Penjelasan dengan langkah-langkah: Mungkin maksudnya [tex]\sqrt{0,7+\sqrt{0,48}}=\sqrt{p}+\sqrt{q}[/tex] Menggunakan bentuk [tex]\because\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q}\therefore\\\displaystyle~~~~\sqrt{0,7+\sqrt{0,48}}=\\~~~~\sqrt{\frac{7}{10}+\sqrt{\frac{48}{100}}}=\\~~~~\sqrt{\frac{7}{10}+\sqrt{\frac{12}{25}}}\rightarrow p+q=\frac{7}{10},\:\:4pq=\frac{12}{25}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~pq=\frac{12}{4(25)}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~p+q=\frac{7}{10},~~pq=\frac{3}{25}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~p+q=\frac{7}{10},~~q=\frac{3}{25p}\\\\~~~~~~~~~~~~~~~~~~~~~~p+\frac{3}{25p}=\frac{7}{10}[/tex] [tex]maka\\\\\displaystyle ~~~~~~~~~~~~~~~~~~~~~~p-\frac{7}{10}=-\frac{3}{25p}\\\\~~~~~~~~~~~~~~~~~~~p(p)-\frac{7p}{10}=-\frac{3p}{25p}\\\\~~~~~~~~~~~~~~~~~~~~~~p^2-\frac{7p}{10}=-\frac{3}{25}\\\\~~p^2-\frac{7p}{10}+\left(\frac{7}{10}\left(\frac{1}{2}\right)\right)^2=-\frac{3}{25}+\left(\frac{7}{10}\left(\frac{1}{2}\right)\right)^2\\\\~~~p^2-2\left(\frac{7p}{20}\right)+\left(\frac{7}{20}\right)^2=-\frac{3}{25}+\left(\frac{7}{20}\right)^2\\\\~~~~~~~~~~~\because x^2-2xy+y^2=(x-y)^2\therefore[/tex] [tex]\displaystyle maka\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\left(\frac{7}{20}\right)^2\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\frac{7^2}{20^2}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\frac{49}{400}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3(16)}{25(16)}+\frac{49}{400}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{48}{400}+\frac{49}{400}[/tex] [tex]\displaystyle maka\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=\frac{1}{400}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\sqrt{\frac{1}{400}}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\frac{\sqrt1}{\sqrt{400}}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\frac{1}{20}[/tex] [tex]\displaystyle untuk \:\: bentuk\:\:\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q},\\dan\:\:p\:\:lebih\:\:kecil\:\:dari\:\:q\:\:dengan\:\:p_1=p, \:\:p_2=q, \\maka\\\\p_1=\frac{7}{20}-\frac{1}{20},\:\:\:p_2=\frac{7}{20}+\frac{1}{20}\\\\p\:\:=\frac{7}{20}-\frac{1}{20},\:\:\:\:\:\:q=\frac{7}{20}+\frac{1}{20}\\\\p\:\:=\frac{6}{20},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:q=\frac{8}{20}\\\\p\:\:=\frac{3}{10},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:q=\frac{2}{5}[/tex]
Ditanya nilai dari [tex]\displaystyle\frac{1}{p}-\frac{3}{q}=\frac{1}{\frac{3}{10}}-\frac{3}{\frac{2}{5}}=\frac{10}{3}-\frac{5(3)}{2}=\frac{10(2)}{3(2)}-\frac{5(3^2)}{2(3)}\\\\=\frac{20}{6}-\frac{5(9)}{6}=\frac{20}{6}-\frac{45}{6}=-\left(\frac{45}{6}-\frac{20}{6}\right)=\\\\\bf-\frac{25}{6}[/tex]
Verified answer
PEMBAHASAN
Aljabar
√(0,7 + √0,48)
= √(0,7 + √(4 × 0,12))
= √(0,7 + 2√0,12)
= √((0,4 + 0,3) + 2√(0,4 × 0,3))
= √0,4 + √0,3
Koreksi soal
√(0,7 + √0,48) = √p + √q
√0,4 + √0,3 = √p + √q
