Намерете корените на уравненията

[Гост]

Моля някой да съдейства, ако може! В прикачения файл има задачи.

[Гост]

mnogo gotini zadachi, zashto ne iskash da gi reshavash? reshi edna pone...

[ammornil]

$(1)\\[6pt]\frac{x-4}{x(x+2)}-\frac{5-2x}{x^{2}-4}=\frac{2}{x(x-2)}\\[6pt]\quad\text{ДМ:}\quad \begin{array}{|l}x(x+2)\ne{}0\\x^{2}-4\ne{}0\\x(x-2)\ne{}0 \end{array} \Rightarrow \begin{array}{|l}x(x+2)\ne{}0\\(x+2)(x-2)\ne{}0\\x(x-2)\ne{}0 \end{array} \Rightarrow \begin{array}{|l}x\ne{}0 \cup x\ne{-2} \\ x\ne{}-2 \cup x\ne{}2 \\x\ne{}0 \cup x\ne{}2 \end{array} \Rightarrow x\in{}(-\infty;-2)\cup{}(-2;0)\cup{}(0;2)\cup{}(2;+\infty)\\[12pt] \underbrace{\frac{\overset{x-2}{x-4}}{x(x+2)}-\frac{\overset{x}{5-2x}}{(x-2)(x+2)}=\frac{\overset{x+2}{2}}{x(x-2)}}_{x(x-2)(x+2)} \\[6pt] (x-2)(x-4)-x(5-2x)-2(x+2)=0 \\[6pt] x^{2}-4x-2x+8-5x+2x^{2}-2x-4=0 \\[6pt] 3x^{2}-13x+4=0 \\[6pt] \quad D=13^{2}-4\cdot{}3\cdot{}4=169-48=121 \\[6pt] \Rightarrow x_{1,2}=\frac{13\pm{}11}{2\cdot{}3} \Rightarrow \begin{cases} x_{1}=4 \in{}\text{ДМ} \\ x_{2}=\dfrac{1}{3}\in{}\text{ДМ} \end{cases} $

[ammornil]

$ (2)\\[6pt] \frac{x^{2}+x-8}{x}+\frac{3x}{x^{2}+x-8}=4\\[6pt]\quad\text{ДМ:}\quad \begin{array}{|l} x\ne{}0 \\ x^{2}+x-8\ne{}0 \end{array} \Rightarrow \begin{array}{|l} x\ne{}0 \\ \dfrac{-1\pm{}\sqrt{1^{2}-4\cdot{}1\cdot{}(-8)}}{2\cdot{}1}\ne{}0 \end{array} \Rightarrow \begin{array}{|l} x\ne{}0 \\ \dfrac{-1\pm{}5}{2}\ne{}0 \end{array} \\ \quad \quad \Rightarrow x\in{}(-\infty;-3)\cup{}(-3;0)\cup{}(0;2)\cup{}(2;+\infty) \\[6pt] \frac{x^{2}+x-8}{x}+3\cdot{}\frac{x}{x^{2}+x-8}=4 \\[6pt] \quad \frac{x^{2}+x-8}{x}=t \ne{}0\quad \forall{x}\in{}\text{ДМ} \Rightarrow \\[6pt] t+3\frac{1}{t}-4=0 \Leftrightarrow t^{2}+3-4t=0 \Leftrightarrow t^{2}-4t+3=0 \\[6pt] \quad t_{1,2}=\dfrac{2\pm{}\sqrt{2^{2}-1\cdot{}3}}{1}=2\pm{}1 \Rightarrow \\[6pt] \begin{array}{lcl} \dfrac{x^{2}+x-8}{x}=1 &\cup&\dfrac{x^{2}+x-8}{x}=3 \end{array} \\[6pt] \begin{array}{lcl} x^{2}+x-8=x &\cup& x^{2}+x-8 =3x \end{array} \\[6pt] \begin{array}{lcl}x^{2}=8 &\cup&x^{2}-2x-8=0 \end{array} \\[6pt] \begin{array}{lcl} x_{1,2}=\pm{}2\sqrt{2}&\cup{}&x_{3,4}=\dfrac{1\pm{}\sqrt{1^{2}-1\cdot{}(-8)}}{1} \end{array} \\[6pt]\begin{array}{lcl} x_{1,2}=\pm{}2\sqrt{2}&\cup{}&x_{3,4}=1\pm{}3 \end{array} \\[6pt] x_{1}=-2\sqrt{2} \in{}\text{ДМ} \\ x_{2}=2\sqrt{2}\in{}\text{ДМ} \\x_{3}=-2\in{}\text{ДМ}\\x_{4}=4\in{}\text{ДМ}$ $$ \text{Отг.:} \quad -2\sqrt{2};\quad 2;\quad 2\sqrt{2};\quad 4 $$

