Giải pt:
a) \(2x^2-8x-3\sqrt{x^2-4x-5}=12\)
b) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
Giải PT:
a) \(\sqrt{11+6\sqrt{2}}\) = \(\sqrt{2x^2-6x\sqrt{2}+9}\)
b) \(\sqrt{4x^2+4x\sqrt{7}+7}\) - \(\sqrt{8-2\sqrt{7}}\) = 0
c) \(\sqrt{x^2}\) = x
d) \(\sqrt{x^2-2x+1}\) = x-1
c: Ta có: \(\sqrt{x^2}=x\)
\(\Leftrightarrow\left|x\right|=x\)
hay \(x\ge0\)
d: Ta có: \(\sqrt{x^2-2x+1}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
hay \(x\ge1\)
giải các pt
1, \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
2, \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
3, \(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
4, \(2x^2+\sqrt{x^2-4x+12}=4x+8\)
5, \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
Câu 1:
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm
- Nhận thấy \(x=-1\) là 1 nghiệm
- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:
\(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)
\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)
\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)
\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm \(x=\pm1\)
Câu 2:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=2\)
- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:
\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)
- Nếu \(1\le x< 2\) pt trở thành:
\(\sqrt{x-1}+1-1+\sqrt{x-1}=2\Leftrightarrow x=2\left(l\right)\)
Vậy nghiệm của pt là \(x\ge2\)
Câu 3:
Bình phương 2 vế ta được:
\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)
\(\Leftrightarrow\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)
Đặt \(x^2+x+1=a>0\) pt trở thành:
\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Câu 5:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Mà \(VT=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)
Vậy nghiệm của pt là \(5\le x\le10\)
Giải PT:
a) -5x+7\(\sqrt{x}\) +12=0
b) \(\dfrac{1}{3}\)\(\sqrt{4x^2-20}\) +2\(\sqrt{\dfrac{x^2-5}{9}}\) -3\(\sqrt{x^2-5}=0\)
c) \(\sqrt{9x+27}+5\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=5\)
d) \(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=3\sqrt{x-2}+8\)
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 2\sqrt{x-2}=8$
$\Leftrightarrow \sqrt{x-2}=4$
$\Leftrightarrow x=4^2+2=18$ (tm)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
1) giải pt:
a) \(\sqrt{3x+10}=4\)
b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\)
c) \(\sqrt{2x+1}=3\)
d) \(\sqrt{2x+1}+1=x\)
giúp mk vs ah
a) \(\sqrt{3x+10}=4\left(đk:x\ge-\dfrac{10}{3}\right)\Leftrightarrow3x+10=16\Leftrightarrow x=2\)
b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\Leftrightarrow\sqrt{\left(3x-1\right)^2}=\sqrt{\left(x+4\right)^2}\Leftrightarrow3x-1=x+4\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)
c) \(\sqrt{2x+1}=3\left(đk:x\ge-\dfrac{1}{2}\right)\Leftrightarrow2x+1=9\Leftrightarrow x=4\)
d) \(\sqrt{2x+1}+1=x\left(đk:x\ge1\right)\Leftrightarrow\sqrt{2x+1}=x-1\Leftrightarrow2x+1=x^2-2x+1\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)\(\Leftrightarrow x=4\)(do \(x\ge1\))
a: Ta có: \(\sqrt{3x+10}=4\)
\(\Leftrightarrow3x+10=16\)
\(\Leftrightarrow3x=6\)
hay x=2
b: Ta có: \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\)
\(\Leftrightarrow\left|3x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+4\\3x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\4x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: Ta có: \(\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow x=4\)
giải pt:
a, \(2x^2-11x+21=3\sqrt[3]{4x-4}\)
b, \(\sqrt{x-3}+\sqrt[3]{x^2+1}+x^2+x-2=0\)
Giải phương trình:
a. \(3\sqrt{8x}-\sqrt{32x}+\sqrt{50x}=21\)
b. \(\sqrt{25x+50}+3\sqrt{4x+8}-2\sqrt{16x+32}=15\)
c. \(\sqrt{\left(x-2\right)^2}=12\)
d. \(\sqrt{x^2-6x+9}-3=5\)
e.\(\sqrt{\left(2x-1\right)^2}-x=3\)
f. \(\sqrt{3x-6}-x=-2\)
h. \(\sqrt{3-2x}-2=x\)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
f.
ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$
$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$
$\Leftrightarrow x=2$ hoặc $x=5$ (tm)
h. ĐKXĐ: $x\leq \frac{3}{2}$
PT $\Leftrightarrow \sqrt{3-2x}=x+2$
\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)
Vậy.......
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
giải pt:
a,\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)
b, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
c,\(\sqrt{2x^2+4x+1}=1-2x-x^2\)
a) ĐKXĐ: \(x^2+3x\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\le-3\end{matrix}\right.\).
PT \(\Leftrightarrow10-\left(x^2+3x\right)=3\sqrt{x^2+3x}\). (*)
Đặt \(\sqrt{x^2+3x}=a\ge0\).
\((*)\Leftrightarrow a^2+3a-10=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+5\right)=0\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\).
Với \(a=2\Rightarrow\sqrt{x^2+3x}=2\Leftrightarrow x^2+3x-4=0\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-4\left(TM\right)\end{matrix}\right.\).
Vậy x = 1; x = -4