Question

Proszę o pomoc w zadaniu liczby rzeczywiste .

Answer 1

Rozwiązanie:

Zadanie 1.

a)

[tex]x=1+1\dfrac23\cdot\left(-\dfrac{12}{35}\right)\\\\x=1+\dfrac{5\!\!\!\!\diagup}{3\!\!\!\!\diagup}\cdot\left(-\dfrac{12\!\!\!\!\!\diagup^4}{35\!\!\!\!\!\diagup_7}\right)\\\\x=1+1\cdot\left(-\dfrac47\right)\\\\x=1-\dfrac47\\\\x=\dfrac37\\\\\boxed{-x=-\dfrac37}[/tex]

b)

[tex]y=\dfrac35:\left(-\dfrac9{50}\right)+4\\\\y=\dfrac{3\!\!\!\!\diagup^1}{5\!\!\!\!\diagup_1}\cdot\left(-\dfrac{50\!\!\!\!\!\diagup^{10}}{9\!\!\!\!\diagup_3}\right)+4\\\\y=1\cdot\left(-\dfrac{10}3\right)+4\\\\y=-\dfrac{10}3+\dfrac{12}3\\\\y=\dfrac23\\\\\dfrac1{y}=1:\dfrac23=1\cdot\dfrac32=\dfrac32=1\dfrac12\\\\\boxed{\dfrac1y=1\dfrac12}[/tex]

Zadanie 2.

[tex]\sqrt{12}-\sqrt{108}+\sqrt{75}=\sqrt{4\cdot 3}-\sqrt{36\cdot 3}+\sqrt{25\cdot 3}=2\sqrt3-6\sqrt3+5\sqrt3=\boxed{\sqrt3}[/tex]

Zadanie 3.

a)

[tex]\sqrt[3]{(-8)\cdot\dfrac1{125}}=\sqrt[3]{-\dfrac8{125}}=\sqrt[3]{\left(-\dfrac25\right)^3}=-\dfrac25[/tex]

b)

[tex]log_264+log_5125=6+3=9[/tex]

c)

[tex]log_{\sqrt[3]2}4-log_2{\sqrt[3]4}=6+\dfrac23=6\dfrac23[/tex]

d)

[tex]log_34-2log_36=log_34-log_336=log_3\left(\dfrac4{36}\right)=log_3\left(\dfrac26\right)^2=2log_3\dfrac13=2\cdot(-1)=\\=-2[/tex]

e)

[tex]\dfrac{\left(15^{-\frac23}\right)^{-\frac38}\cdot\left(\dfrac13\right)^{\frac14}}{\left(5^{-\frac12}\right)^{1,5}}=\dfrac{15^{\frac14}\cdot\left(\dfrac13\right)^{\frac14}}{5^{-\frac34}}=\dfrac{\left(15\cdot \dfrac13\right)^{\frac14}}{5^{-\frac34}}=\dfrac{5^{\frac14}}{5^{-\frac34}}=5^{\frac14+\frac34}=5^1=5[/tex]

Zadanie 4.

[tex]3^{-\frac34}\cdot\sqrt{3^3}\cdot\sqrt[4]9=3^{-\frac34}\cdot(3^3)^{\frac12}\cdot (3^2)^{\frac14}=3^{-\frac34}\cdot3^{\frac32}\cdot 3^{\frac12}=3^{-\frac34+\frac32+\frac12}=3^{\frac54}[/tex]

Zadanie 5.

[tex]95zl \to 1\\76 \to 1-x\\\\76=95(1-x)\\76=95-95x\\95x=95-76\\95x=19 |:{95}\\x=0.2 = \boxed{20\%}[/tex]

Zadanie 6.

[tex]\dfrac5{100}x=2^4\cdot\sqrt[3]{64}\cdot 2^{-5}\cdot 2^0\\\\\dfrac1{20}x=2^4\cdot (2^6)^{\frac13}\cdot 2^{-5}\cdot 1\\\\\dfrac1{20}x=2^{4+2-5}\\\\\dfrac1{20}x=2^{1}\\\\\dfrac1{20}x=2 |\cdot 20\\\\\boxed{x=40}[/tex]

Zadanie 7.

[tex]1,5x+200=1700 |-200\\1,5x=1500 |\cdot 10\\15x=15000|:15\\\boxed{x=1000}[/tex]