Câu 1:
\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)
\(\Leftrightarrow x^2-4x+1=4x-2x\sqrt{x+\frac{1}{x}}\)
\(\Leftrightarrow x^2-4x+1=2x(2-\sqrt{x+\frac{1}{x}})\)
\(\Leftrightarrow x^2-4x+1=2x.\frac{2^2-\left(x+\frac{1}{x}\right)}{2+\sqrt{x+\frac{1}{x}}}\)
\(\Leftrightarrow x^2-4x+1=2x.\frac{4x-x^2-1}{x\left(2+\sqrt{x+\frac{1}{x}}\right)}\)
\(\Leftrightarrow (x^2-4x+1)\left(1+\frac{2}{2+\sqrt{x+\frac{1}{x}}}\right )=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn 0, do đó
\(x^2-4x+1=0\)
\(\Leftrightarrow x=2\pm \sqrt{3}\)
Câu 2:
Vì \(x+y+z=0\Leftrightarrow x=-(y+z)\)
\(\Rightarrow x^2=(y+z)^2=y^2+z^2+2yz\)
\(\Rightarrow y^2+z^2-x^2=-2yz\)
\(\Rightarrow \frac{x^2}{y^2+z^2-x^2}=\frac{x^2}{-2yz}=\frac{x^3}{-2xyz}\)
Hoàn toàn tương tự. ta có:
\(\frac{y^2}{z^2+x^2-y^2}=\frac{y^3}{-2xyz}; \frac{z^2}{x^2+y^2-z^2}=\frac{z^3}{-2xyz}\)
Do đó:
\(P=\frac{x^3+y^3+z^3}{-2xyz}\)
Ta biết rằng:
\(x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)\)
\(=-3(x+y)(y+z)(x+z)\)
\(=-3(-z)(-x)(-y)=3xyz\)
Suy ra \(P=\frac{3xyz}{-2xyz}=\frac{-3}{2}\)
Câu 1 ): ĐKXĐ \(x>0\) ; Phương trình tương đương
\(x^2-8x+1+2\sqrt{\dfrac{x^2\left(x^2+1\right)}{x}}=0\)
\(\Leftrightarrow x^2-8x+1+2\sqrt{x\left(x^2+1\right)}=0\)
Đặt \(u=\sqrt{x^2+1}\Rightarrow u^2=x^2+1\left(u>0\right)\)
\(v=\sqrt{x}\Rightarrow v^2=x\left(v>0\right)\)
Phương trình trở thành
\(u^2-8v^2+2uv=0\Leftrightarrow u^2-2uv+4uv-8v^2=0\)\(\Leftrightarrow u\left(u-2v\right)+4v\left(u-2v\right)=0\)\(\Leftrightarrow\left(u-2v\right)\left(u+4v\right)=0\)
do \(u>0\) ; \(v>0\) nên \(u\ne-4v\)
\(\Rightarrow u=2v\)\(\Rightarrow\sqrt{x^2+1}=2\sqrt{x}\)\(\Rightarrow x^2+1=4x\Leftrightarrow x^2-4x+1=0\Leftrightarrow x=2\pm\sqrt{3}\)
vậy phương trình có 2 nghiệm \(x=2+\sqrt{3}\) và \(x=2-\sqrt{3}\)
Bài 1:
\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\)
ĐK:\(x\ge 0\)
\(\Leftrightarrow4x^2\left(x+\dfrac{1}{x}\right)=x^4-16x^3+66x^2-16x+1\)
\(\Leftrightarrow -x^4+20x^3-66x^2+20x-1=0\)
\(\Leftrightarrow -(x^2-16x+1)(x^2-4x+1)=0\)
\(\Rightarrow x=\dfrac{4\pm\sqrt{12}}{2}\) (thỏa mãn)
Bài 2:
Từ \(x+y+z=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow y^2+z^2=-x^2-2\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{x^2}{y^2+z^2-x^2}=\dfrac{x^2}{-2x^2-2\left(xy+yz+xz\right)}=-\dfrac{x^2}{2\left(x+y\right)\left(x+z\right)}\)
\(\Rightarrow P=-\dfrac{1}{2}\left(\dfrac{x^2}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2}{\left(x+y\right)\left(y+z\right)}+\dfrac{z^2}{\left(x+z\right)\left(y+z\right)}\right)\)
\(=-\dfrac{1}{2}\cdot\dfrac{3xy\left(x+y\right)}{xy\left(x+y\right)}=-\dfrac{3}{2}\)
Bài 1:
\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\)
ĐK: \(x\ge 0\)
\(\Leftrightarrow4x^2\left(x+\dfrac{1}{x}\right)=x^4-16x^3+66x^2-16x+1\)
\(\Leftrightarrow -x^4+20x^3-66x^2+20x-1=0\)
\(\Leftrightarrow -(x^2-16x+1)(x^2-4x+1)=0\)
\(\Rightarrow x=\dfrac{4\pm\sqrt{12}}{2}\) ( thỏa mãn )
Bài 2:
Từ \(x+y+z=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow y^2+z^2=-x^2-2\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{x^2}{y^2+z^2-x^2}=\dfrac{x^2}{-2x^2-2\left(xy+yz+xz\right)}=-\dfrac{x^2} {2\left(x+y\right)\left(x+z\right)}\)
\(\Rightarrow P=-\dfrac{1}{2}\left(\dfrac{x^2}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2} {\left(x+y\right)\left(y+z\right)}+\dfrac{z^2}{\left(x+z\right)\left(y+z\right)}\right)\)
\(=-\dfrac{1}{2}\cdot\dfrac{3xy\left(x+y\right)}{xy\left(x+y\right)}=-\dfrac{3}{2}\)
