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a, Đặt \(x^2-4x+8=a\left(a>0\right)\)
\(\Rightarrow a-2=\frac{21}{a+2}\)
\(\Leftrightarrow a^2-4=21\Rightarrow a^2=25\Rightarrow a=5\)
Thay vào là ra
b) ĐK: \(y\ne1\)
bpt <=> \(\frac{4\left(1-y\right)}{1-y^3}+\frac{1+y+y^2}{1-y^3}+\frac{2y^2-5}{1-y^3}\le0\)
<=> \(\frac{3y^2-3y}{1-y^3}\le0\)
\(\Leftrightarrow\frac{y\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}\ge0\)
\(\Leftrightarrow\frac{y}{y^2+y+1}\ge0\)
vì \(y^2+y+1=\left(y+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
nên bpt <=> \(y\ge0\)
ĐKXĐ : \(x\ne\pm1\)
Pt \(\Leftrightarrow\frac{2\cdot3\cdot\left(y+1\right)+\left(y-1\right)}{3\left(y-1\right)\left(y+1\right)}-\frac{1}{6}=0\)
\(\Leftrightarrow\frac{7y+5}{3\left(y-1\right)\left(y+1\right)}-\frac{3\left(y-1\right)\left(y+1\right)}{6\left(y-1\right)\left(y+1\right)}=0\)
\(\Leftrightarrow14y+10-3y^2+3=0\)
\(\Leftrightarrow3y^2+14y+13=0\)
\(\Leftrightarrow y=-\frac{7}{3}\pm\frac{\sqrt{10}}{3}\)
1/ Ta có
\(x^2+9x+20=x^2+4x+5x+20=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
Tương tự
\(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
Đk: x khác 4, 5, 6, 7
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\frac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}+\frac{\left(x+7\right)-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\) EM tự làm tiếp nhé
\(\frac{x^8-1}{\left(x^4+1\right)\left(x^2-1\right)}\)
\(=\frac{\left(x^2-1\right)\left(x^4+x^2+1\right)}{\left(x^4+1\right)\left(x^2-1\right)}\)
\(=\frac{x^4+x^2+1}{x^4+1}\)
\(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\)
\(=\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)
\(=\frac{\left(x+y-2\right)\left(x+y+2\right)}{\left(x+2-y\right)\left(x+2+y\right)}\)
\(=\frac{x+y-2}{x+2-y}\)
\(\frac{4x^2+12x+9}{2x^2-x-6}\)
\(=\frac{\left(2x+3\right)^2}{2x^2-4x+3x-6}\)
\(=\frac{\left(2x+3\right)^2}{2x\left(x-2\right)+3\left(x-2\right)}\)
\(=\frac{\left(2x+3\right)^2}{\left(2x+3\right)\left(x-2\right)}\)
\(=\frac{2x+3}{x-2}\)
\(\frac{25-10x+x^2}{xy-5y}\)
\(=\frac{\left(5-x\right)^2}{-y\left(5-x\right)}\)
\(=-\frac{5-x}{y}\)
\(\frac{\left|x\right|-3}{x^2-9}\)
\(=\frac{x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{1}{x+3}\)
\(\frac{3\left|x-4\right|}{3x^2-3x-36}\)
\(=\frac{3\left(x-4\right)}{3\left(x^2-x-12\right)}\)
\(=\frac{x-4}{x^2-4x+3x-12}\)
\(=\frac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\frac{x-4}{\left(x-4\right)\left(x+3\right)}\)
\(=\frac{1}{x+3}\)
\(\frac{-1}{y-1}+\frac{24}{y+2}=13\) ĐKXĐ: y khác 1; y khác 2
=> -1(y+2) + 24(y-1) = 13( y + 2 )(y-1 )
<=> -y - 2 + 24y - 24 = 13(y2 - y + 2y - 2 )
<=> -y - 2 + 24y - 24 - 13y2 + 13y-26y + 26 = 0
<=> -13y2 + 10y = 0
<=> y( -13y + 10 ) = 0
<=> y = 0 hoặc -13y + 10 = 0
<=> y = 0 hoặc y = 10/13
Vậy S = { 0; 10/13 }
Bài làm
\(\frac{-1}{y-1}+\frac{24}{y+2}=13\) ĐKXĐ: y khác 1; y khác -2
\(\Rightarrow-1\left(y+2\right)+24\left(y-1\right)=13\)
\(\Leftrightarrow-y-2+24y-24-13=0\)
\(\Leftrightarrow23y-39=0\)
\(\Leftrightarrow y=\frac{39}{23}\)
Vậy y = 39/23 là nghiệm phương trình.