Cho mik cái này đề bài vs

e: \(=\dfrac{1}{6}-\dfrac{4}{9}+\dfrac{5}{18}=\dfrac{3-8+5}{18}=0\)

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

ĐKXĐ: \(x>=-\dfrac{1.96}{3}=-\dfrac{196}{300}=\dfrac{-49}{75}\)

\(\sqrt{\dfrac{1.96+3x}{4}}=\sqrt{0,04}+\dfrac{1}{4}\cdot\sqrt{\dfrac{256}{25}}\)

=>\(\sqrt{\dfrac{3x+1.96}{4}}=0.2+\dfrac{1}{4}\cdot\dfrac{16}{5}=1\)

=>\(\dfrac{3x+1,96}{4}=1\)

=>3x+1,96=4

=>3x=2,04

=>\(x=\dfrac{2.04}{3}=0.68\left(nhận\right)\)

b: =3/4x(-1)=-3/4

a) 4.25-12.5+170:10

=100-60+17

=40+17

=57

b) (7+3³:3²).4-3

=(7+3).4-3

=10.4-3

=40-3

=37

c) 12:{400:[500-(125+25.7)]}

=12:{400:[500-(125+175)]}

=12:{400:[500-300]}

=12:{400:200}

=12:2

=6

d) 168+{[2.(2⁴+3²)-256⁰]:7²}

=168+{[2.(16+9)-1]:49}

=168+{[2.25-1]:49}

=168+{[50-1]:49}

=168+{49:49}

=168+1

=169

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

2.

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

⇒ \(5x+1=\pm\frac{6}{7}\)

⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

a. \(\dfrac{-5}{4}\) x⁴ . \(\dfrac{8}{15}\) x = \(\dfrac{-40}{60}\) x⁵ = \(\dfrac{-2}{3}\) x⁵

b. -2x\(\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) = -\(\dfrac{-3}{2}\) x³ + 2x³ - x

c. \(x\left(x-\dfrac{1}{2}\right)\) - (x - 2)(x + 3) 

= x² - \(\dfrac{1}{2}\) x - x² - 3x - 2x - 6

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{3\cdot2}-\dfrac{1}{5}\cdot\dfrac{-6}{9}=\dfrac{1}{6}+\dfrac{6}{45}=\dfrac{45+36}{270}=\dfrac{81}{270}=\dfrac{3}{10}\)