right)^2\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4+2x+1\right)\times\left(7x-4-2x-1\right)=\left(9x-3\right)\times\left(5x-5\right)\)

\(=3\times5\times\left(3x-1\right)\times\left(x-1\right)=15\times\left(3x-1\right)\times\left(x-1\right)\)

c)\(\left(x-2\right)^2-4y^2=\left(x-2-2y\right)\left(x-2+2y\right)\)

d) \(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)