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\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow25x=-25\Rightarrow x=-1\)
\(\text{x.(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12}\\ \Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\\ \Leftrightarrow4x=-38\\ \Leftrightarrow x=-\dfrac{19}{2}\)
a) ĐK : x khác 2/3 ; x khác 0
\(\frac{x+5}{3x-2}=\frac{A}{x\left(3x-2\right)}\)
\(\Leftrightarrow\frac{x\left(x+5\right)}{x\left(3x-2\right)}=\frac{A}{x\left(3x-2\right)}\)
\(\Leftrightarrow A=x^2+5x\)
b) \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{2\left(2-x\right)}{\left(x+2\right)}\)
\(=\frac{-5}{2}\)
\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)
\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Leftrightarrow x^3-25x-x^3-8=42\)
\(\Leftrightarrow-25x-8=42\)
\(\Leftrightarrow-25x=42+8\)
\(\Leftrightarrow-25x=50\)
\(\Leftrightarrow x=-\dfrac{50}{25}=-2\)