a, \(x^2-5=0\Leftrightarrow\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\Leftrightarrow x=\pm\sqrt{5}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{\pm\sqrt{5}\right\}\)

b, \(x^2-2\sqrt{11}x+11=0\Leftrightarrow x^2-2\sqrt{11}x+\left(\sqrt{11}\right)^2=0\)

\(\Leftrightarrow\left(x-\sqrt{11}\right)^2=0\Leftrightarrow x=\sqrt{11}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\sqrt{11}\right\}\)

x² - 5 = 0

Δ = b² - 4ac = 0 + 20 = 20

Δ > 0, áp dụng công thức nghiệm thu được x = ±√5

x² - 2√11x + 11 = 0

Δ = b² - 4ac = 44 - 44 = 0

Δ = 0 => phương trình có nghiệm kép x₁ = x₂ = -b/2a = √11

a, \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

b, \(x^2-6=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

c, \(x^2+2\sqrt{3}+3=x^2+2\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d, \(x^2-2\sqrt{5}x+5=x^2-2\sqrt{5}x+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)

a) \(x^2\) - 3 = (x-\(\sqrt{3}\))(x+\(\sqrt{3}\))

b)\(x^2\)-6=(x-\(\sqrt{6}\))(x+\(\sqrt{6}\))

c) \(x^2+2\sqrt{3}x+3\)= \(\left(x+\sqrt{3}\right)^2\)

d) \(x^2-2\sqrt{5}x+5\)=\(\left(x-\sqrt{5}\right)^2\)

a, \(2\sqrt{a^2}-5a=2\left|a\right|-5a\)do a < 0 

\(=-2a-5a=-7a\)

b, \(\sqrt{25a^2}+3a=\sqrt{\left(5a\right)^2}+3a=\left|5a\right|+3a\)do \(a\le0\)

TH1 : \(-5a+3a=-2a\)với \(a< 0\)

hoặc TH2 : \(5+3=8\)

c, \(\sqrt{9a^4}+3a^2=\sqrt{\left(3a^2\right)^2}+3a^2=\left|3a^2\right|+3a^2\)

\(=3a^2+3a^2=6a^2\)do \(3>0;a^2\ge0\forall a\Rightarrow3a^2\ge0\forall a\)

d, \(5\sqrt{4a^6}-3a^3=5\sqrt{\left(2a^3\right)^2}-3a^3\)

\(=5\left|2a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)do \(a< 0\Rightarrow a^3< 0\)

a) \(2\sqrt{a^2}-5a\)=2\(|a|\)-5a = -2a-5a=-7a

b) \(\sqrt{25a^2}\) +3a = 5\(|a|\) + 3a=5a+3a=8a.

c) \(\sqrt{9a^4}\) + 3\(a^2\)=6\(a^2\)

d) \(5\sqrt{4a^6}\) - 3\(a^3\)=-13\(a^3\)

a) √2x+7

Để √2x+7 có nghĩa⇔2x+7≥0

⇔2x≥-7

⇔x≥−7/2

b) √−3x+4

Để √−3x+4 có nghĩa ⇔-3x+4≥≥0

⇔-3x≥-4

⇔x≤4/3

c)√1/−1+x1

Để √1/−1+x có nghĩa ⇔1/−1+x≥0

⇔-1+x>0

⇔x>1

d) √1+x21+x2

Ta có x²+1≥≥1>0;∀x∈R

Vậy x∈R

+a) \(\sqrt{2x+7}\) co nghia khi 2x+7≥0⇒x≥\(\dfrac{-7}{2}\)

b) \(\sqrt{-3x+4}\) co nghia khi -3x+4≥0⇒x≤\(\dfrac{4}{3}\)

c) \(\sqrt{\dfrac{1}{-1+x}}\) cp nghia khi \(\dfrac{1}{-1+x}\)≥0 ⇒-1+x>0⇒x>1

d) \(\sqrt{1+x^2}\) co nghia khi 1+x² ≥0 ma \(x^2\)≥0⇒\(x^2\) + 1≥1>0 vs moi x 

a) \(\sqrt{16}\).\(\sqrt{25}\)+\(\sqrt{196}\):\(\sqrt{49}\)

=4.5+14/7

=20+2

=22

a) \(\sqrt{16}\).\(\sqrt{25}\) + \(\sqrt{196}\) : \(\sqrt{49}\) = 4.5+14:9=22

b) 36:\(\sqrt{2.3^2.18}\) - \(\sqrt{169}\)= 36 :  \(\)18 - 13 = -11

c) \(\sqrt{\sqrt{81}}\) = 3

d) \(\sqrt{3^2+4^2}\)= \(\sqrt{25}\)=5

a) (\(\sqrt{3}\)-1)²=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)

b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0

Bình phương 2 vế, ta có:

4-2\(\sqrt{3}\)=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)

a)  \(\left(\sqrt{3}-1\right)^2\)=\(\left(\sqrt{3}\right)^2\)- 2\(\sqrt{3}\) +1= 3- 2\(\sqrt{3}\) +1=4-2\(\sqrt{3}\)

b)  \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)= \(|\sqrt{3}-1|\)-\(\sqrt{3}\)=\(\sqrt{3}\)-1-\(\sqrt{3}\)=-1

a) \(\sqrt{x^2}\)=7

=> x²=49

=> x={-7;7}

b) \(\sqrt{x^2}\)=|-8|=8

=> x²=64

=>x={-8;8}

c) \(\sqrt{4x^2}\)=6

4x²=36

=>x²=9

=> x={-3;3}

d)\(\sqrt{9x^2}\)=|-12|=12

=> 9x²=144

=> x²=16

=> x={-4;4}

a)x=+7 hoặc x= -7

b) x=8 hoặc x= -8

c)x=3 hoặc x =-3

d) x=4 hoặc x= -4

what đề bài đâu

a, \(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)do \(2-\sqrt{3}>0\)

b, \(\sqrt{\left(3-\sqrt{11}\right)^2}=\sqrt{\left(\sqrt{11}-3\right)^2}=\left|\sqrt{11}-3\right|=\sqrt{11}-3\)

do \(\sqrt{11}-3>0\)

c, \(2\sqrt{a^2}=2\left|a\right|=2a\)do \(a\ge0\)

d, \(3\sqrt{\left(a-2\right)^2}=3\sqrt{\left(2-a\right)^2}=3\left|2-a\right|=3\left(2-a\right)=6-3a\)

do \(a< 2\)

a, \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)do \(0,1>0\)

b, \(\sqrt{\left(-0,3\right)^2}=\sqrt{\left(0,3\right)^2}=\left|0,3\right|=0,3\)do \(0,3>0\)

c, \(-\sqrt{\left(-1,3\right)^2}=-\sqrt{\left(1,3\right)^2}=-\left|1,3\right|=-1,3\)do \(1,3>0\)

d, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\sqrt{\left(0,4\right)^2}=-0,4.\left|0,4\right|=-0,4.0,4=-0,14\)

do \(0,4>0\)

\(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)

\(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)

\(-\sqrt{\left(-1,3\right)^2}=-\left|-1,3\right|=-1,3\)

\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,16\)