Question

Bài 1 :

a) Tính B = \(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.35}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.\left(\sqrt{196}\right)^3}\)

b)Tìm x biết : \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3,2+\sqrt{\dfrac{4}{25}}\right|\)

c)Tính \(\left|3x+1\right|>4\)

Answer 1

Ta có: \(\left|x-\dfrac{1}{3}\right|\)+0,8 = \(\left|-3,2+0,4\right|\)

---> \(\left|x-\dfrac{1}{3}\right|\)+0,8 =2,8 ---> \(\left|x-\dfrac{1}{3}\right|\)= 2,8-0,8=2 --> TH1: x-\(\dfrac{1}{3}\) = 2 --> x= 2+ \(\dfrac{1}{3}\)= \(\dfrac{7}{3}\) TH2: x-\(\dfrac{1}{3}\) = -2---> x= -2+\(\dfrac{1}{3}\)=\(\dfrac{-5}{3}\) Vạy x =\(\dfrac{-5}{3}\) hoặc x= \(\dfrac{7}{3}\) Tick cho mik nha!!