x²+y²+z²=xy+yz+zx

<=>2(x²+y²+z²)=2(xy+yz+zx)

<=>2x²+2y²+2z²=2xy+2yz+2zx

<=>2x²+2y²+2z²-2xy-2yz-2zx=0

<=>(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)(đpcm)

\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)

\(\Rightarrow\left(3x+3-1\right)\left(x-1\right)-3\left(x+1\right)\left(x-1-1\right)=4\)

\(\Rightarrow\left(3x+3\right)\left(x-1\right)-\left(x-1\right)-3\left(x+1\right)\left(x-1\right)+3\left(x+1\right)=4\)

\(=3\left(x+1\right)\left(x-1\right)-3\left(x+1\right)\left(x-1\right)+3x+3-x+1=4\)

\(\Rightarrow2x+4=4\Rightarrow2x=0\Rightarrow x=0\)

\(\left(3x+2\right).\left(x-1\right)-3.\left(x+1\right).\left(x-2\right)=4\Leftrightarrow3x^2-x-2-3.\left(x^2-x-2\right)-4=0\)

\(\Leftrightarrow3x^2-x-2-3x^2+3x+6-4=0\Leftrightarrow2x=0\Leftrightarrow x=0\)

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