[ammornil]

$ (3)\\[6pt] \sqrt{x^{2}+x-6}=1-x \\[6pt] \quad \text{ДМ:} \quad \begin{array}{|l} x^{2}+x-6\ge{}0 \\ 1-x\ge{}0 \end{array} \Rightarrow \begin{array}{|l} \left(x+3\right)\left(x-2 \right)\ge{}0 \\ x\le{}1 \end{array} \Rightarrow x\in{}(-\infty;-3] \\[6pt] \sqrt{x^{2}+x-6}=1-x \quad |(\cdots{})^{2} \\[6pt] x^{2}+x-6=1^{2}-2x+x^{2} \\[6pt] 3x=7 \\[6pt]x=\frac{7}{3} \notin\text{ДМ} $ $$ \text{Отг.:}\quad \text{няма решения} $$

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$ x^{2}+x-6 = 0 \Rightarrow x_{1,2}=\frac{-1\pm{}\sqrt{1^{2}-4\cdot{}1\cdot{}(-6)}}{2\cdot{}1}=\frac{-1\pm{}5}{2} $

[ammornil]

$(4) \\[6pt] \sqrt{x+2}-\sqrt{x-6}=2 \\[6pt]\quad \text{ДМ:} \quad \begin{array}{|l} x+2\ge{}0 \\ x-6\ge{}0 \end{array} \Rightarrow \begin{array}{|l} x\ge{}-2 \\ x\ge{}6 \end{array} \Rightarrow x\in{}[6;+\infty) \\[6pt] \sqrt{x+2}=2+\sqrt{x-6} \quad |(\cdots{})^{2} \\[6pt] x+2=2^{2}+4\sqrt{x-6}+x-6 \\[6pt] 4\sqrt{x-6} = 4 \quad |\div{}4 \\[6pt]\sqrt{x-6}=1 \quad |(\cdots{})^{2} \\[6pt] x-6=1 \\[6pt] x=7 \in\text{ДМ} \\[12pt] \text{Проверка за придобити корени от повдигания на втора степен}:\\[6pt] \sqrt{7+2}-\sqrt{7-6}=2 \\[6pt] 3-1=2 \quad \text{вярно равенство} $ $$\text{Отг.:}\quad 7 $$

[ammornil]

$(5)\\[6pt] x^{2}-3x+\sqrt{x^{2}-3x+5}=7 \\ \quad \text{ДМ:} \quad x^{2}-3x+5\ne{}0 \cdots \\[6pt] x^{2}-3x\red{+5}+\sqrt{x^{2}-3x+5}=7\red{+5} \\[6pt]\quad t=x^{2}-3x+5 \\[6pt] \sqrt{t}=12-t $
Намирате [tex]t[/tex]. Проверявате дали получените отговори за [tex]t[/tex] са решения на уравнението за [tex]t[/tex], замествате обратно в полагането и после проверявате намерените [tex]x[/tex] дали са в [tex]\text{ДМ}[/tex].

[ammornil]

$(6)\\[6pt]3^{x-\sqrt{3x-5}}=27\\[6pt]\quad \text{ДМ:} \quad 3x-5\ge{}0 \\[6pt] 3^{x-\sqrt{3x-5}}=3^{3} \\[6pt] x-\sqrt{3x-5}=3 \\[6pt] x-3=\sqrt{3x-5} \quad |(\cdots)^{2} \\[6pt] \cdots{}$

Проверете дали получените стойности за [tex][/tex] са от дефиниционното множество и заместете с всяка в условоето за да отхвърлите придобити корени от повдигането на втора степен.

[ammornil]

[tex](7)\\[6pt] \log_{2}{\left(x+\frac{5}{3}\right)}=1-\log_{2}{3x} \\[6pt]\quad\text{ДМ:}\quad\begin{array}{|l}x+\dfrac{5}{3} >0 \\ 3x>0 \end{array} \\[6pt] \log_{2}{\left(x+\frac{5}{3}\right)}+\log_{2}{3x}=\log_{2}{2} \\[6pt]\log_{2}{\left[3x\cdot{}\left(x+\frac{5}{3}\right)\right]}=\log_{2}{2} \\[6pt]3x\cdot{}\left(x+\frac{5}{3}\right)=2 \\[6pt] 3x^{2}+5x-2=0 \\ \cdots[/tex]

[ammornil]

