Câu 2 : D
Câu 3: C
Câu 4 : D
Câu 5 : C
Câu 6 : A
Câu 7 : A
Câu 8 : C
Phần II:
Bài 1 :
\(a,A=\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{3}{7}\)
\(A=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{4}\)
\(A=1+\dfrac{3}{4}\)
\(A=\dfrac{7}{4}\)
\(b,B=\dfrac{4}{12}+\dfrac{-18}{45}+\dfrac{6}{9}+\dfrac{21}{35}+\dfrac{-6}{30}\)
\(B=\dfrac{1}{3}+\dfrac{-2}{5}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{-1}{5}\)
\(B=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{5}+\dfrac{3}{5}+\dfrac{-1}{5}\right)\)
\(B=1+0\)
\(B=1\)
Bài 2 :
\(x-1\dfrac{2}{3}=0,2\)
\(x-\dfrac{5}{3}=0,2\)
\(x-\dfrac{5}{3}=\dfrac{2}{10}\)
\(x=\dfrac{2}{10}+\dfrac{5}{3}\)
\(x=\dfrac{6}{30}+\dfrac{50}{30}\)
\(x=\dfrac{56}{30}=\dfrac{28}{15}\)
\(\dfrac{7}{5}+\dfrac{5}{6}:x=\dfrac{5}{6}\)
\(\dfrac{5}{6}:x=\dfrac{5}{6}+\dfrac{7}{5}\)
\(\dfrac{5}{6}:x=\dfrac{25}{30}+\dfrac{42}{30}\)
\(\dfrac{5}{6}:x=\dfrac{67}{30}\)
\(x=\dfrac{5}{6}:\dfrac{67}{30}\)
\(x=\dfrac{5}{6}\times\dfrac{30}{67}\)
\(x=\dfrac{5}{1}\times\dfrac{5}{67}\)
\(x=\dfrac{25}{67}\)
Câu 2 : D
Câu 3: C
Câu 4 : D
Câu 5 : C
Câu 6 : A
Câu 7 : A
Câu 8 : C
Phần II:
Bài 1 :
\(a,A=\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{3}{7}\)
\(A=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{4}\)
\(A=1+\dfrac{3}{4}\)
\(A=\dfrac{7}{4}\)
\(b,B=\dfrac{4}{12}+\dfrac{-18}{45}+\dfrac{6}{9}+\dfrac{21}{35}+\dfrac{-6}{30}\)
\(B=\dfrac{1}{3}+\dfrac{-2}{5}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{-1}{5}\)
\(B=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{5}+\dfrac{3}{5}+\dfrac{-1}{5}\right)\)
\(B=1+0\)
\(B=1\)
Bài 2 :
\(x-1\dfrac{2}{3}=0,2\)
\(x-\dfrac{5}{3}=0,2\)
\(x-\dfrac{5}{3}=\dfrac{2}{10}\)
\(x=\dfrac{2}{10}+\dfrac{5}{3}\)
\(x=\dfrac{6}{30}+\dfrac{50}{30}\)
\(x=\dfrac{56}{30}=\dfrac{28}{15}\)
\(\dfrac{7}{5}+\dfrac{5}{6}:x=\dfrac{5}{6}\)
\(\dfrac{5}{6}:x=\dfrac{5}{6}+\dfrac{7}{5}\)
\(\dfrac{5}{6}:x=\dfrac{25}{30}+\dfrac{42}{30}\)
\(\dfrac{5}{6}:x=\dfrac{67}{30}\)
\(x=\dfrac{5}{6}:\dfrac{67}{30}\)
\(x=\dfrac{5}{6}\times\dfrac{30}{67}\)
\(x=\dfrac{5}{1}\times\dfrac{5}{67}\)
\(x=\dfrac{25}{67}\)
Bài 3:
giá vốn của chiếc TV đó là :
3 : 25 % = 12 (triệu đồng)
Đ/S:
Bài 4:
Gợi ý
a) So sánh các đoạn thẳng và dựa vào tính chất của trung điểm đoạn thẳng
b) Từ (a) = > Tính CD và so sánh.
c) Từ (a) , (b) suy ra.
