Bài 1: Căn bậc hai
Các câu hỏi tương tự
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, \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\sqrt{\dfrac{x+2}{4}}+\sqrt{25x+50}-2\sqrt{x+2}=14\) ; \(\sqrt{2x+3}=x\) ; \(\sqrt{25x^2+20x+4}=1\) ; \(\sqrt{\dfrac{x+1}{2x-1}}=2\) ; \(\dfrac{\sqrt{x-2}}{\sqrt{3x+1}}=6\)
Tìm x
Giải các pt sau:
1, \(\sqrt{x^2+x+1}=2x+\sqrt{x^2-x+1}\)
2, \(2x^2+2x+6=2x\sqrt{x^2-x+1}+4\sqrt{3x+1}\)
3, \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)
4, \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2-2x+3}+\sqrt{x^2-x+2}\)
5, \(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
tìm x:
\(\sqrt{x^2+x+1}=1\)
\(\sqrt{x^2+1}=-3\)
\(\sqrt{x^2-10x+25}=7-2x\)
\(\sqrt{2x+5}=5\)
\(\sqrt{x^2-4x+4}-2x+5=0\)
câu1 ; gp trình
a, \(\sqrt{x-x}+x=3\)
b, \(\sqrt{x^2-4x+4}=x-2\)
câu 2 rút gọn
A = \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
B= \(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}\)
Giải pt:
\(a)x^{4}-2\sqrt{2}x^{2}+2=\sqrt{2}+x \\b)(2x+3)\sqrt{x^{2}-2}=2x^{2}+3x-4 \\c)2x^{2}+2(x+1)\sqrt{x^{2}-1}-6x+1=0\)
a)\(\sqrt{x^2-2x+4}=2x-2\)
b) \(\sqrt{-x^2+x+4}=x-3\)
c) \(\sqrt{x^2-2x}=\sqrt{2-3x}\)
d) \(\sqrt{x-3}-2\sqrt{x^2-9}=0\)
Rút gọn:
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
27 tháng 12 2020 lúc 16:00
\(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)}\)
giải các phương trình sau:
\(\sqrt{x^2+3x}=3x-1\)
\(x-\sqrt{4x-3}=2\)
\(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
\(\sqrt{2x^2-2x+4}=\sqrt{x^2-x+2}\)