$\text{a)}$  

Wyznaczamy  $l$ .

$p=srl+s{r}^2\mid -s{r}^2$  

$p-s{r}^2=srl\mid :sr,\mathbf{s}\ne 0,\mathbf{r}\ne 0$  

$\frac{p-s{r}^2}{sr}=l\mid :sr$  

Wyznaczamy  $s$ .

$p=srl+s{r}^2$  

$p=s\left(rl+r^2\right)\mid :\left(rl+r^2\right),\mathbf{r}\ne 0,\mathbf{l}\ne -\mathbf{r}$  

$\frac{p}{rl+r^2}=s$  

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$\text{b)}$  

Wyznaczamy  $b$ .

$a=\sqrt{2}bc-bd$  

$a=b\left(\sqrt{2}c-d\right)\mid :\left(\sqrt{2}c-d\right),\mathbf{d}\ne \sqrt{2}\mathbf{c}$  

$\frac{a}{\sqrt{2}c-d}=b$  

Wyznaczamy  $d$ .

$a=\sqrt{2}bc-bd\mid +bd$  

$bd+a=\sqrt{2}bc\mid -a$  

$bd=\sqrt{2}bc-a\mid :b,\mathbf{b}\ne 0$  

$d=\frac{\sqrt{2}bc-a}{b}$  

---

$\text{c)}$  

Wyznaczamy  $b$ .

$P=\frac{\left(a+b\right)\cdot h}{2}\mid \cdot 2$  

$2P=\left(a+b\right)\cdot h$  

$2P=ah+bh\mid -ah$  

$2P-ah=bh\mid :h,\mathbf{h}\ne 0$  

$\frac{2P-ah}{h}=b$  

$\frac{2P}{h}-a=b$  

Wyznaczamy  $h$ .

$P=\frac{\left(a+b\right)\cdot h}{2}\mid \cdot 2$  

$2P=\left(a+b\right)\cdot h\mid :\left(a+b\right),\mathbf{a}\ne -\mathbf{b}$  

$\frac{2P}{a+b}=h$  

---

$\text{d)}$  

Wyznaczamy  $q$ .

$S=\frac{a_1}{1-q}\mid \cdot \left(1-q\right),\mathbf{q}\ne 1$  

$S\left(1-q\right)=a_1$  

$S-Sq=a_1\mid +Sq$  

$S=a_1+Sq\mid -a_1$  

$S-a_1=Sq\mid :S\mathbf{S}\ne 0$  

$\frac{S-a_1}{S}=q$  

---

$\text{e)}$  

Wyznaczamy  $l$ .

$P=\pi r\left(r+l\right)$  

$P=\pi r^2+\pi rl\mid -\pi r^2$  

$P-\pi r^2=\pi rl\mid :\pi r,\mathbf{r}\ne 0$  

$\frac{P-\pi r^2}{\pi r}=l$  

---

$\text{f)}$  

Wyznaczamy  $r$ .

$a_n=a_1+\left(n-1\right)\cdot r\mid -a_1$  

$a_n-a_1=\left(n-1\right)\cdot r\mid :\left(n-1\right),\mathbf{n}\ne 1$  

$\frac{a_n-a_1}{n-1}=r$  

Wyznaczamy  $n$ .

$a_n=a_1+\left(n-1\right)\cdot r$  

$a_n=a_1+nr-r\mid -a_1$  

$a_n-a_1=nr-r\mid +r$  

$a_n-a_1+r=nr\mid :r,\mathbf{r}\ne 0$  

$\frac{a_n-a_1+r}{r}=n$  

$\frac{a_n-a_1}{r}+1=n$  

---

$\text{g)}$  

Wyznaczamy  $a_1$ .

$S_n=\frac{2a_1+\left(n-1\right)r}{2}\cdot n\mid \cdot 2$  

$2S_n=\left(2a_1+\left(n-1\right)r\right)\cdot n\mid :n,\mathbf{n}\ne 0$  

$\frac{2S_n}{n}=2a_1+\left(n-1\right)r\mid -\left(n-1\right)r$  

$\frac{2S_n}{n}-\left(n-1\right)r=2a_1\mid :2$  

$\frac{S_n}{n}-\frac{\left(n-1\right)r}{2}=a_1$  

Wyznaczamy  $r$ .

$S_n=\frac{2a_1+\left(n-1\right)r}{2}\cdot n\mid \cdot 2$  

$2S_n=\left(2a_1+\left(n-1\right)r\right)\cdot n\mid :n,\mathbf{n}\ne 0$  

$\frac{2S_n}{n}=2a_1+\left(n-1\right)r\mid -2a_1$  

$\frac{2S_n}{n}-2a_1=\left(n-1\right)r\mid :\left(n-1\right),\mathbf{n}\ne 1$  

$\frac{2S_n}{n\left(n-1\right)}-\frac{2a_1}{n-1}=r$  

$\frac{2S_n-2n{a}_1}{n\left(n-1\right)}=r$  

---

$\text{h)}$  

Wyznaczamy  $a$ .

$q=\frac{4ac-b^2}{4a}\mid \cdot 4a,\mathbf{a}\ne 0$  

$4aq=4ac-b^2\mid -4ac$  

$4aq-4ac=-b^2$  

$a\left(4q-4c\right)=-b^2\mid :\left(4q-4c\right),\mathbf{q}\ne \mathbf{c}$  

$a=-\frac{b^2}{4q-4c}$  

Wyznaczamy  $b$ .