p < q
p = 0,3 = 3/10
q = 0,4 = 4/10 = 2/5
1/p - 3/q
= 10/3 - 3 × 5/2
= 10/3 - 15/2
= (10 × 2 - 15 × 3)/(3 × 2)
= (20 - 45)/6
= -25/6
Jawab:
[tex]\displaystyle\bf-\frac{25}{6}[/tex]
Penjelasan dengan langkah-langkah:
Mungkin maksudnya
[tex]\sqrt{0,7+\sqrt{0,48}}=\sqrt{p}+\sqrt{q}[/tex]
Menggunakan bentuk
[tex]\because\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q}\therefore\\\displaystyle~~~~\sqrt{0,7+\sqrt{0,48}}=\\~~~~\sqrt{\frac{7}{10}+\sqrt{\frac{48}{100}}}=\\~~~~\sqrt{\frac{7}{10}+\sqrt{\frac{12}{25}}}\rightarrow p+q=\frac{7}{10},\:\:4pq=\frac{12}{25}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~pq=\frac{12}{4(25)}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~p+q=\frac{7}{10},~~pq=\frac{3}{25}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~p+q=\frac{7}{10},~~q=\frac{3}{25p}\\\\~~~~~~~~~~~~~~~~~~~~~~p+\frac{3}{25p}=\frac{7}{10}[/tex]
[tex]maka\\\\\displaystyle ~~~~~~~~~~~~~~~~~~~~~~p-\frac{7}{10}=-\frac{3}{25p}\\\\~~~~~~~~~~~~~~~~~~~p(p)-\frac{7p}{10}=-\frac{3p}{25p}\\\\~~~~~~~~~~~~~~~~~~~~~~p^2-\frac{7p}{10}=-\frac{3}{25}\\\\~~p^2-\frac{7p}{10}+\left(\frac{7}{10}\left(\frac{1}{2}\right)\right)^2=-\frac{3}{25}+\left(\frac{7}{10}\left(\frac{1}{2}\right)\right)^2\\\\~~~p^2-2\left(\frac{7p}{20}\right)+\left(\frac{7}{20}\right)^2=-\frac{3}{25}+\left(\frac{7}{20}\right)^2\\\\~~~~~~~~~~~\because x^2-2xy+y^2=(x-y)^2\therefore[/tex]
[tex]\displaystyle maka\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\left(\frac{7}{20}\right)^2\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\frac{7^2}{20^2}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3}{25}+\frac{49}{400}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{3(16)}{25(16)}+\frac{49}{400}\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=-\frac{48}{400}+\frac{49}{400}[/tex]
[tex]\displaystyle maka\\\\~~~~~~~~~~~~~~~~~~\left(p-\frac{7}{20}\right)^2=\frac{1}{400}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\sqrt{\frac{1}{400}}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\frac{\sqrt1}{\sqrt{400}}\\\\~~~~~~~~~~~~~~~~~~~~~~\left|p-\frac{7}{20}\right|=\frac{1}{20}[/tex]
[tex]\displaystyle untuk \:\: bentuk\:\:\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q},\\dan\:\:p\:\:lebih\:\:kecil\:\:dari\:\:q\:\:dengan\:\:p_1=p, \:\:p_2=q, \\maka\\\\p_1=\frac{7}{20}-\frac{1}{20},\:\:\:p_2=\frac{7}{20}+\frac{1}{20}\\\\p\:\:=\frac{7}{20}-\frac{1}{20},\:\:\:\:\:\:q=\frac{7}{20}+\frac{1}{20}\\\\p\:\:=\frac{6}{20},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:q=\frac{8}{20}\\\\p\:\:=\frac{3}{10},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:q=\frac{2}{5}[/tex]
Ditanya nilai dari
[tex]\displaystyle\frac{1}{p}-\frac{3}{q}=\frac{1}{\frac{3}{10}}-\frac{3}{\frac{2}{5}}=\frac{10}{3}-\frac{5(3)}{2}=\frac{10(2)}{3(2)}-\frac{5(3^2)}{2(3)}\\\\=\frac{20}{6}-\frac{5(9)}{6}=\frac{20}{6}-\frac{45}{6}=-\left(\frac{45}{6}-\frac{20}{6}\right)=\\\\\bf-\frac{25}{6}[/tex]
(xcvi)