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Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
Sủa đề : CM : \(\sqrt{1+\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}}=1+\frac{1}{x}-\frac{1}{x+1}\)
Ta có :\(\frac{2}{x}-\frac{2}{\left(x+1\right)x}-\frac{2}{x+1}=2\left(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x+1}\right)=0\) nên
\(\sqrt{1+\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}}=\sqrt{1+\left(\frac{1}{x}\right)^2+\left(\frac{1}{x+1}\right)^2+\frac{2}{x}-\frac{2}{\left(x+1\right)x}-\frac{2}{x+1}}\)
\(=\sqrt{\left(1+\frac{1}{x}-\frac{1}{x+1}\right)^2}=1+\frac{1}{x}-\frac{1}{x+1}\)(đpcm)
Mk chua hc cai nay