Tolong bantu ya teman teman dari no. 3-5 thanks ya.... Question from @RayaAndreini05

Takamori37 Nomor 3.
Manipulasikan:
$$\begin{align}n&=5^x+5^{-x} \\ n^2&=(5^x+5^{-x})^2 \\ n^2&=5^{2x}+2.5^x.5^{-x}+5^{-2x} \\ n^2&=(5^{2x}+5^{-2x})+2 \\ n^2&=34+2 \\ n^2&=36 \\ n&=\sqrt{36} \\ n&=6 \\ 5^x+5^{-x}&=6\ \ [B]\end{align}$

Nomor 4.
Selesaikan dan sederhanakan.
$$\begin{align}\sqrt{\frac{(\frac2{10})^{3-2x}}{(\frac4{100})^{3x-1}}}&=0,008 \\ \sqrt{\frac{(\frac15)^{3-2x}}{(\frac1{25})^{3x-1}}}&=\frac8{1000} \\ \sqrt{\frac{(5^{-1})^{3-2x}}{(5^{-2})^{3x-1}}}&=\frac1{125} \\ \sqrt{\frac{5^{2x-3}}{5^{-6x+2}}}&=5^{-3} \\ \sqrt{5^{2x-3-(-6x+2)}}&=5^{-3} \\ \sqrt{5^{8x-5}}&=5^{-3} \\ 5^{8x-5}&=5^{2\times-3} \\ 8x-5&=-6 \\ 8x&=-1 \\ x&=-\frac18\ \ [D] \end{align}$

Nomor 5.
$$\begin{align}\left[\frac49\right]^{x^2-3}.\left[\frac8{27}\right]^{1-x}&=\frac32 \\ \left[\frac23\right]^{2(x^2-3)}.\left[\frac23\right]^{3(1-x)}&=\left[\frac23\right]^{-1} \\ \left[\frac23\right]^{2x^2-6+3-3x}&=\left[\frac23\right]^{-1} \\ 2x^2-3x-3&=-1 \\ 2x^2-3x-2&=0 \\ (2x+1)(x-2)&=0 \end{align}$
Sehingga:
x₁ = -1/2
x₂ = 2
Didapat:
$$\begin{align}(x_1-x_2)^2&=\left[-\frac12-2\right]^2 \\ &=\left[-\frac52\right]^2 \\ &=\frac{25}4\ \ [B]\end{align}$

2 votes Thanks 1

hendrisyafa 3. misal : $5^{x}=p$ maka $5^{2x}=p^{2}$

persamaan diubah :
$p^{2}+ \frac{1}{p^{2}}=34$

$(p+ \frac{1}{p})^{2}-2.p \frac{1}{p}=34$

$(p+ \frac{1}{p})^{2}-2=34$

$(p+ \frac{1}{p})^{2}=36$

$p+ \frac{1}{p}= \sqrt{36} =6$

4. $\sqrt{ \frac{ (\frac{2}{10})^{3-2x} }{ (\frac{4}{100})^{3x-1} } } =0,008$

$\sqrt{ \frac{ (\frac{1}{5})^{3-2x} }{ (\frac{1}{25})^{3x-1} } } = \frac{1}{125}$

$\sqrt{ \frac{ (5)^{2x-3} }{ (5)^{2-6x} } } = 5^{-3}$

$\sqrt{ (5)^{8x-5}} = 5^{-3}$

$5^{8x-5}= 5^{-6}$

$8x-5=-6$

$8x=-1$

$x=- \frac{1}{8}$

5. $\frac{2^{2 x^{2} -6}}{3^{2 x^{2} -6}}. \frac{2^{3-3x}}{3^{3-3x}}= \frac{3}{2}$

$\frac{2^{2 x^{2} -6+3-3x}}{3^{2 x^{2} -6+3-3x}}= \frac{3}{2}$

$\frac{2^{2 x^{2} -3x-3}}{3^{2 x^{2} -3x-3}}= \frac{3}{2}$

$\frac{2^{2 x^{2} -3x-3}.2}{3^{2 x^{2} -3x-3}.3}= 1$

$
\frac{2^{2 x^{2} -3x-2}}{3^{2 x^{2} -3x-2}}= \frac{2^{0}}{3^{0}}$

$2 x^{2} -3x-2=0$

(2x + 1)(x - 2) = 0

2x+1 = 0 dan x - 2 = 0
x = - 1/2 x = 2

(x₁ - x₂)² = (2 - (-1/2) )²
= (2 + 1/2 )²
= (5/2)²
= 25/4

1 votes Thanks 1

Bantuin ya, teman-teman. Aku mau mastiin jawabanku betul atau enggak.....

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