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\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+4x^2+3}\left(ĐKXĐ:x\in R\right)\).
\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\).
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\).
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\).
\(M=\frac{x^4+2+\left(x^2-1\right)\left(x^2+1\right)-x^4+x^2-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\).
\(M=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^4+x^2}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)
\(M=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2}{x^4-x^2+1}\).
Vậy với \(x\in R\)thì \(M=\frac{x^2}{x^4-x^2+1}\).
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
a) \(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}+\frac{x^2+3}{x^4+4x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+3x^2+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^2\left(x^2+3\right)+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)
\(M=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{0+x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2}{x^4-x^2+1}\)
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
\(ĐKXĐ:x\ne0;x\ne\pm2\)
a) \(M=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow M=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(\Leftrightarrow M=\frac{3x^2-6x\left(x+2\right)+3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(\Leftrightarrow M=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(\Leftrightarrow M=\frac{-18x\left(x+2\right)}{18x\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow M=-\frac{1}{x-2}\)
\(\Leftrightarrow M=\frac{1}{2-x}\)
b) Để M đạt giá trị lớn nhất
\(\Leftrightarrow2-x\)đạt giá trị nhỏ nhất
\(\Leftrightarrow x\)đạt giá trị lớn nhất
Vậy để M đạt giá trị lớn nhất thì x phải đạt giá trị lớn nhất \(\left(x\inℤ\right)\)
玉明, bạn làm sai rồi. Dấu ngoặc vuông là dấu phần nguyên không phải dấu ngoặc thường
\(a,\)
\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)
\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{-\left(x+1\right)}{x^2+x+1}\)
\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)
\(B=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(B=\frac{x+2}{x^2+x+1}\)
\(b,\)
Ta có:
\(A+B=\frac{-\left(x+1\right)}{x^2+x+1}+\frac{x+2}{x^2+x+1}\)
\(=\frac{-x-1+x+2}{x^2+x+1}\)
\(=\frac{1}{x^2+x+1}\)
\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)
Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)
\(\Rightarrow A+B\le\frac{4}{3}\)
\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)
Vậy........