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a, \(16x^2-5=0\)
\(\Rightarrow16x^2=5\)
\(\Rightarrow x^2=\frac{5}{16}\)
\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)
b, \(2\sqrt{x-3}=4\)
\(\Rightarrow\sqrt{x-3}=4:2\)
\(\Rightarrow\sqrt{x-3}=2\)
\(\Rightarrow x-3=4\)
\(\Rightarrow x=4+3\)
\(\Rightarrow x=7\)
c, \(\sqrt{4x^2-4x+1}=3\)
\(\Rightarrow\sqrt{\left(2x-1\right)^2}=3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
d, \(\sqrt{x+3}\ge5\)
\(\Rightarrow x+3\ge25\)
\(\Rightarrow x\ge22\)
e, \(\sqrt{3x-1}< 2\)
\(\Rightarrow3x-1< 4\)
\(\Rightarrow3x< 5\)
\(\Rightarrow x< \frac{5}{3}\)
g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)
\(\Rightarrow\sqrt{x-3}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) \(16x^2-5=0\)
\(\Leftrightarrow16x^2=5\)
\(\Leftrightarrow x^2=\frac{5}{16}\)
\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)
b) \(2\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=2\)
\(\Leftrightarrow x-3=4\)
\(\Leftrightarrow x=7\)
c) \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
d) \(\sqrt{x+3}\ge5\)
\(\Leftrightarrow x+3\ge25\)
\(\Leftrightarrow x\ge22\)
e) \(\sqrt{3x-1}< 2\)
\(\Leftrightarrow3x-1< 4\)
\(\Leftrightarrow3x< 5\)
\(\Leftrightarrow x< \frac{5}{3}\)
g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)
\(\Leftrightarrow\sqrt{x-3}=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
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Bạn nên viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo). Viết thế này khó dịch quá.
![](https://rs.olm.vn/images/avt/0.png?1311)
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1) thay m=1 vào pt: \(x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
2) theo định lí viets, ta có: x1+x2=2(m+1)
x1x2=2m
\(\sqrt{x_1}+\sqrt{x_2}=\sqrt{2}\Leftrightarrow\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=2\)
\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=2\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{2m}=2\)
tới đây bạn làm tiếp nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
cộng thêm 2 vế vs 3x cho thành hằng đẳng thức