HOTS Zone : Aljabar

Berikut ini adalah kumpulan soal mengenai Aljabar. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Diberikan a dan b bilangan real positif yang memenuhi,

\dfrac1a-\dfrac1b=\dfrac1{a+b}

Nilai dari ekspresi \left(\dfrac{b}a+\dfrac{a}b\right)^2 adalah ....

1. 7
2. 5

3. 3
4. 1

Grup WhatsApp

$$\begin{array}{rl}{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}& ={\displaystyle \frac{1}{a+b}}\\ {\displaystyle \frac{a+b}{a}}-{\displaystyle \frac{a+b}{b}}& =1\\ 1+{\displaystyle \frac{b}{a}}-({\displaystyle \frac{a}{b}}+1)& =1\\ {\displaystyle \frac{b}{a}}-{\displaystyle \frac{a}{b}}& =1\\ {({\displaystyle \frac{b}{a}}-{\displaystyle \frac{a}{b}})}^2& =1^2\\ {\left({\displaystyle \frac{b}{a}}\right)}^2+{\left({\displaystyle \frac{a}{b}}\right)}^2-2\left({\displaystyle \frac{b}{a}}\right)\left({\displaystyle \frac{a}{b}}\right)& =1\\ {\left({\displaystyle \frac{b}{a}}\right)}^2+{\left({\displaystyle \frac{a}{b}}\right)}^2-2& =1\\ {\left({\displaystyle \frac{b}{a}}\right)}^2+{\left({\displaystyle \frac{a}{b}}\right)}^2& =3\end{array}$$

$$\begin{array}{rl}{({\displaystyle \frac{b}{a}}+{\displaystyle \frac{a}{b}})}^2& ={\left({\displaystyle \frac{b}{a}}\right)}^2+{\left({\displaystyle \frac{a}{b}}\right)}^2+2\left({\displaystyle \frac{b}{a}}\right)\left({\displaystyle \frac{a}{b}}\right)\\ & =3+2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$$

Jadi, \left(\dfrac{b}a+\dfrac{a}b\right)^2=5. JAWAB: B

No.

Benar atau salah?
Jika x adalah bilangan real dan {\dfrac1x\lt 8488}, maka {x\gt\dfrac1{8488}}.

Brilliant.org

Ambil x=-1
\dfrac1{-1}=-1\lt 8488 (benar)
-1\gt\dfrac1{8488} (salah)

Jadi, Pernyataan di atas adalah salah.

No.

Jika x+\dfrac1y=32, y+\dfrac1z=126, z+\dfrac1x=130, maka xyz+\dfrac1{xyz}= ....

$$\begin{array}{rl}xyz+{\displaystyle \frac{1}{xyz}}& =(x+{\displaystyle \frac{1}{y}})(y+{\displaystyle \frac{1}{z}})(z+{\displaystyle \frac{1}{x}})-(x+{\displaystyle \frac{1}{y}}+y+{\displaystyle \frac{1}{z}}+z+{\displaystyle \frac{1}{x}})\\ & =(32)(126)(130)-(32+126+130)\\ & =524160-288\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 523872}}}}\end{array}$$

Jadi, xyz+\dfrac1{xyz}=523872.

No.

Dalam sekelompok wanita dan pria, jika 12 orang pria keluar, maka setiap pria mendapat pasangan 9 orang wanita. Tetapi jika 22 orang wanita keluar, maka setiap wanita mendapat pasangan 3 orang pria. Berapa banyak pria dalam kelompok mula-mula?

Misal p adalah jumlah pria mula-mula, dan w adalah jumlah wanita mula-mula.
$$\begin{array}{rl}w& =9(p-12)\\ w& =9p-108\end{array}$$

$$\begin{array}{rl}p& =3(w-22)\\ p& =3(9p-108-22)\\ p& =3(9p-130)\\ p& =27p-390\\ 26p& =390\\ p& ={\displaystyle \frac{390}{26}}\\ p& =15\end{array}$$

Jadi, banyak pria dalam kelompok mula-mula adalah 15 orang.

No.

Suatu operasi * selalu memenuhi {(a*b)*b=a}, dan {a*(a*b)=b}. Jika diketahui {8*3=7}, maka berapakah 3*7?

