Giải pt
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
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PH
1
25 tháng 11 2018
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}\right)^2+2\sqrt{2x-3}\cdot1+1^2}+\sqrt{\left(\sqrt{2x-3}\right)+2\sqrt{2x-3}\cdot4+4^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)
\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
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)\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
\(\Leftrightarrow2\sqrt{2x-3}+5=5\)\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
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 ,
\(\sqrt{x^2-10x+25}=3-19x\)
<=>\(\sqrt{\left(x-5\right)^2}=3-19x\)
<=> \(\left|x-5\right|=3-19x\)
<=> \(x-5=3-19x\)
\(\Leftrightarrow x+19x=3+5\)
\(\Leftrightarrow20x=8\Leftrightarrow x=\dfrac{8}{20}=\dfrac{2}{5}\)
AV
2
15 tháng 7 2021
Ta có: \(\sqrt{2x-2+2\sqrt{2x-3}+\sqrt{2x+13+8\sqrt{2x-3}}}=5\)
\(\Leftrightarrow\sqrt{2x-2+2\sqrt{2x-3}+2\sqrt{2x-3}+4}=5\)
\(\Leftrightarrow\sqrt{2x+2+4\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot2+4+1}=5\)
\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2+1=25\)
\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2=24\)
\(\Leftrightarrow\sqrt{2x-3}+2=2\sqrt{6}\)
\(\Leftrightarrow2x-3=\left(2\sqrt{6}-2\right)^2\)
\(\Leftrightarrow2x-3=28-8\sqrt{6}\)
\(\Leftrightarrow2x=31-8\sqrt{6}\)
hay \(x=\dfrac{31-8\sqrt{6}}{2}\)
AV
1
16 tháng 7 2021
`\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8sqrt{2x-3}}=5(x>=3/2)`
`<=>\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5`
`<=>\sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5`
`<=>\sqrt{2x-3}+1+\sqrt{2x-3}+4=5`
`<=>2\sqrt{2x-3}=0`
`<=>\sqrt{2x-3}=0<=>2x-3=0<=>x=3/2(tmdk)`
Vậy `S={3/2}`
DC
1
4 tháng 9 2021
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13-8\sqrt{2x-3}}=5\\ \Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3-8\sqrt{2x-3}+16}=5\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}-4\right)^2}=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}-4\right|=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|=5\)
Có \(\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|\ge\left|\sqrt{2x-3}+1+4-\sqrt{2x-3}\right|=\left|5\right|=5\)
Dấu "=" xảy ra ⇔ Đẳng thức ban đầu xảy ra \(\Leftrightarrow\left(\sqrt{2x-3}+1\right)\left(4-\sqrt{2x-3}\right)=0\\ \Leftrightarrow4\sqrt{2x-3}-2x+3+4-\sqrt{2x-3}=0\\ \Leftrightarrow3\sqrt{2x-3}=2x-7\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{2x-7}{3}\left(ĐK:x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow2x-3=\dfrac{\left(2x-7\right)^2}{9}\\ \Leftrightarrow\left(2x-7\right)^2=9\left(2x-3\right)\\ \Leftrightarrow4x^2-28x+49-18x+27=0\\ \Leftrightarrow4x^2-40x+76=0\\ \Leftrightarrow x^2-10x+19=0\\ \Leftrightarrow\left(x^2-10x+25\right)-6=0\\ \Leftrightarrow\left(x-5\right)^2-\left(\sqrt{6}\right)^2=0\\ \Leftrightarrow\left(x-5-\sqrt{6}\right)\left(x-5+\sqrt{6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{6}\left(tmđk\right)\\x=5-\sqrt{6}\left(ktmđk\right)\end{matrix}\right.\)
Vậy \(x=5+\sqrt{6}\) là nghiệm của pt.
24 tháng 8 2021
\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
Lời giải:
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)
b.
ĐKXĐ: $x\geq \frac{3}{2}$
PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)
\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)
\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)
\(\Leftrightarrow \sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
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
ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2.4.\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)
\(\Leftrightarrow2\sqrt{2x-3}=0\)
\(\Leftrightarrow2x-3=0\Rightarrow x=\dfrac{3}{2}\)