K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DT
2
7 tháng 11 2018
đề sai ko vậy bạn
nếu đề đúng thì mình nghỉ là
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
7 tháng 11 2018
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-13+8\sqrt{2x-13}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-13}+4\right)^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-13}+4=5\)
\(\Leftrightarrow\sqrt{2x-3}+\sqrt{2x-13}=0\left(vl\right)\)
suy ra pt vô nghiệm
theo tôi là vậy
PH
1
25 tháng 11 2018
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}\right)^2+2\sqrt{2x-3}\cdot1+1^2}+\sqrt{\left(\sqrt{2x-3}\right)+2\sqrt{2x-3}\cdot4+4^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)
\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
PH
1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 11 2018
ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2.4.\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)
\(\Leftrightarrow2\sqrt{2x-3}=0\)
\(\Leftrightarrow2x-3=0\Rightarrow x=\dfrac{3}{2}\)
NT
0
TN
8 tháng 9 2017
a)\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|1-x\right|+\left|x-2\right|=3\)
Có: \(VT=\left|1-x\right|+\left|x-2\right|\)
\(\ge\left|1-x+x-2\right|=3=VP\)
Khi \(x=0;x=3\)
b)\(\sqrt{x^2-10x+25}=3-19x\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=3-19x\)
\(\Leftrightarrow\left|x-5\right|=3-19x\)
\(\Leftrightarrow x^2-10x+25=361x^2-114x+9\)
\(\Leftrightarrow-360x^2+104x+16=0\)
\(\Leftrightarrow-5\left(5x-2\right)\left(9x+1\right)=0\)
\(\Rightarrow x=\frac{2}{5};x=-\frac{1}{9}\)
c)\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
\(\Leftrightarrow2\sqrt{2x-3}+5=5\)\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
NT
1
CV
11 tháng 8 2018
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(|\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)
roi xet cac truong hop cua gia tri tuyet doi roi giai
NN
1
14 tháng 9 2018
ĐKXĐ: \(x\ge\frac{3}{2}\)
PT (=) \(\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)
(=) \(\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
(=) \(2\sqrt{2x-3}=2\) (=) \(\sqrt{2x-3}=1\)(=) 2x = 4 (=) x = 2 ( Thỏa mãn điều kiện )
Vậy x=2
PG
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 7 2020
ĐKXĐ: \(x\ge\frac{3}{2}\)
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)
\(\Leftrightarrow2\sqrt{2x-3}=0\)
\(\Rightarrow2x-3=0\Rightarrow x=\frac{3}{2}\)
9 tháng 9 2017
\(\sqrt{x^2-2x+1}\) + \(\sqrt{x^2-4x+4}\) = 3
<=> \(\sqrt{\left(x-1\right)^2}\)+ \(\sqrt{\left(x-2\right)^2}\)= 3
<=> \(\left|x-1\right|\)+\(\left|x-2\right|\)=3
<=> x - 1 + x - 2 = 3
<=> 2x - 3 = 3
<=> x = \(\dfrac{6}{2}\)= 3
b ,
\(\sqrt{x^2-10x+25}=3-19x\)
<=>\(\sqrt{\left(x-5\right)^2}=3-19x\)
<=> \(\left|x-5\right|=3-19x\)
<=> \(x-5=3-19x\)
\(\Leftrightarrow x+19x=3+5\)
\(\Leftrightarrow20x=8\Leftrightarrow x=\dfrac{8}{20}=\dfrac{2}{5}\)
ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=7\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=7\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
\(\Leftrightarrow2\sqrt{2x-3}=2\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow x=2\)
ĐK: \(x\ge\dfrac{3}{2}\)
Ta có: \(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=7\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=7\)
Vì \(\sqrt{2x-3}\ge0\) \(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
\(\Leftrightarrow2\sqrt{2x-3}=2\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow2x-3=1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)