[tex]\frac{x(x-3)}{x+2} = 0\\\\\frac{3x+2}{x-1} = 3\\\\x+2 = \frac{-1}{x}\\\\\frac{3x+3}{x+2} = 3-x\\\\\frac{x+1}{2x-1} = \frac{2}{x}[/tex]Równania wymierne
Najlepiej jakby byly pełne rozwiązania abym się lepiej nauczyl. Prosze mam to na jutro i od tego zalezy czy zdam czy nie.
Najlepiej jakby byly pełne rozwiązania abym się lepiej nauczyl. Prosze mam to na jutro i od tego zalezy czy zdam czy nie.
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[tex]W\ rownaniach \ wymiernych\ pierwsze\ co\ robimy\ to\ okreslamy\ dziedzine ,\\\\ mianownik\ musi\ byc\ rozny\ od\ zera.Nastepnie\ mnozymy\ na\ krzyz.\\\\1.)\\\\\frac{x(x-3)}{x+2}=0\\\\x+2\neq 0\\\\x\neq -2\\\\ D=R\setminus \left \{ -2 \right \}\\\\x(x-3)=0\\\\x=0\ \ \vee\ \ x-3=0\\\\x=0\ \ \vee\ \ x=3[/tex]
[tex]2.\\\\\frac{ 3x+2}{x-1}=3\\\\x-1\neq 0\\\\x\neq 1\\\\ D=R\setminus \left \{1 \right \}\\\\ 3x+2=3*(x-1)\\\\3x+2=3x-3\\\\3x-3x=-3-2\\\\0=-5\\\\rownanie\ sprzeczne\ brak\ rozwiazania[/tex]
[tex]3.\\\\x+2=\frac{ -1}{x } \\\\x \neq 0\\\\ D=R\setminus \left \{0 \right \}\\\\ x(x+2)=-1\\\\x^2+2x+1=0\\\\(x+1)^2=0\\\\x+1=0\\\\x=-1[/tex]
[tex]4.\\\\\frac{3x+3}{x+2}=3-x\\\\x +2\neq \\\\x\neq -2\\\\ D=R\setminus \left \{-2 \right \}\\\\ 3x+3=(3-x)(x+2)\\\\3x+3=3x+6-x^2-2x\\\\3x+3=-x^2+x+6\\\\x^2-x-6+3x+3=0\\\\x^2+2x-3=0\\\\ a=1,\ \ b=2,\ \ c=-3\\\\\Delta=b^2-4ac=2^2-4*1*(-3)=4+12=16\\\\\sqrt{\Delta}=\sqrt{16}=4[/tex]
[tex]x_ {1}=\frac{-b-\sqrt{\Delta}}{2a} =\frac{-2-4}{2*1}=\frac{-6}{2}=-3\\\\x_ {2}=\frac{-b+\sqrt{\Delta}}{2a} =\frac{-2+4}{2*1}=\frac{2}{2}=1[/tex]
[tex]5.\\\\\frac{ x+1}{ 2x-1}= \frac{2}{x}\\\\2x -1\neq 0\ \ \wedge \ \ x\neq 0\\\\ 2x\neq 1\ \ |:2\\\\x \neq \frac{1}{2}\\\\ D=R\setminus \left \{ 0,\frac{1}{2} \right \}[/tex]
[tex]x(x+1)=2( 2x-1)\\\\x^2+x=4x-2\\\\x^2+x-4x+2=0\\\\x^2-3x+2=0\\\\a=1,\ \ b=-3,\ \ c=2\\\\\Delta=b^2-4ac=(-3)^2-4*1*2=9-8=1\\\\\sqrt{\Delta}=\sqrt{1}=1[/tex]
[tex]x_ {1}=\frac{-b-\sqrt{\Delta}}{2a} =\frac{- (-3)-1}{2*1}=\frac{ 2}{2}= 1\\\\x_ {2}=\frac{-b+\sqrt{\Delta}}{2a} =\frac{3+1}{2*1}=\frac{4}{2}=2[/tex]
Odpowiedź:
[tex]\huge\boxed {a)~~x=0~~\lor~~x=3~~}[/tex]
[tex]\huge\boxed {b)~~x\in \varnothing~~}[/tex]
