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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, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Rightarrow\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

7 tháng 7 2019

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 \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Leftrightarrow\sqrt{x-3}=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Y
17 tháng 6 2019

a) \(\sqrt{9x^2}-2x\) \(=-3x-2x\) ( do x < 0 )

\(=-5x\)

b) \(3\sqrt{\left(x-2\right)^2}=3\left(2-x\right)\) ( do x - 2 < 0 )

\(=6-3x\)

c) \(x-4+\sqrt{16-8x+x^2}\)

\(=x-4+\sqrt{\left(x-4\right)^2}\)

\(x-4+x-4=2x-8\)

17 tháng 6 2019

\(a)\sqrt{9\times^2}-2\times\)

\(=\sqrt{3^2\times^2}-2\times\)

\(=\sqrt{(3\times)^2}-2\times\)

\(=3\times-2\times\)

\(=\times\)

17 tháng 6 2019

\(b)3\cdot\sqrt{(\times-2)^2}\)

\(=3\cdot(\times-2)\)

12 tháng 7 2018

a. \(x^4-4x^3+8x^2-16x+16\)

\(=x^4-2x^3-2x^3+4x^2+4x^2-8x-8x+16\)

\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)

\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)

\(=\left(x-2\right)^2\left(x^2+4\right)\)

b. \(x^4-25x^2+20x-4\)

\(=x^4+5x^3-5x^3-25x^2+2x^2-2x^2+10x+10x-4\)

\(=\left(x^4+5x^3-2x^2\right)-\left(5x^3+25x^2-10x\right)+\left(2x^2+10x-4\right)\)

\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)

\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)

27 tháng 12 2018

Hn thi mx ra câu này T.T

7 tháng 7 2017

\(A=\sqrt{4x^2-16x+20}=2\sqrt{x^2-4x+5}=2\sqrt{\left(x-2\right)^2+1}\ge2\)

Dấu "=" xảy ra khi x = 2

Min A = 2 <=> x = 2

Câu b) bạn ghi rõ đề hơn nhé.