\(\sqrt{-2x+3}=0\)
<=> \(\left(\sqrt{-2x+3}\right)^2=0\)
<=> \(-2x+3=0\)
<=> \(-2x=-3\)
<=> \(x=\dfrac{3}{2}\)
Vậy x= \(\dfrac{3}{2}\) (=1,5)
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Bài 1: Căn bậc hai
Các câu hỏi tương tự
Giải các pt sau:\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\sqrt{x^2-10x+25}=3-19x\)
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
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, \(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)
\(3\sqrt{2x-1}-\sqrt{x-2}=\sqrt{x+3}\)
\(\sqrt{2x-1}-\sqrt{x-1}=x\)
tìm x:
\(\sqrt{x^2+x+1}=1\)
\(\sqrt{x^2+1}=-3\)
\(\sqrt{x^2-10x+25}=7-2x\)
\(\sqrt{2x+5}=5\)
\(\sqrt{x^2-4x+4}-2x+5=0\)
\(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
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, \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
Cho biểu thức:
\(M=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
a/ Rút gọn M
b/ Tính M khi \(x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)
Cho biểu thức:
\(M=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right)\)\(:\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
a/ Rút gọn M
b/ Tính M khi \(x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)
giải hệ pt sau
\(\left\{{}\begin{matrix}y^3+\sqrt{8x^4-2y}=2\left(2x^4+3\right)\\\sqrt{2x^2+x+y}+2\sqrt{x+2y}=\sqrt{9x-2x^2+19y}\end{matrix}\right.\)