Vương Ngọc Như Quỳnh
01/09/2017 at 17:16-
VTK-VangTrangKhuyet 01/09/2017 at 17:33Condition \(x\ge4\)
a) When m = 1 we have equation :
\(\sqrt{x-3-2\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=1\)
\(\Leftrightarrow\sqrt{\left(x-4\right)-2\sqrt{x-4}+1}+\sqrt{4-4\sqrt{x-4}+x-4}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}-1\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-4}-1\right|+\left|2-\sqrt{x-4}\right|=1\)
Using inequality \(\left|a+b\right|\le\left|a\right|+\left|b\right|\) we have :
L.S \(\ge\left|\sqrt{x-4}-1+2-\sqrt{x-4}=1\right|\)
So the given equation is the same with
\(\left(\sqrt{x-4}-1\right)\left(2-\sqrt{x-4}\right)\ge0\)
\(\Leftrightarrow1\le\sqrt{x-4}\le2\Leftrightarrow5\le x\le8\)
b) The equation has given is the same with \(\left|\sqrt{x-4}-1\right|+\left|2-\sqrt{x-4}\right|=m\)
Using checklist (Trans : Bang Xet Dau) : You make the list by your own.
\(+\) If \(0\le\sqrt{x-4}\le1\Leftrightarrow4\le x\le5\)
The equation becomes : \(3-2\sqrt{x-4}=m\Leftrightarrow\sqrt{x-4}=\dfrac{3-m}{2}\)
The equation has solution \(\Leftrightarrow0\le\dfrac{3-m}{2}\le1\Leftrightarrow1\le m\le3\)
+ If \(1< \sqrt{x-4}< 2\Leftrightarrow5< x< 8\)
The equation becomes \(1=m\)
All cases \(m\ne1\) the equation has no solution.
+ If \(\sqrt{x-4}\ge2\Leftrightarrow x\ge8\)
The equation becomes \(2\sqrt{x-4}-3=m\Leftrightarrow\sqrt{x-4}=\dfrac{m+3}{2}\)
The equation has solution \(\Leftrightarrow\dfrac{m+3}{2}\ge2\Leftrightarrow m\ge1\)
Answer :
- If \(m\ge1\) the equation has solution.
- If \(m< 1\) the equation has no solution.
(If you don't understand anything in my answer please inbox me :))
Selected by MathYouLike -
Faded 26/01/2018 at 12:43Condition x≥4
a) When m = 1 we have equation :
√x−3−2√x−4+√x−4√x−4=1
⇔√(x−4)−2√x−4+1+√4−4√x−4+x−4=1
⇔√(√x−4−1)2+√(2−√x−4)2=1
⇔∣∣√x−4−1∣∣+∣∣2−√x−4∣∣=1
Using inequality |a+b|≤|a|+|b|
we have :
L.S ≥∣∣√x−4−1+2−√x−4=1∣∣
So the given equation is the same with
(√x−4−1)(2−√x−4)≥0
⇔1≤√x−4≤2⇔5≤x≤8
b) The equation has given is the same with ∣∣√x−4−1∣∣+∣∣2−√x−4∣∣=m
Using checklist (Trans : Bang Xet Dau) : You make the list by your own.
+
If 0≤√x−4≤1⇔4≤x≤5
The equation becomes : 3−2√x−4=m⇔√x−4=3−m2
The equation has solution ⇔0≤3−m2≤1⇔1≤m≤3
+ If 1<√x−4<2⇔5<x<8
The equation becomes 1=m
All cases m≠1
the equation has no solution.
+ If √x−4≥2⇔x≥8
The equation becomes 2√x−4−3=m⇔√x−4=m+32
The equation has solution ⇔m+32≥2⇔m≥1
Answer :
- If m≥1
the equation has solution.
- If m<1
the equation has no solution.
(If you don't understand anything in my answer please inbox me :))
-
hochiminh 07/09/2017 at 19:48x ≥ 4x≥4
a) Khi m = 1 ta có phương trình:
√ x - 3 - 2 √ x - 4 + √ x - 4 √ x - 4 =1x−3−2x−4+x−4x−4=1
⇔ √ ( x - 4 ) - 2 √ x - 4 + 1 + √ 4 - 4 √ x - 4 + x - 4 = 1⇔(x−4)−2x−4+1+4−4x−4+x−4=1
⇔ √ ( √ x - 4 - 1 ) 2 + √ ( 2 - √ x - 4 ) 2 = 1⇔(x−4−1)2+(2−x−4)2=1
⇔ | | √ x - 4 - 1 | | + | | 2 - √ x - 4 | | = 1⇔|x−4−1|+|2−x−4|=1
| a + b | ≤ | a | + | b ||a+b|≤|a|+|b|
≥ | | √ x - 4 - 1 + 2 - √ x - 4 = 1 | |≥|x−4−1+2−x−4=1|
Vì vậy, phương trình đã cho là như nhau với
( √ x - 4 - 1 ) ( 2 - √ x - 4 ) ≥ 0(x−4−1)(2−x−4)≥0
⇔ 1 ≤ √ x - 4 ≤ 2 ⇔ 5 ≤ x ≤ 8⇔1≤x−4≤2⇔5≤x≤8
| | √ x - 4 - 1 | | + | | 2 - √ x - 4 | | =m|x−4−1|+|2−x−4|=m
Sử dụng danh sách kiểm tra (Trans: Bang Xet Dau): Bạn tự lập danh sách.
++0 ≤ √ x - 4 ≤ 1 ⇔ 4 ≤ x ≤ 50≤x−4≤1⇔4≤x≤5
3 - 2 √ x - 4 = m ⇔ √ x - 4 = 3 - m 23−2x−4=m⇔x−4=3−m2
⇔ 0 ≤ 3 - m 2≤ 1 ⇔ 1 ≤ m ≤ 3⇔0≤3−m2≤1⇔1≤m≤3
1 < √ x - 4 < 2 ⇔ 5 < x < 81<x−4<2⇔5<x<8
1 = m1=m
m ≠ 1m≠1
√ x - 4 ≥2⇔x≥8x−4≥2⇔x≥8
2 √ x - 4 - 3 = m ⇔ √ x - 4 = m + 3 22x−4−3=m⇔x−4=m+32
⇔ m + 3 2≥ 2 ⇔ m ≥ 1⇔m+32≥2⇔m≥1
Câu trả lời :
m ≥ 1m≥1
m < 1
-
Saiki_Kusuo 1 01/09/2017 at 20:25yeah it probaly hard
-
Vương Ngọc Như Quỳnh 01/09/2017 at 17:17
I have some more questions please help me today and tomorrow !!