$ (8)\\[6pt] \lg{10^{\lg{(x^{2}+21)}}}-1=\lg{x} \\[6pt]\quad\text{ДМ:}\quad \begin{array}{|l} x^{2}+21>0 \\ x>0 \end{array} \Rightarrow x\in{}(0;+\infty) \\[6pt] \lg{(x^{2}+21)}-\lg{x}=1 \\[6pt] \lg{\frac{x^{2}+21}{x}}=\lg{10} \\[6pt] x^{2}+21=10x \\ \cdots$

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[tex]\lg{10^{A}}=A \Rightarrow \lg{10^{\lg{(x^{2}+21)}}}=\lg{(x^{2}+21)}[/tex]

[ammornil]

$(9)\\[6pt] \sin{(3x)}=\sin{(2x)}+\sin{(x)} \\[6pt]\quad\text{ДМ:}\quad \forall{x}\in{}\mathbb{R} \\[6pt] \sin{(3x)}-\sin{(x)} =\sin{(2x)} \\[6pt]2\cos{(2x)}\sin{(x)}=2\sin{(x)}\cos{(x)} \\[6pt] 2\sin{(x)}[\cos{(2x)}-\cos{(x)}]=0 \\[6pt] \begin{array}{lclcl}\sin{(x)}=0&\cup&2\cos^{2}{(x)}-1-\cos{(x)}=0 \\[6pt] x=\pm{}k\pi &\cup& \cos_{1,2}{(x)}=\dfrac{1\pm{}\sqrt{(-1)^{2}-4\cdot{2}\cdot{(-1)}}}{2\cdot{2}} \\[6pt]x=\pm{}k\pi &\cup&\cos{(x)}=\dfrac{1-3}{4} \quad \cup \quad \cos{(x)}=\dfrac{1+3}{4} \\[6pt] x=\pm{}k\pi &\cup& x=\pm\dfrac{2\pi}{3}\pm{}2k\pi \quad \cup \quad x=\pm{}k\pi \end{array}$ $$ \text{Отг.:}\quad x=\pm{}k\pi \quad\cup\quad x=\pm\dfrac{2\pi}{3}\pm{}2k\pi \\ k \in \mathbb{N_{0}}$$

[tex]---\\[24pt]\sin{(3x)}-\sin{(x)}=2\cos{\frac{3x+x}{2}}\sin{\frac{3x-x}{2}}=2\cos{(2x)}\sin{(x)}[/tex]

[Гост]

Много благодаря!

[Гост]

ostana 10-ta...

[ammornil]

$(10)\\[12pt] \sin^{2}{(x)}+\sin^{2}{(2x)}=\sin^{2}{(3x)} \\[6pt] \quad \text{ДМ:}\quad \forall{x}\in{}\mathbb{R} \\[6pt] \frac{1-\cos{(2x)}}{2}+\frac{1-\cos{(4x)}}{2}=\frac{1-\cos{(6x)}}{2} \\[6pt] \quad 2x=u \Leftrightarrow x=\frac{u}{2} \\[6pt] 1-\cos{(u)}+1-\cos{(2u)}-1+\cos{(3u)}=0 \\[6pt] \cos{(3u)}-\cos{(u)}-\cos{(2u)}+1=0 \\[6pt] -2\sin{(2u)}\sin{(u)}-\cos{(2u)}+1=0 \\[6pt] -2\cdot{}2\sin{(u)}\cos{(u)}\cdot{}\sin{(u)}-(1-2\sin^{2}{(u)})+1=0 \\[6pt] -4\sin^{2}{(u)}\cos{(u)}-1+2\sin^{2}{(u)}+1=0 \\ -2\sin^{2}{(u)}\cdot{}[2\cos{(u)}-1]=0 \\[6pt] \begin{array}{lcl} \sin{(u)}=0 & \cup & 2\cos{(u)}-1=0 \\[6pt] u=\pm{}k\pi & \cup & \cos{(u)}=\dfrac{1}{2} \\[6pt] u=\pm{}k\pi & \cup & u=\pm{}\dfrac{\pi}{3}\pm{}2k\pi \\[12pt] x=\pm{}k\dfrac{\pi}{2} & \cup & x=\pm{}\dfrac{\pi}{6}\pm{}k\pi \\ & \forall{}k\in{}\mathbb{N_{0}}\end{array} $

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$$ \sin^{2}{(\alpha)}=\frac{1-\cos{(2a)}}{2} \\[12pt] \cos{(3\alpha)}-\cos{(\alpha)}=-2\cdot{}\sin{\dfrac{3\alpha+\alpha}{2}}\cdot{}\sin{\dfrac{3\alpha-\alpha}{2}}=-2\sin{(2\alpha)}\sin{(\alpha)} \\[12pt] \sin{(2\alpha)}= 2\sin{(\alpha)}\cos{(\alpha)} \\[12pt] \cos{(2\alpha)}=1-2\sin^{2}{(\alpha)} $$