Tìm x biết :a) \(\sqrt{x}\) = \(x\)
b) \(x-2\sqrt{x}=0\)
c) \(\sqrt{x+1}=1-x\)
d) \(\sqrt{3x^2+4}+\sqrt{2004x^2+1}=3-4x^2\)
e) \(\sqrt{3x^2+4}+\sqrt{2007x^2+25}=7-69x^2\)
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)\(\dfrac{9x^2-4}{\sqrt{5x^2-1}}\le3x+2\)
Câu 2: Tìm x biết:
a. \(\sqrt{x-1}=2\)
b. \(\sqrt{3x+1}=\sqrt{4x-3}\)
c. \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
d. \(\sqrt{x^2-4x+4}=\sqrt{6+2\sqrt{5}}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
Giải phương trình
a, \(x+1+2\sqrt{7-x}-2\sqrt{x+1}=\sqrt{7+6x-x^2}\)
b, \(4x^2+3x+3=4\sqrt{x^3+3x^2}+2\sqrt{2x-1}\)
c, \(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
d, \(3x^2+4x+10=2\sqrt{14x^2-7}\)
Giải phương trình
a, \(x+1+2\sqrt{7-x}-2\sqrt{x+1}=\sqrt{7+6x-x^2}\)
b, \(4x^2+3x+3=4\sqrt{x^3+3x^2}+2\sqrt{2x-1}\)
c, \(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
d, \(3x^2+4x+10=2\sqrt{14x^2-7}\)
a,đk -1<x<7
x+1+2 căn 7-x-2 căn x+1=căn (x+1)(7-x)
giải các pt sau:
a, \(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
b, \(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
c, \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
d, \(2\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{2x-1}\)
e, \(\sqrt{2x-1}+x^2-3x+1=0\)
f, \(\sqrt{4x^2-7x-2}=2\sqrt{x^2-1+1}-1\)
g, \(\sqrt{x^2+2x-15}+\sqrt{x^2-8x+15}=\sqrt{4x^2-18x+18}\)
h,\(\sqrt{2-x}+\sqrt{3-x}=2\sqrt{4-x}\)
a) \(\sqrt{4x^2-4x+1}=3\)
b)\(\sqrt{x^2-10x+25}+2-x\)
c)\(\sqrt{x^2-6x+9}+x=11\)
d)\(\sqrt{3x+19}=x+3\)
e)\(\sqrt{x^2+x+5}-1=x\)
a: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>2x=-2 hoặc 2x=4
=>x=2 hoặc x=-1
c: \(\Leftrightarrow\left|x-3\right|=11-x\)
=>x<=11 và (x-3)^2=(11-x)^2
=>x<=11 và x^2-6x+9=x^2-22x+121
=>x<=11 và 16x=112
=>x=7
d:
ĐKXĐ: 3x+19>=0
=>x>=-19/3
PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9
=>x>=-3 và x^2+6x+9-3x-19=0
=>x>=-3 và (x+5)(x-2)=0
=>x=2
e: =>\(\sqrt{x^2+x+5}=x+1\)
=>x>=-1 và x^2+x+5=x^2+2x+1
=>x>=-1 và 2x+1=x+5
=>x=4
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
bài 1 : giải phương trình:
a. \(\sqrt{x+2\sqrt{ }x-1}=2\)
b. \(\sqrt{x^2-4x+4}=\sqrt{4x^212x+9}\)
c.\(\sqrt{x+4\sqrt{ }x-4}=2\)
d. \(\sqrt{x^2-6x+9}=2\)
e. \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
f. \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}
\(a,x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(b,\sqrt{x^2+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(c,\sqrt{x+3+4\sqrt{x-1}}+\sqrt{8-6\sqrt{x-1}+x}=1\)
\(d,\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)
\(e,\sqrt{4x-9}+2\sqrt{3x^2-5x+2}=\sqrt{3x-2}+\sqrt{x-1}\)