[tex]\huge\boxed {c)~~x=-1~~}[/tex]
[tex]\huge\boxed {d)~~x=-3~~\lor ~~x=1~~}[/tex]
[tex]\huge\boxed {e)~~x=1~~\lor ~~x=2~~}[/tex]
Szczegółowe wyjaśnienie:
Dla przypomnienia:
Rozwiązujemy:
[tex]a)\\\\\dfrac{x(x-3)}{x+2} =0\\\\zal.\\\\x+2\neq 0~~\Rightarrow ~~x\neq -2\\\\\boxed {~~D=\ma{R}-\{-2\}~~}\\\\\dfrac{x(x-3)}{x+2} =0\\\\~~~~~~~~\Downarrow \\\\x(x-3)=0\\\\x=0~~\lor~~x-3=0\\\\(x=0~~\lor~~x=3)~~\land~~x\in D\\\\~~~~~~~~~~~~~~~\Downarrow \\\\\huge\boxed {~~x=0~~\lor~~x=3~~}[/tex]
[tex]b)\\\\\dfrac{3x+2}{x-1} =3\\\\zal.\\\\x-1\neq 0~~\Rightarrow ~~x\neq 1\\\\\boxed {~~D=\mathbb {R}-\{1\}~~}\\\\\dfrac{3x+2}{x-1} =3\\\\\\\dfrac{3x+2}{x-1} =\dfrac{3}{1} \\\\1\cdot (3x+2)=3\cdot (x-1)\\\\3x+2=3x-3\\\\2\neq -3~~\Rightarrow ~~ x\in \varnothing\\\\\huge\boxed {~~x\in \varnothing~~}[/tex]
[tex]c)\\\\x+2=\dfrac{-1}{x} \\\\zal.\\\\x\neq 0\\\\\boxed {~~D=\mathbb{R}-\{0\}~~}\\\\x+2=\dfrac{-1}{x} \\\\\dfrac{x+2}{1} =\dfrac{-1}{x} \\\\x(x+2)=-1\cdot 1\\\\x^{2} +2x=-1\\\\x^{2} +2x+1=0\\\\x^{2} +2\cdot 1 \cdot x +1^2=0\\\\(x+1)^2=0\\\\x+1=0\\\\x=-1~~\land ~~x\in D\\\\~~~~~~~~~~~~~\Downarrow \\\\\huge\boxed {~~x=-1~~}[/tex]
[tex]d)\\\\\dfrac{3x+3}{x+2} =3-x\\\\zal.\\\\x+2\neq 0~~\Rightarrow ~~x\neq -2\\\\D=\mathbb {R}-\{-2\}\\\\\dfrac{3x+3}{x+2} =3-x\\\\\dfrac{3x+3}{x+2} =\dfrac{3-x}{1} \\\\1\cdot (3x+3)=(x+2)(3-x)\\\\3x+3=3x-x^{2} +6-2x\\\\x^{2} +2x-3=0\\\\a=1,~b=2,~c=-3\\\\\Delta =2^2-4\cdot 1\cdot (-3)=4+12=16\\\\\sqrt{\Delta} =4\\\\(x_{1} =\dfrac{-2-4}{2} =-3~~\lor~~x_{2} =\dfrac{-2+4}{2}=1)~\lan~~x\in D\\\\~~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\\huge\boxed {~~x=-3~~\lor ~~x=1~~}[/tex]
[tex]e)\\\\\dfrac{x+1}{2x-1} =\dfrac{2}{x} \\\\zal.\\\\2x-1\neq 0~~\Rightarrow ~~x\neq \frac{1}{2} \\\\x\neq 0\\\\D=\mathbb {R}-\{0,\frac{1}{2} \}\\\\\\\dfrac{x+1}{2x-1} =\dfrac{2}{x} \\\\x(x+1)=2(2x-1)\\\\x^{2} +x=4x-2\\\\x^{2} -3x+2=0\\\\a=1,~b=-3,~c=2\\\\\Delta =(-3)^2-4\cdot 1\cdot 2=9-8=1\\\\\sqrt{\Delta} =1\\\\(x_{1} =\dfrac{3-1}{2} =1~~\lor ~~x_{2} =\dfrac{3+1}{2} =2)~~\land~~x\in D\\\\~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\\huge\boxed {~~x=1~~\lor ~~x=2~~}[/tex]