\(\sqrt{2x+1}+\sqrt{x}=\sqrt{2x^2+4x-23}\)
1. Giải các phương trình sau:
a)\(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt[]{x+\sqrt{x^2-1}}=2\)
b)\(x^2-x-\sqrt{x^2-x+13}=7\)
c)\(x^2+2\sqrt{x^2-3x+1}=3x+4\)
d)\(2x^2+5\sqrt{x^2+3x+5}=23-6x\)
e)\(\sqrt{x^2+2x}=-2x^2-4x+3\)
f)\(\sqrt{\left(x+1\right)\left(x+2\right)}=x^2+3x+4\)
2. Giải các bất phương trình sau:
1)\(\sqrt{x^2-4x+5}\ge2x^2-8x\)
2)\(2x^2+4x+3\sqrt{3-2x-x^2}>1\)
3)\(\dfrac{\sqrt{-3x+16x-5}}{x-1}\le2\)
4)\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
5)\(\dfrac{9x^2-4}{\sqrt{5x^2-1}}\le3x+2\)
a) \(\sqrt{x+10}-\sqrt{3+x}=\sqrt{4x-23}\)
b) \(\sqrt{3x+4}-\sqrt{1+2x}=\sqrt{x+3}\)
a.
ĐKXĐ: \(x\ge\dfrac{23}{4}\)
\(\sqrt{x+10}=\sqrt{x+3}+\sqrt{4x-23}\)
\(\Leftrightarrow x+10=5x-20+2\sqrt{\left(x+3\right)\left(4x-23\right)}\)
\(\Leftrightarrow\sqrt{4x^2-11x-69}=15-2x\) \(\left(x\le\dfrac{15}{2}\right)\)
\(\Leftrightarrow4x^2-11x-69=\left(15-2x\right)^2\)
\(\Leftrightarrow49x-294=0\)
\(\Leftrightarrow x=6\) (thỏa mãn)
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(\sqrt{3x+4}=\sqrt{2x+1}+\sqrt{x+3}\)
\(\Leftrightarrow3x+4=3x+4+2\sqrt{\left(2x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)\left(x+3\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\left(loại\right)\end{matrix}\right.\)
Giải phương trình:(Nhớ tìm điều kiện)
a) \(\sqrt{2x-1}=\sqrt{5}\)
b)\(\sqrt{x-5}\) = 3
c)\(\sqrt{4x^2+4x+1}=6\)
d)\(\sqrt{\left(x-3\right)^2}=3-x\)
e)\(\sqrt{2x+5}=\sqrt{1-x}\)
f)\(\sqrt{x^2-x}=\sqrt{3-x}\)
g)\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
h)\(\sqrt{2x-5}=\sqrt{x-3}\)
i)\(\sqrt{x^2-x+6}=\sqrt{x^2+3}\)
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)
1) \(\sqrt{x^2}=2x-5\)
2) \(\sqrt{25x^2-10x+1}=2x-6\)
3) \(\sqrt{25-10x+x^2}=2x-5\)
4) \(\sqrt{1-2x+x^2}=2x-1\)
5) \(\sqrt{4x^2+4x+1}=-x-3\)
1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
2) ĐKXĐ: \(x\ge3\)
\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)
4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
Giải các phương trình
a) \(\sqrt{2x+9}=\sqrt{5-4x}\)
b) \(\sqrt{2x-1}=\sqrt{x-1}\)
c) \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)
d) \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)
\(\Leftrightarrow2x+9=5-4x\)
\(\Leftrightarrow6x=-4\)
hay \(x=-\dfrac{2}{3}\left(nhận\right)\)
b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)
\(\Leftrightarrow2x-1=x-1\)
hay x=0(loại)
c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2+3x=x\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
a. \(\sqrt{2x+9}=\sqrt{5-4x}\)
<=> 2x + 9 = 5 - 4x
<=> 2x + 4x = 5 - 9
<=> 6x = -4
<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)
d: Ta có: \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\Leftrightarrow2x^2-3=4x-3\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Giải các phương trình sau:
a) \(\sqrt{x+4\sqrt{x-4}}=2\)
b) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
c) \(\sqrt{2x^2-2x+1}=2x-1\)
Lời giải:
a. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
b. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
c.
PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)
1) \(\left|x^2-4x-5\right|=x-1\)
2) \(\sqrt{2x^2+2x+9}=x-3\)
3) \(\sqrt{x+1}+1=4x^2+\sqrt{3x}\)
4) \(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-3\)
1) Ta có: \(\left|x^2-4x-5\right|=x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x-1\left(\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\right)\\-x^2+4x+5=x-1\left(-1< x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5-x+1=0\\-x^2+4x+5-x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=0\\-x^2+3x+6=0\end{matrix}\right.\Leftrightarrow x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{41}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{41}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{41}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}+5}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}+5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{41}+5}{2}\right\}\)
giải phương trình
a) \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
b) \(2x-8\sqrt{2x-3}+9=0\)
c)\(\sqrt{x-2}+\sqrt{y+2000}+\sqrt{z-2001}=\frac{1}{2}\left(x+y+z\right)\)
d) \(x+y+z+23=4\sqrt{x-1}+6\sqrt{y-2}+8\sqrt{z-3}\)
e)\(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
b/ \(2x-8\sqrt{2x-3}+9=0\)
\(\Leftrightarrow\left(2x-3-2.4.\sqrt{2x-3}+16\right)-4=0\)
\(\Leftrightarrow\left(4-\sqrt{2x-3}\right)^2-4=\)
\(\Leftrightarrow\left(2-\sqrt{2x-3}\right)\left(6-\sqrt{2x-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2=\sqrt{2x-3}\\6=\sqrt{2x-3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{39}{2}\end{cases}}}\)
Giải phương trình:
1. \(5x^2+2x+10=7\sqrt{x^4+4}\)
2. \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
3. \(\sqrt{x^2+2x}=\sqrt{3x^2+4x+1}-\sqrt{3x^2+4x+1}\)