\(\sqrt{2x-3}\) + 13 = 16
Giai cac pt:
a, \(2x^2-8x+\sqrt{x^2-4x-5}=13\)
b, \(\sqrt{1-x}+\sqrt{4+x}=3\)
c, \(x^3+4x+5=2\sqrt{2x+3}\)
d, \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2-16}\)
e, \(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-13}}=5\)
đề sai ko vậy bạn
nếu đề đúng thì mình nghỉ là
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-13+8\sqrt{2x-13}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-13}+4\right)^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-13}+4=5\)
\(\Leftrightarrow\sqrt{2x-3}+\sqrt{2x-13}=0\left(vl\right)\)
suy ra pt vô nghiệm
theo tôi là vậy
a)\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)
b)\(x^2+4x+5=2\sqrt{2x+3}\)
b/ \(\Rightarrow2x+3+2\sqrt{2x+3}-x^2-6x-8=0\)
Đặt \(a=\sqrt{2x+3}\left(a\ge0\right)\)
\(\left(1\right)\Rightarrow a^2+2a-x^2-6x-8=0\)
Có: \(\Delta=1+x^2+6x+8=x^2+6x+9=\left(x+3\right)^2\)
\(\Rightarrow\sqrt{\Delta}=x+3\)
\(\Rightarrow a=\frac{-1+x+3}{1}=x+2\)
hoặc \(a=\frac{-1-x-3}{1}=-x-4\)
+) Với a = x + 2 \(\Leftrightarrow\sqrt{2x+3}=x+2\left(x\ge-2\right)\)
......... tự giải ra x
+) Với a = -x - 4 \(\Leftrightarrow\sqrt{2x+3}=-x-4\left(x\le-4\right)\)
.........tự giải ra x
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=7\)
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 \(\sqrt{2x-3}\ge0\) \(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
\(\Leftrightarrow2\sqrt{2x-3}=2\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow2x-3=1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)
a)\(\sqrt{x-5}-\frac{x-14}{3+\sqrt{x-5}}=3\)
b)\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)
c) \(3x^2+2x=2\sqrt{x^2+x}+1-x\)
a) Đk: x \(\ge\) 5
\(\sqrt{x-5}-\frac{x-14}{3x+\sqrt{x-5}}=3\)
\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\frac{x-14}{3\sqrt{x-3}}\left(3+\sqrt{x-5}\right)=3\left(3+\sqrt{x-5}\right)\)
\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\left(x-14\right)=3\left(3+\sqrt{x-5}\right)\)
\(3\sqrt{x-5}+9-\left(3\sqrt{x-5}+9\right)=9+3\sqrt{x-5}-\left(3\sqrt{x-5}+9\right)\)
=> Luôn đúng với x \(\ge\) 5
chúc bạn học tốt
Giải phương trình:
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
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 \(x=\dfrac{31-8\sqrt{6}}{2}\)
Giải phương trình:
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
`\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}`
Giải phương trình
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13-8\sqrt{2x-3}}=5\)
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13-8\sqrt{2x-3}}=5\\ \Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3-8\sqrt{2x-3}+16}=5\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}-4\right)^2}=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}-4\right|=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|=5\)
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.
1) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
2) \(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
1: \(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
=>căn x-3=0
=>x-3=0
=>x=3
2: =>\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot4+16}=5\)
=>\(\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
=>2*căn 2x-3+5=5
=>2x-3=0
=>x=3/2
Giải phương trình:
1/ \(13\sqrt{x-1}+9\sqrt{x+1}=16.x\)
2/ \(x^4+\frac{8}{x^2}+y^2=4+2x^2\)
3/ \(\sqrt{9-x^2}+\sqrt{3+x}+\sqrt{3-x}\ge3+2\sqrt{3}\)