Biết x+ y+ z= 2020 Tính
P=\(\frac{\text{x^3+y^3+z^3-3xyz}}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
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P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Với \(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)
a. \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Rightarrow10x^2-35x+16x-10x^2=5\)
\(\Rightarrow-19x=5\)
\(\Rightarrow x=-\frac{5}{19}\)
b. \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)
\(\Rightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)
\(\Rightarrow\frac{7}{6}x=\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{14}\)
c. \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Rightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Rightarrow-14x+5=x-2\)
\(\Rightarrow-14x-x=-2-5\)
\(\Rightarrow-15x=-7\)
\(\Rightarrow x=\frac{7}{15}\)
a, \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Leftrightarrow10x^2-35x+16x-10x^2=5\)
\(\Leftrightarrow-19x=5\Leftrightarrow x=-\frac{5}{19}\)
b, \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)
\(\Leftrightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{7}{6}x=\frac{1}{4}\Leftrightarrow x=\frac{3}{14}\)
c, \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-15x+7=0\Leftrightarrow x=\frac{7}{15}\)
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
Bài làm:
Ta có: \(xy=5\)\(\Rightarrow x=\frac{5}{y}\)
Thay vào ta được:
\(x^2+y^2=26\)
\(\Leftrightarrow\frac{25}{y^2}+y^2=26\)
\(\Leftrightarrow\frac{25+y^4}{y^2}=26\)
\(\Leftrightarrow y^4-26y^2+25=0\)
\(\Leftrightarrow\left(y^4-y^2\right)-\left(25y^2-25\right)=0\)
\(\Leftrightarrow\left(y^2-1\right)\left(y^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y^2-1=0\\y^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=\pm1\\y=\pm5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm5\\x=\pm1\end{cases}}\)
Vậy ta có các cặp số (x;y) thỏa mãn: \(\left(1;5\right);\left(-1;-5\right);\left(5;1\right);\left(-5;-1\right)\)
Ta có :
\(x^2+y^2=26\Rightarrow x^2+y^2+2xy=26+2.5\)
\(\Rightarrow\left(x+y\right)^2=36\Leftrightarrow x+y=6\left(1\right)\)
\(x^2+y^2=26\Rightarrow x^2+y^2-2xy=26-2.5\)
\(\Rightarrow\left(x-y\right)^2=16\Leftrightarrow x-y=4\left(2\right)\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow x=\frac{6+4}{2}=5\)
\(\Rightarrow y=5-4=1\)
Vậy x = 5 ; y = 1
\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{x^2-1}.\frac{x^2-1}{5}=\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)
\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\left(x\ne\pm1\right)\)
\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{\left(x^2+5x+8\right)\cdot\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)5}=\frac{x^2+5x+8}{5}\)
Bài làm:
Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
và
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)
\(=2\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Từ đó thay vào P rút ra:
\(P=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{2020}{2}=1010\)
Vậy P = 1010