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a: \(\Leftrightarrow1-x+3x+3=2x+3\)
=>2x+4=2x+3(vô lý)
b: \(\Leftrightarrow\left(x+2\right)^2-2x+3=x^2+10\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
=>4x+7=10
hay x=3/4
d: \(\Leftrightarrow\left(-2x+5\right)\left(3x-1\right)+3\left(x-1\right)\left(x+1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+2x+15x-5+3\left(x^2-1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+17x-5+3x^2-3=x-3x^2+2-6x\)
\(\Leftrightarrow-3x^2+17x-8=-3x^2-5x+2\)
=>22x=10
hay x=5/11
a: \(\Leftrightarrow5x-2+\left(2x-1\right)\left(1-x\right)=2-2x-2x^2-2x+6\)
\(\Leftrightarrow5x-2+2x-2x^2-1+x=-2x^2-4x+8\)
=>8x-3=-4x+8
=>-4x=11
hay x=-11/4
b: \(\Leftrightarrow\left(-2x+5\right)\left(3x-1\right)+3\left(x^2-1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+2x+15x-5+3x^2-3=x-3x^2+2-6x\)
\(\Leftrightarrow17x-8=-5x+2\)
=>22x=10
hay x=5/11
d. ĐKXĐ: x khác 1, x khác 3
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+1+8=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\) (loại)
Vậy pt vô nghiệm
\(a\)) \(2\left(x+1\right)=4x+2\)
\(\Leftrightarrow4x+2=4x+2\)
\(\Leftrightarrow0x=0\)
Vậy phương trình đã cho có vô số nghiệm .
b) \(x^2-9x+8=0\)
\(\Leftrightarrow x^2-9x+9-1=0\)
\(\Leftrightarrow\left(x^2-9x+9\right)^2-1=0\)
\(\Leftrightarrow\left(x-3\right)^2-1=0\)
\(\Leftrightarrow\left(x-3+1\right)\left(x-3-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 2 hoặc x = 4 .
c) \(\dfrac{x+2}{x+3}-\dfrac{1}{x}=\dfrac{-3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)
\(\Rightarrow x\left(x+2\right)-x-3+3=0\)
\(\Leftrightarrow x^2+2x-x=0\)
\(\Leftrightarrow x\left(x+2-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy phương trình có nghiệm x = -1 .
Tick cho mình nha Hanako Aki
a/ \(2\left(x+1\right)=4x+2\)
\(\Leftrightarrow2x+2=4x+2\)
\(\Leftrightarrow2x-4x=2-2\)
\(\Leftrightarrow-2x=0\)
\(\Rightarrow x=0\)
b/ dễ => tự lm hoặc vào link:
https://hoc24.vn/hoi-dap/question/205563.html
c/ \(\dfrac{x+2}{x+3}-\dfrac{1}{x}=\dfrac{-3}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{1}{x}-\dfrac{-3}{x\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)-1\left(x+3\right)+3}{x\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+2x-x-3+3}{x\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+x}{x\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{x\left(x+3\right)\ne0}=0\)
\(\Rightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\Rightarrow x=-1\end{matrix}\right.\) Vậy pt có 2 ngiệm là:\(\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)