Question

[Nguyễn Duy Long]

 gguúgguúp ttớ vvới

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 3:

a: \(2\sqrt{6}\cdot\left(-\sqrt{6}\right)=-2\sqrt{6\cdot6}=-2\sqrt{36}=-2\cdot6=-12\)

b: \(\sqrt{1,44}-2\cdot\left(\sqrt{0,6}\right)^2=1,2-2\cdot0,6=1,2-1,2=0\)

c: \(\left|0,1\right|\cdot\left(\sqrt{7}\right)^2+\sqrt{1,69}\)

\(=0,1\cdot7+1,3=1,3+0,7=2\)

d: \(\left|-0,1\right|\cdot\left(\sqrt{120}\right)^2-\dfrac{1}{4}\cdot\left(\sqrt{20}\right)^2\)

\(=0,1\cdot120-\dfrac{1}{4}\cdot20\)

=12-5

=7

Bài 3:

\(\dfrac{-7}{8}=-1+\dfrac{1}{8};\dfrac{-2}{3}=-1+\dfrac{1}{3}\)

\(\dfrac{-3}{4}=-1+\dfrac{1}{4};-\dfrac{18}{19}=-1+\dfrac{1}{19};-\dfrac{27}{28}=-1+\dfrac{1}{28}\)

mà \(\dfrac{1}{28}< \dfrac{1}{19}< \dfrac{1}{8}< \dfrac{1}{4}< \dfrac{1}{3}\)

nên \(-\dfrac{27}{28}< -\dfrac{18}{19}< -\dfrac{7}{8}< -\dfrac{3}{4}< -\dfrac{2}{3}\)

Bài 1:

a: \(x\cdot\dfrac{-3}{7}=\dfrac{5}{21}\)

=>\(x=-\dfrac{5}{21}:\dfrac{3}{7}=-\dfrac{5}{21}\cdot\dfrac{7}{3}=-\dfrac{5}{9}\)

b: \(3x-\dfrac{5}{3}=x-\dfrac{1}{4}\)

=>\(3x-x=-\dfrac{1}{4}+\dfrac{5}{3}\)

=>\(2x=-\dfrac{3}{12}+\dfrac{20}{12}=\dfrac{17}{12}\)

=>\(x=\dfrac{17}{12}:2=\dfrac{17}{24}\)

c: \(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5+1=6\\x=-5+1=-4\end{matrix}\right.\)

d: \(x+\dfrac{1}{5}=-0,4\)

=>\(x+\dfrac{1}{5}=-\dfrac{2}{5}\)

=>\(x=-\dfrac{2}{5}-\dfrac{1}{5}=-\dfrac{3}{5}\)

e: \(2^{x-1}=16\)

=>\(2^{x-1}=2^4\)

=>x-1=4

=>x=4+1=5

f: |2x-1|=5

=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Answer 2

[Nguyễn Đình Đạt]

làm mấy bài rứa bạn