Title: Irratsionaalavaldiste lihtsustamine/TEOORIA.NÄITED

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Irratsionaalavaldiste lihtsustamine

Irratsionaalsete muutujaid sisaldavate avaldiste teisendamisel jääb kehtima kõik,  mis on seotud ratsionaalavaldiste teisendamisega. Samuti jääb kehtima kõik astendamise ja juurimisega seotu.

Oluliseks erinevuseks irratsionaalavaldiste puhul on vaid see, et ilma avaldistele tingimusi juurde lisamata eeldatakse tavaliselt, et muutujad avaldistes omavad vaid selliseid väärtusi, mille korral kõik juuritavad on positiivsed. Seega, kui pole eraldi nõutud, siis ei kirjutata näiteks

 $$\sqrt{ax^2}=a\left|x\right|$$, vaid $$\sqrt{ax^2}=ax$$.

Näide 1

Lihtsustada avaldis $$\left(\sqrt{c}-1\right)\left(\sqrt{c}+2\right)-\left(2\sqrt{c}+3\right)^2$$.

$$\left(\sqrt{c}-1\right)\left(\sqrt{c}+2\right)-\left(2\sqrt{c}+3\right)^2=\sqrt{c}\cdot \sqrt{c}+2\cdot \sqrt{c}-1\cdot \sqrt{c}-1\cdot 2-\left(\left(2\sqrt{c}\right)^2+2\cdot 2\sqrt{c}\cdot 3+3^2\right)=$$

$$=c+2\sqrt{c}-\sqrt{c}-2-4c-12\sqrt{c}-9=-3c-11\sqrt{c}-11$$

Harjuta!

Q: Lihtsusta avaldised.

a) Kui $$\left(\sqrt{5}+\sqrt{m}\right)^2-\sqrt{m}\left(2\sqrt{5}+\sqrt{5}\right)=m-\sqrt{\color{blue} {A}}+\color{blue} {B}$$, siis $$\color{blue} {A}=$$[...], $$\color{blue} {B}=$$[...].

b) Kui $$\left(\sqrt{7}-3a\right)\left(\sqrt{7}-1\right)+\sqrt{7}\left(3a-\sqrt{7}\right)=\color{blue} {X}-\sqrt{\color{blue} {Y}}$$, siis $$\color{blue} {X}=$$[...], $$\color{blue} {Y}=$$[...].
A: 5m; 5; 3a; 7

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Näide 2

Lihtsustada avaldis $$\left(3-\dfrac{\sqrt{x}-3}{3+\sqrt{x}}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\right):\left(x-9\right)^{-1}$$.

Lihtsustame tehete kaupa.

1) $$3^{(x-9}-\dfrac{\sqrt{x}-3}{3+\sqrt{x}}^{^{(\sqrt{x}-3}}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}^{^{(\sqrt{x}+3}}=\dfrac{3\left(x-9\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=$$

$$=\dfrac{3\left(x-9\right)-\left(\sqrt{x}-3\right)^{^2}-\left(\sqrt{x}+3\right)^2}{x-9}=\require{cancel}\dfrac{3x-27-x+\cancel{6\sqrt{x}}-9-x-\cancel{6\sqrt{x}}-9}{x-9}=\dfrac{x-45}{x-9}$$

2) $$\dfrac{x-45}{x-9}:\left(x-9\right)^{-1}=\dfrac{x-45}{x-9}:\dfrac{1}{x-9}=\require{cancel}\dfrac{(x-45)\cdot \cancel{(x-9)}^1}{\cancel{(x-9)}^1\cdot 1}=x-45$$

Harjuta!

