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\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Leftrightarrow x^3-25x-x^3-8=42\)
\(\Leftrightarrow-25x-8=42\)
\(\Leftrightarrow-25x=42+8\)
\(\Leftrightarrow-25x=50\)
\(\Leftrightarrow x=-\dfrac{50}{25}=-2\)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow25x=-25\Rightarrow x=-1\)
\(\text{x.(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12}\\ \Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\\ \Leftrightarrow4x=-38\\ \Leftrightarrow x=-\dfrac{19}{2}\)
\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)
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) \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{2\left(2-x\right)}{\left(x+2\right)}\)
\(=\frac{-5}{2}\)
Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)
\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)
\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)
Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(2\sqrt{2}x+1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\\ < =>x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\\ < =>x^3-25x-x^3-8=3\\ < =>-25x=11\\ < =>x=-\dfrac{11}{25}\)
\(x\left(x^2-25\right)-x^3-8-3=0\\ x^2-25x-x^3-8-3=0\\ -25x-11=0\\ x=-\dfrac{11}{25}\)