tolong dong kaaa,pake caranya makasih..yg dibuletin ya ka

Soal No. 1-b
[tex] \sqrt{61}\ \text{berada di antara}\ \sqrt{49}\ \text{dan}\ \sqrt{64} \\ \\ \sqrt{49}=7\ \text{sedangkan}\ \sqrt{64}=8 \\ \\ \text{Jarak angka antara 49 dan 64 adalah 15 angka}\\ \text{Sedangkan jarak angka antara 49 dan 61 adalah 12 angka}\\ \text{Maka :} \\ \\ \sqrt{61}\approx \sqrt{49}+ \frac{61-49}{64-49}\\ \\ \sqrt{61}\approx 7+ \frac{12}{15}\\ \\ \sqrt{61}\approx 7+ \frac{4}{5}\\ \\ \sqrt{61}\approx 7+ 0,8\\ \\ \sqrt{61}\approx 7,8[/tex]

Soal No. 1-c
[tex]\sqrt{136}\ \text{berada di antara}\ \sqrt{121}\ \text{dan}\ \sqrt{144} \\ \\ \sqrt{121}=11\ \text{sedangkan}\ \sqrt{144}=12\\ \\ \text{Jarak angka antara 121 dan 144 adalah 23 angka}\\ \text{Sedangkan jarak angka antara 121 dan 136 adalah 15 angka}\\ \text{Maka :} \\ \\ \sqrt{136}\approx \sqrt{121}+ \frac{136-121}{144-121}\\ \\ \sqrt{136}\approx 11+ \frac{15}{23}\\ \\ \sqrt{136}\approx 11+ 0,65\\ \\ \sqrt{136}\approx 11,65[/tex]

Soal No. 2-b

[tex]= \sqrt{0,16}+ \sqrt{0,25}+ \sqrt{0,01} \\ \\ = \sqrt{16\times10^{-2}}+\sqrt{25\times10^{-2}}+\sqrt{10^{-2}}\\ \\ = \sqrt{4^2\times10^{-2}}+\sqrt{5^2\times10^{-2}}+\sqrt{10^{-2}}\\ \\ = 4\times10^{-1}+5\times10^{-1}+10^{-1} \\ \\ = 0,4+0,5+0,1 \\ \\ =1,0[/tex]

Soal No. 2-c

[tex]= \sqrt[3]{-8}+ \sqrt[3]{-27}+ \sqrt[3]{125} \\ \\ = \sqrt[3]{(-2)^3}+ \sqrt[3]{(-9)^3}+ \sqrt[3]{5^3} \\ \\ =(-2)^{ \frac{3}{3} } + (-9)^{ \frac{3}{3} } + 5^{ \frac{3}{3} }\\ \\ =(-2) + (-9) + 5 \\ \\ =-11+5 \\ \\ =-6 [/tex]

Soal No. 3-b
a√x - b√x = (a - b)√x

[tex] =7\sqrt[3]{5}-6\sqrt[3]{5} \\ \\ =(7 - 6) \sqrt[3]{5} \\ \\=\sqrt[3]{5} [/tex]

Soal No. 3-c

[tex]= (\sqrt{2}+ \sqrt{5})-(\sqrt{2}- \sqrt{5}) \\ \\ = \sqrt{2}+ \sqrt{5}-\sqrt{2}+ \sqrt{5} \\ \\= (\sqrt{2}-\sqrt{2})+ (\sqrt{5}+ \sqrt{5}) \\ \\ =2 \sqrt{5} [/tex]

Soal No. 4-b

[tex]=q^{ (-\frac{2}{3}) } \\ \\ = \frac{1}{q^{(\frac{2}{3})}} \\ \\ = \frac{1}{ \sqrt[3]{q^2} } \\ \\= \frac{1}{ \sqrt[3]{q^2} }\times\frac{(\sqrt[3]{q^2})^2 }{(\sqrt[3]{q^2})^2} \\ \\= \frac{(\sqrt[3]{q^2})^2}{(\sqrt[3]{q^2})^3} \\ \\ =\frac{\sqrt[3]{q^4}}{\sqrt[3]{q^6}}\\ \\ =\frac{q\sqrt[3]{q}}{q^2} \\ \\=\frac{1}{q} \sqrt[3]{q} [/tex]

Soal No. 4-c

[tex]= \frac{1}{m^{(- \frac{1}{3}) }} \\ \\= m^{\frac{1}{3} } \\ \\ = \sqrt[3]{m} [/tex]

***Semoga Terbantu***