Q: Leia puuuduvad suurused avaldise $$\left(\dfrac{a+1}{\sqrt{a}}-2\right):\dfrac{\sqrt{a}-1}{a+\sqrt{a}}$$ lihtsustamise käigus.

$$\left(\dfrac{a+1}{\sqrt{a}}-2\right):\dfrac{\sqrt{a}-1}{a+\sqrt{a}}=\left(\dfrac{a+1-2\sqrt{a}}{\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+\color{blue} {M}\right)}=\dfrac{\left(\sqrt{a}-\color{blue} {L}\right)^2}{\sqrt{a}}:\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+\color{blue} {M}\right)}=$$

$$=\dfrac{\left(\sqrt{a}-\color{blue} {L}\right)^2\cdot \sqrt{a}\left(\sqrt{a}+\color{blue} {M}\right)}{\sqrt{a}\cdot \left(\sqrt{a}-1\right)}=\dfrac{\left(\sqrt{a}-\color{blue} {L}\right)\cdot \left(\sqrt{a}+\color{blue} {M}\right)}{\color{blue} {N}}=\color{blue} {K}$$

$$\color{blue} {M}=$$[...], $$\color{blue} {K}=$$[...], $$\color{blue} {L}=$$[...], $$\color{blue} {N}=$$[...]
A: 1; a-1/-1+a; 1; 1

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Näide 3

Lihtsustada avaldis $$\dfrac{\:1}{\:m^2-\sqrt{m}}:\dfrac{\sqrt{m}+1}{m\sqrt{m}+m+\sqrt{m}}+\dfrac{m^2-m-1}{m-1}$$.

Lihtsustame tehete kaupa.

1) $$\dfrac{\:1}{\:m^2-\sqrt{m}}:\dfrac{\sqrt{m}+1}{m\sqrt{m}+m+\sqrt{m}}=\dfrac{1}{\sqrt{m}\left(\sqrt{m^3}-1\right)}:\dfrac{\sqrt{m}+1}{\sqrt{m}\left(m+\sqrt{m}+1\right)}=$$

$$\require{cancel}\dfrac{1}{\sqrt{m}\left(\sqrt{m}-1\right)\left(m+\sqrt{m}+1\right)}:\dfrac{\sqrt{m}+1}{\sqrt{m}\left(m+\sqrt{m}+1\right)}=\dfrac{1\cdot \cancel{\sqrt{m}}^1\cancel{(m+\sqrt{m}+1)}^1}{\cancel{\sqrt{m}}^1(\sqrt{m}-1)\cancel{(m+\sqrt{m}+1)}^1\cdot (\sqrt{m}+1)}=$$

$$=\dfrac{1}{\left(\sqrt{m}-1\right)\cdot \left(\sqrt{m}+1\right)}=\dfrac{1}{m-1}$$

2) $$\dfrac{1}{m-1}+\dfrac{m^2-m-1}{m-1}=\dfrac{1+m^2-m-1}{m-1}=\dfrac{m^2-m}{m-1}=\require{cancel}\dfrac{m\cancel{(m-1)}^1}{\cancel{m-1}^1}=m$$

Harjuta!

Q: Leia puuuduvad suurused avaldise $$\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)$$ lihtsustamise käigus.

$$\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)=$$

$$=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot \left(\dfrac{\sqrt{b}^{(\sqrt{a}+\sqrt{b}}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}^{(\sqrt{a}-\sqrt{b}}}{\sqrt{\color{blue} {M}}\left(\sqrt{a}+\sqrt{b}\right)}\right)=$$

$$=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{\color{blue} {M}}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)=$$

$$=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot \dfrac{\color{blue} {K}\sqrt{\color{blue} {L}}}{\sqrt{\color{blue} {M}}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=$$

$$=\dfrac{\sqrt{a}+\sqrt{b}-1+1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\color{blue} {R}}{\sqrt{\color{blue} {P}}}$$

$$\color{blue} {K}=$$[...], $$\color{blue} {L}=$$[...], $$\color{blue} {M}=$$[...], $$\color{blue} {P}=$$[...], $$\color{blue} {R}=$$[...]
A: 2; ab/ba; a; a; 1