$q=\frac{4ac-b^2}{4a}\mid \cdot 4a,\mathbf{a}\ne 0$  

$4aq=4ac-b^2\mid -4ac$  

$4aq-4ac=-b^2\mid \cdot \left(-1\right)$      

(liczba -b² jest ujemna, więc liczba 4aq-4ac również)

$4ac-4aq=b^2$      

(liczba b² jest dodatnia, więc liczba 4ac-4aq również)

$4a\left(c-q\right)=b^2$  

$b=\sqrt{4a\left(c-q\right)}$  

$b=\sqrt{4}\cdot \sqrt{a\left(c-q\right)},\mathbf{a}\left(\mathbf{c}-\mathbf{q}\right)\ge 0$  

$b=2\sqrt{a\left(c-q\right)}$  

---

$\text{i)}$  

Wyznaczamy  $a$ .

$x=\frac{2\pi }{a-c^2}+\frac{d}{2}\mid \cdot 2$  

$2x=\frac{4\pi }{a-c^2}+d\mid \cdot \left(a-c^2\right),\mathbf{a}\ne {\mathbf{c}}^2$  

$2x\cdot \left(a-c^2\right)=4\pi +d\cdot \left(a-c^2\right)$  

$2xa-2x{c}^2=4\pi +da-d{c}^2\mid +2x{c}^2$  

$2xa=4\pi +da-d{c}^2+2x{c}^2\mid -da$  

$2xa-da=4\pi -d{c}^2+2x{c}^2$  

$a\left(2x-d\right)=4\pi -d{c}^2+2x{c}^2\mid :\left(2x-d\right),\mathbf{d}\ne 2\mathbf{x}$  

$a=\frac{4\pi -d{c}^2+2x{c}^2}{2x-d}$  

$a=\frac{4\pi +2x{c}^2-d{c}^2}{2x-d}$  

$a=\frac{4\pi +c^2\left(2x-d\right)}{2x-d}$  

$a=\frac{4\pi }{2x-d}+\frac{c^2\left(2x-d\right)}{2x-d}$  

$a=\frac{4\pi }{2x-d}+c^2$  

Wyznaczamy  $d$ .

$x=\frac{2\pi }{a-c^2}+\frac{d}{2}\mid \cdot 2$  

$2x=\frac{4\pi }{a-c^2}+d\mid -\frac{4\pi }{a-c^2}$  

$2x-\frac{4\pi }{a-c^2}=d$  

---

$\text{j)}$  

Wyznaczamy  $a$ .

$f=\frac{4{\pi }^{2}a}{2\left(a-b\right)}\mid \cdot 2$  

$2f=\frac{4{\pi }^{2}a}{a-b}\mid \cdot \left(a-b\right),\mathbf{a}\ne \mathbf{b}$  

$2f\left(a-b\right)=4{\pi }^{2}a$  

$2fa-2fb=4{\pi }^{2}a\mid -4{\pi }^{2}a$  

$2fa-4{\pi }^{2}a-2fb=0\mid +2fb$  

$2fa-4{\pi }^{2}a=2fb$  

$a\left(2f-4{\pi }^2\right)=2fb\mid :\left(2f-4{\pi }^2\right),\mathbf{f}\ne 2{\pi }^2$  

$a=\frac{2fb}{2f-4{\pi }^2}$  

Wyznaczamy  $b$ .

$f=\frac{4{\pi }^{2}a}{2\left(a-b\right)}\mid \cdot 2$  

$2f=\frac{4{\pi }^{2}a}{a-b}\mid \cdot \left(a-b\right),\mathbf{a}\ne \mathbf{b}$  

$2f\left(a-b\right)=4{\pi }^{2}a$  

$2fa-2fb=4{\pi }^{2}a\mid +2fb$  

$2fa=4{\pi }^{2}a+2fb\mid -4{\pi }^{2}a$  

$2fa-4{\pi }^{2}a=2fb\mid :2f,\mathbf{f}\ne 0$  

$\frac{2fa-4{\pi }^{2}a}{2f}=b$  

$\frac{fa-2{\pi }^{2}a}{f}=b$  

$\frac{fa}{f}-\frac{2{\pi }^{2}a}{f}=b$  

$a-\frac{2{\pi }^{2}a}{f}=b$  

---

$\text{k)}$  

Wyznaczamy  $a$ .

$s=a^2+\frac{4\left(a^2-1\right)}{2}+5b$  

$s=a^2+2\left(a^2-1\right)+5b$  

$s=a^2+2a^2-2+5b$  

$s=3a^2-2+5b\mid +2$  

$s+2=3a^2+5b\mid -5b$  

$s-5b+2=3a^2\mid :3$  

$\frac{s-5b+2}{3}=a^2$  

$a=\sqrt{\frac{s-5b+2}{3}}\text{lub}a=-\sqrt{\frac{s-5b+2}{3}},\mathbf{s}\ge 5\mathbf{b}-2$  

---

$\text{l)}$  

Wyznaczamy  $h$ .

$E=\frac{m{v}^2}{2}+mgh\mid -\frac{m{v}^2}{1}$  

$E-\frac{m{v}^2}{2}=mgh\mid :mg,\mathbf{m}\ne 0,\mathbf{g}!-0$  

$\frac{E}{mg}-\frac{m{v}^2}{2}mg=h$  

$\frac{E}{mg}-\frac{v^2}{2g}=h$  

Wyznaczamy  $m$ .

$E=\frac{m{v}^2}{2}+mgh\mid \cdot 2$  

$2E=m{v}^2+2mgh$  

$2E=m\left(v^2+2gh\right)\mid :\left(v^2+2gh\right),\frac{{\mathbf{v}}^2}{2}\ne \mathbf{g}\mathbf{h}$  

$\frac{2E}{v^2+2gh}=m$