a) \(\dfrac{1}{2}x=\dfrac{3}{2}-\dfrac{5}{3}=-\dfrac{1}{6}\)

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

b) \(-\dfrac{1}{6}+2x=\dfrac{1}{2}\)

\(x=\left(\dfrac{1}{2}+\dfrac{1}{6}\right):2=\dfrac{1}{3}\)

Số Hs nam là : 40 - 24 = 16 (Hs)

Tỉ số phần trăm giữa số Hs nam và số Hs cả lớp là:

           16 : 40 x 100 = 40%

                            Đáp số: 40%

`Answer:`

`(1-5/21).(1-5/32).(1-5/45)...(1-5/320)`

`=(21-5)/21 . (32-5)/32 . (45-5)/45 ... (320-5)/320`

`=16/21 . 27/32 . 40/45 ... 315/320`

`=(16.27.40...315)/(21.32.45...320)`

`=((2.8).(3.9).(4.10)...(15.21))/((3.7).(4.8).(5.9)...(16.20))`

`=(2.8.3.9.4.10...15.21)/(3.7.4.8.5.9...16.20)`

`=((2.3.4...15).(8.9.10...21))/((3.4.5...16).(7.8.9...20))`

`=(2.3.4...15)/(3.4.5...16) . (8.9.10...21)/(7.8.9...20)`

`=2/16 . 21/7`

`=3/8`

`Answer:`

a) \(x+\frac{2}{5}=-\frac{11}{15}\)

\(\Leftrightarrow x=-\frac{11}{15}-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{17}{15}\)

b) \(\left(x-\frac{7}{18}\right).\frac{18}{29}=-\frac{12}{29}\)

\(\Leftrightarrow x-\frac{7}{18}=-\frac{12}{29}:\frac{18}{29}\)

\(\Leftrightarrow x-\frac{7}{18}=-\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{2}{3}+\frac{7}{18}\)

\(\Leftrightarrow x=-\frac{5}{18}\)

Bài 3:

Số học sinh Giỏi của lớp `6B` là:

`35.40%=14` học sinh

Số học sinh Khá của lớp `6B` là:

`14. 9/7=18` học sinh

Số học sinh Trung bình của lớp `6B` là:

`35-14-18=3` học sinh

`Answer:`

Bài 1:

a. \(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)

\(\Rightarrow x:\frac{3}{4}=-\frac{2}{3}-\frac{1}{4}\)

\(\Rightarrow x:\frac{3}{4}=-\frac{11}{12}\)

\(\Rightarrow x=-\frac{11}{12}.\frac{3}{4}\)

\(\Rightarrow x=-\frac{11}{16}\)

b. \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\pm\frac{3}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

c. \(\frac{1}{3}x+\frac{2}{5}.\left(x-1\right)=0\)

\(\Rightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Rightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Rightarrow x=\frac{2}{5}:\frac{11}{15}\)

\(\Rightarrow x=\frac{6}{11}\)

d. \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\-\frac{1}{2}x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=1\\-\frac{1}{2}x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}}\)

e. \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{18}\)

Bài 2:

a. \(\frac{2}{5}.\frac{1}{3}-\frac{2}{25}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)

\(=\frac{2}{15}-\frac{2}{3}+\frac{3}{15}\)

\(=\frac{5}{15}-\frac{2}{3}\)

\(=-\frac{1}{3}\)

b. \(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)

\(=\left(\frac{10}{3}+2,5\right):\left(\frac{19}{6}-\frac{21}{5}\right)-\frac{11}{31}\)

\(=\frac{35}{6}:\left(-\frac{31}{30}\right)-\frac{11}{31}\)

\(=-6\)

c. \(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)

\(=\left(-8\right).\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)

\(=-\frac{48}{13}\)

d. \(E=-\frac{11}{12}.\frac{7}{8}+\frac{11}{12}.\left(-\frac{15}{17}\right)-\frac{11}{12}\)

\(=\frac{11}{12}.\left(-1\right).\frac{7}{8}+\frac{11}{12}.\left(-\frac{15}{16}\right)-\frac{11}{12}\)

\(=\frac{11}{12}.\left(-\frac{7}{8}+-\frac{15}{16}-1\right)\)

\(=\frac{11}{12}.-\frac{45}{16}\)

\(=-\frac{165}{64}\)

e. \(F=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=\frac{1004}{1005}\)

g. \(\left(1-\frac{5}{21}\right)\left(1-\frac{5}{32}\right)\left(1-\frac{5}{45}\right)...\left(1-\frac{5}{320}\right)\)

\(=\frac{21-5}{21}.\frac{32-5}{32}.\frac{45-5}{45}...\frac{320-5}{320}\)

\(=\frac{16}{21}.\frac{27}{32}.\frac{40}{45}...\frac{315}{320}\)

\(=\frac{\left(2.8\right)\left(3.9\right)\left(4.10\right)...\left(15.21\right)}{\left(3.7\right)\left(4.8\right)\left(5.9\right)...\left(16.20\right)}\)

\(=\frac{2.8.3.9.4.10...15.21}{3.7.4.8.5.9...16.20}\)

\(=\frac{\left(2.3.4...15\right)\left(8.9.10...21\right)}{\left(3.4.5...16\right)\left(7.8.9...20\right)}\)

\(=\frac{2.3.4...15}{3.4.5...16}.\frac{8.9.10...21}{7.8.9...20}\)

\(=\frac{2}{16}.\frac{21}{7}\)

\(=\frac{3}{8}\)

có nè k ikkk

đang lấy tạm nick ngta":>

381/280

\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\)

\(\Rightarrow\sqrt{a^2+3b^2}\ge\sqrt{\dfrac{\left(a+3b\right)^2}{4}}=\dfrac{a+3b}{2}\)

Tương tự:

\(\sqrt{b^2+3c^2}\ge\dfrac{b+3c}{2}\) ; \(\sqrt{c^2+3a^2}\ge\dfrac{c+3a}{2}\)

 Cộng vế \(\Rightarrow VT\ge\dfrac{4\left(a+b+c\right)}{2}=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

toán lớp 6 đây á