\(\sqrt{\left(0,1\right)^2}=0,1=\dfrac{1}{10}\)

\(\sqrt{\left(0.1\right)^2}=0.1\)

a)1/10

b)3/10

c)-13/10

d)-4/25

Rất tiếc! Bạn Freya làm sai rồi...

tính thành căn hay p/ số hay giải nghiệm

\(a,\sqrt{0,1^2}=0,1\)

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

\(c,-\sqrt{\left(-1,7\right)^2}=-|-1,7|=-1,7\)

\(d,-0,5\sqrt{\left(-0,5\right)^4}=\frac{-1}{2}\sqrt{[\left(\frac{-1}{2}\right)^2]^2}=-\frac{1}{2}.\left(\frac{1}{2}\right)^2=\frac{-1}{2}.\frac{1}{4}=\frac{-1}{8}\)

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

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

a là 0,1

b là 0,4

a: \(=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)

b: =8

c: \(=2-\sqrt{3}\)

d: \(=3-2\sqrt{2}\)

e: \(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)

a: \(=\dfrac{-4}{5}\cdot\left|-\dfrac{5}{4}\right|=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)

b: \(=\left|\left(-2\right)^3\right|=8\)

c: \(=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)

d: \(=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

e: \(=\left|\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\right|=\dfrac{\sqrt{2}-1}{2}\)

f: \(=\left|\dfrac{1}{10}-\dfrac{\sqrt{10}}{10}\right|=\dfrac{\sqrt{10}-1}{10}\)

\(A=0,1.3.\left(\frac{6}{3}-\left(2-\sqrt{3}\right)\right)^2\)

\(A=0,3.\left(\sqrt{3}\right)^2\)

\(A=0,3.3=0,9\)

Vậy ...

a) Phương trình 1,5x² – 1,6x + 0,1 = 0

Có a + b + c = 1,5 – 1,6 + 0,1 = 0 nên x₁ = 1; x₂ = \(\dfrac{0,1}{15}\)

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

Có \(a+b+c=2-\sqrt{3}+2\sqrt{3}-\left(2+\sqrt{3}\right)=0\)

Nên x₁ = 1, x₂ = \(\dfrac{-\left(2+\sqrt{3}\right)}{2-\sqrt{3}}\) = -(2 + \(\sqrt{3}\))² = -7 - 4\(\sqrt{3}\)

d) (m – 1)x² – (2m + 3)x + m + 4 = 0

Có a + b + c = m – 1 – (2m + 3) + m + 4 = 0

Nên x₁ = 1, x₂ = \(\dfrac{m+4}{m-1}\)

a) Phương trình 1,5x2 – 1,6x + 0,1 = 0

Có a + b + c = 1,5 – 1,6 + 0,1 = 0 nên x1 = 1; x2 =

b) Phương trình √3x2 – (1 - √3)x – 1 = 0

Có a – b + c = √3 + (1 - √3) + (-1) = 0 nên x1 = -1, x2 = =

c) (2 - √3)x2 + 2√3x – (2 + √3) = 0

Có a + b + c = 2 - √3 + 2√3 – (2 + √3) = 0

Nên x1 = 1, x2 = = -(2 + √3)2 = -7 - 4√3

d) (m – 1)x2 – (2m + 3)x + m + 4 = 0

Có a + b + c = m – 1 – (2m + 3) + m + 4 = 0

Nên x1 = 1, x2 =