SNBT Zone : Pertidaksamaan Bentuk Aljabar
Table of Contents
Tipe:
No. 1
Nilai p yang memenuhi pertidaksamaan \(\dfrac{x^2+px-2}{x^2-x+1}\gt-3\) adalah ....- −7 > p > 1
- −1 > p > 7
- 1 > p > 7
- p < −7 atau p > 1
- p < −1 atau p > 7
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{x^2+px-2}{x^2-x+1}&\gt-3\\[4pt]
\dfrac{x^2+px-2}{x^2-x+1}+3&\gt0\\[4pt]
\dfrac{x^2+px-2+3(x^2-x+1)}{x^2-x+1}&\gt0\\[4pt]
\dfrac{x^2+px-2+3x^2-3x+3}{x^2-x+1}&\gt0\\[4pt]
\dfrac{4x^2+(p-3)x+1}{x^2-x+1}&\gt0
\end{aligned}\)
x2 − x + 1 selalu positif sehingga4x2 + (p − 3)x + 1 juga harus selalu positif atau diskriminannya lebih dari 0.
\(\begin{aligned} D&\gt0\\ b^2-4ac&\gt0\\ (p-3)^2-4(4)(1)&\gt0\\ p^2-6p+9-16&\gt0\\ p^2-6p-7&\gt0\\ (p+1)(p-7)&\gt0 \end{aligned}\)
p < −1 atau p > 7
x2 − x + 1 selalu positif sehingga
\(\begin{aligned} D&\gt0\\ b^2-4ac&\gt0\\ (p-3)^2-4(4)(1)&\gt0\\ p^2-6p+9-16&\gt0\\ p^2-6p-7&\gt0\\ (p+1)(p-7)&\gt0 \end{aligned}\)
p < −1 atau p > 7
Jadi, p < −1 atau p > 7.
JAWAB: E
JAWAB: E
No. 2
Jika himpunan penyelesaian pertidaksamaan \(\dfrac3{1+x}\lt2\) dan \(\dfrac3{1-x}\gt2\) adalah- 0
- 1
- 2
- 3
- 4
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac3{1+x}&\lt2\\[4pt]
\dfrac3{x+1}-2&\lt0\\[4pt]
\dfrac{3-2(x+1)}{x+1}&\lt0\\[4pt]
\dfrac{3-2x-2}{x+1}&\lt0\\[4pt]
\dfrac{-2x+1}{x+1}&\lt0\\[4pt]
\dfrac{2x-1}{x+1}&\gt0
\end{aligned}\)
x\lt\dfrac-1 , dan x\gt\dfrac12
\(\begin{aligned} \dfrac3{1-x}&\gt2\\[4pt] \dfrac3{1-x}-2&\gt0\\[4pt] \dfrac{3-2(1-x)}{1-x}&\gt0\\[4pt] \dfrac{3-2+2x}{1-x}&\gt0\\[4pt] \dfrac{2x+1}{1-x}&\gt0\\[4pt] \dfrac{2x+1}{x-1}&\lt0 \end{aligned}\)
Pembuat nol:
\(-\dfrac12\lt x\lt1\)
\(\dfrac12\lt x\lt 1\)
\(p=\dfrac12\)
q = 1
\(\begin{aligned} 4p-q&=4\left(\dfrac12\right)-1\\ &=2-1\\ &=\boxed{\boxed{1}} \end{aligned}\)
\(\begin{aligned} \dfrac3{1-x}&\gt2\\[4pt] \dfrac3{1-x}-2&\gt0\\[4pt] \dfrac{3-2(1-x)}{1-x}&\gt0\\[4pt] \dfrac{3-2+2x}{1-x}&\gt0\\[4pt] \dfrac{2x+1}{1-x}&\gt0\\[4pt] \dfrac{2x+1}{x-1}&\lt0 \end{aligned}\)
Pembuat nol:
\(-\dfrac12\lt x\lt1\)
\(\dfrac12\lt x\lt 1\)
\(p=\dfrac12\)
q = 1
\(\begin{aligned} 4p-q&=4\left(\dfrac12\right)-1\\ &=2-1\\ &=\boxed{\boxed{1}} \end{aligned}\)
Jadi, 4p − q = 1 .
JAWAB: B
JAWAB: B
No. 3
Himpunan semua bilangan real- −1
- −2
- −3
- 1
- 0
ALTERNATIF PENYELESAIAN
|x| > 1
x < −1 atau x > 1
\(\begin{aligned} \dfrac{x^2-2x-6}{-x+2}&\gt x\\[4pt] \dfrac{x^2-2x-6}{-x+2}-x&\gt0\\[4pt] \dfrac{x^2-2x-6-x(-x+2)}{-x+2}&\gt0\\[4pt] \dfrac{x^2-2x-6+x^2-2x}{-x+2}&\gt0\\[4pt] \dfrac{2x^2-4x-6}{-x+2}&\gt0\\[4pt] \dfrac{x^2-2x-3}{-x+2}&\gt0\\[4pt] \dfrac{(x+1)(x-3)}{x-2}&\lt0 \end{aligned}\)
x < −1 atau 2 < x < 3
a = −1, b = 2, c = 3
a − b + c = −1 − 2 + 3 = 0
x < −1 atau x > 1
\(\begin{aligned} \dfrac{x^2-2x-6}{-x+2}&\gt x\\[4pt] \dfrac{x^2-2x-6}{-x+2}-x&\gt0\\[4pt] \dfrac{x^2-2x-6-x(-x+2)}{-x+2}&\gt0\\[4pt] \dfrac{x^2-2x-6+x^2-2x}{-x+2}&\gt0\\[4pt] \dfrac{2x^2-4x-6}{-x+2}&\gt0\\[4pt] \dfrac{x^2-2x-3}{-x+2}&\gt0\\[4pt] \dfrac{(x+1)(x-3)}{x-2}&\lt0 \end{aligned}\)
x < −1 atau 2 < x < 3
a = −1, b = 2, c = 3
a − b + c = −1 − 2 + 3 = 0
Jadi, a − b + c = 0.
JAWAB: E
JAWAB: E
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