1. 7
2. 8

3. 9
4. 10

$$\begin{array}{rl}8\ast (8\ast 3)& =3\\ 8\ast 7& =3\end{array}$$

$$\begin{array}{rl}(8\ast 7)\ast 7& =8\\ 3\ast 7& =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$$

Jadi, 3*7=8. JAWAB: B

No.

Jika \dfrac{a+b}c=\dfrac65 dan \dfrac{b+c}a=\dfrac92, maka nilai \dfrac{a+c}b= ....

$$\begin{array}{rl}{\displaystyle \frac{a+b}{c}}& ={\displaystyle \frac{6}{5}}\\ 5a+5b& =6c\\ -5a+6c& =5b\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{b+c}{a}}& ={\displaystyle \frac{9}{2}}\\ 2b+2c& =9a\\ 9a-2c& =2b\end{array}$$

$$\begin{array}{rl}-5a+6c& =5b\\ 9a-2c& =2b& +\\ 4a+4c& =7b\\ 4(a+c)& =7b\\ {\displaystyle \frac{a+c}{b}}& ={\displaystyle \frac{7}{4}}\end{array}$$

Jadi, \dfrac{a+c}b=\dfrac74.

No.

a dan b adalah bilangan real. Jika {\dfrac{a}b=a\times b=a+b}, maka {a-b=} ....

Brilliant.org

$$\begin{array}{rl}{\displaystyle \frac{a}{b}}& =ab\\ {\displaystyle \frac{1}{b}}& =b\\ b^2& =1\\ b& =\pm 1\end{array}$$

Jika b=1
$$\begin{array}{rl}ab& =a+b\\ a& =a+1\end{array}$$
tidak ada nilai a yang memenuhi

Jika b=-1
$$\begin{array}{rl}ab& =a+b\\ -a& =a-1\\ 2a& =1\\ a& ={\displaystyle \frac{1}{2}}\end{array}$$

$$\begin{array}{rl}a-b& ={\displaystyle \frac{1}{2}}-(-1)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{2}}}}}}\end{array}$$

Jadi, a-b=\dfrac32.

No.

Diketahui x^5y^4=512 dan xy=4. Maka nilai \dfrac{8x}{x-1} adalah ....

$$\begin{array}{rl}{x}^5{y}^4& =512\\ x{x}^4{y}^4& =512\\ x(xy{)}^4& =512\\ x(4{)}^4& =512\\ x(256)& =512\\ x& =2\end{array}$$

\dfrac{8x}{x-1}=\dfrac{8(2)}{2-1}=16

Jadi, \dfrac{8x}{x-1}=16.

No.

Diketahui {a^2 +b^2 = 1} dan {c^2 + d^2 = 1}. Nilai minimum dari {ac+bd- 2} adalah ...

Learncy.net

$$\begin{array}{rl}(a+c{)}^2+(b+d{)}^2& \ge 0\\ a^2+2ac+c^2+b^2+2bd+d^2& \ge 0\\ 2ac+2bd+a^2+b^2+c^2+d^2& \ge 0\\ 2ac+2bd+1+1& \ge 0\\ 2ac+2bd+2& \ge 0\\ 2ac+2bd& \ge -2\\ ac+bd& \ge -1\\ ac+bd-2& \ge -3\end{array}$$

Jadi, nilai minimum dari ac+bd- 2 adalah -3.

No.

Jika x merupakan bilangan riil sehingga {x\sqrt{x}=4\sqrt{x}+\sqrt3}, maka nilai {x-\sqrt{3x}=} ....

$$\begin{array}{rl}x\sqrt{x}& =4\sqrt{x}+\sqrt{3}\\ x\sqrt{x}-4\sqrt{x}& =\sqrt{3}\\ (x-4)\sqrt{x}& =\sqrt{3}\\ (x-4{)}^{2}x& =3\\ (x^2-8x+16)x-3& =0\\ x^3-8x^2+16x-3& =0\\ (x-3)(x^2-5x+1)& =0\end{array}$$

• x-3=0
  x=3
  Tidak memenuhi
• x^2-5x+1=0
  $$\begin{array}{rl}{x}^2-2x-3x+1& =0\\ x^2-2x+1& =3x\\ (x-1{)}^2& =3x\\ x-1& =\sqrt{3x}\\ x-\sqrt{3x}& =1\end{array}$$

Jadi, nilai x-\sqrt{3x}=1.