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\(a\frac{x^2-49}{x+5}:\left(x-7\right)\)
\(=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{\left(x-7\right)}\)
\(=\frac{x+7}{x+5}\)
\(b,\frac{2x+7}{x+2}-\frac{x+8}{2x+4}\)
\(=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{4x+14-x+8}{2\left(x+2\right)}\)
\(=\frac{3x+22}{2\left(x+2\right)}\)
a) \(\frac{x^2-49}{x+5}\div\left(x-7\right)=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{x-7}=\frac{x+7}{x+5}\)
b) \(\frac{2x+7}{x+2}-\frac{x+8}{2x+4}=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{\left(4x+14\right)-\left(x+8\right)}{2\left(x+2\right)}\)
\(=\frac{4x+14-x-8}{2\left(x+2\right)}=\frac{3x+6}{2\left(x+2\right)}=\frac{3\left(x+2\right)}{2\left(x+2\right)}=\frac{3}{2}\)
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
1) a) \(\frac{x}{x+1}+\frac{x^3-2x^2}{x^3+1}=\frac{x}{x+1}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3-x^2+x+x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{2x^3-3x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b) \(\frac{x+1}{2x-2}+\frac{3}{x^2-1}+\frac{x+3}{2x+2}=\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{x+3}{2\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x^2+4x+2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)
2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)
Thay x = 14 ; y = -15 vào biểu thức ta được
\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)
a) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\frac{2x+4}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{x^2-4}=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{\left(x+2\right)}=\frac{-2x+4}{x+2}\)
b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
+) x = 0 \(\Rightarrow A=\frac{-2.0+4}{0+2}=\frac{4}{2}=2\)
+) x = 3 \(\Rightarrow A=\frac{-2.3+4}{3+2}=\frac{-2}{5}\)
Đề sai ! Sửa nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow A=\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x-2}\right)\)
\(\Leftrightarrow A=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow A=\frac{2x+4-4}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(\Leftrightarrow A=\frac{2x\left(x-2\right)}{-x\left(x+2\right)}\)
\(\Leftrightarrow A=-\frac{2\left(x-2\right)}{x+2}\)
b) Để \(A\le-2\)
\(\Leftrightarrow-\frac{2\left(x-2\right)}{x+2}\le-2\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{x+2}\ge2\)
\(\Leftrightarrow\frac{x-2}{x+2}\ge1\)
\(\Leftrightarrow x-2\ge x+2\)
\(\Leftrightarrow-2\ge2\)(ktm)
Vậy để \(A\le-2\Leftrightarrow x\in\varnothing\)
a.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)
\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)
\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)
\(\left(x^2+\frac{2}{5}y\right)\left(x^2+\frac{2}{5}y\right)=x^4+\frac{4}{5}x^2y+\frac{4}{25}y\)
b\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2-\frac{1}{3}\right)^3-x^2-\frac{1}{3}\)
\(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\frac{2x+4-2x+4}{x^2-4}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{8}{x^2-4}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
Ta có:
\(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2x+4}{x^2-4}-\frac{2x-4}{x^2-4}\right).\frac{x^2+4x+4}{8}\)
\(=\frac{0}{x^2-4}.\frac{x^2+4x+4}{8}\)
\(=0.\frac{x^2+4x+4}{8}\)
\(=0\)