\(S=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{1}{2}=\dfrac{3}{16}\left(m^2\right)\)

Số kg gạo ngày 2 bán được: \(85-10=75\left(kg\right)\)

Số kg gạo trước khi bán cửa hàng có: \(46+85+75=206\left(kg\right)\)

b) \(\Rightarrow\left|5-3x\right|-\dfrac{7}{4}=\dfrac{1}{4}\Rightarrow\left|5-3x\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}5-3x=2\\5-3x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=3\\3x=7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{3}\end{matrix}\right.\)

c) Áp dụng t/c dtsbn:

\(\dfrac{x-2}{5}=\dfrac{2-y}{3}=\dfrac{x-y-2+2}{5+3}=\dfrac{32}{8}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.5+2=22\\y=2-4.3=-10\end{matrix}\right.\)

b) \(\Rightarrow\dfrac{3}{2}-11x=\dfrac{2}{5}\Rightarrow11x=\dfrac{3}{2}-\dfrac{2}{5}=\dfrac{11}{10}\Rightarrow x=\dfrac{1}{10}\)

c) ĐKXĐ: \(x\ne-\dfrac{1}{3}\)

\(\Rightarrow\left(3x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}3x+1=4\\3x+1=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)

Gọi số cây tổ 1,2,3 trồng lần lượt là a,b,c(cây)(a,b,c∈N*,a,b,c<108)

Áp dụng t/c dtsbn:

\(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{12}=\dfrac{a+b+c}{7+8+12}=\dfrac{108}{27}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a=7.4=28\\b=4.8=32\\c=4.12=48\end{matrix}\right.\)(nhận)

Vậy...

Cái này là số vô tỉ nhé bạn nên nó khá dài

Gọi số HS nam và số HS nữ lần lượt là a,b(HS)(a,b∈N*,a>4)

Áp dụng t/c dtsbn:

\(\dfrac{a}{6}=\dfrac{b}{5}=\dfrac{a-b}{6-5}=\dfrac{4}{1}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a=4.6=24\\b=4.5=20\end{matrix}\right.\)(nhận)

Vậy...

a) \(A=\dfrac{y^2\left(x-2\right)\left(x+2\right)}{4xy}.\dfrac{x^2y}{xy\left(2-x\right)}=\dfrac{y\left(x+2\right)}{-4}=-4xy-8y\)

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

c) \(C=\dfrac{\left(7y+1\right)\left(y+7\right)+\left(7y-1\right)\left(y-7\right)}{y\left(y-7\right)\left(y+7\right)}.\dfrac{\left(y-7\right)\left(y+7\right)}{y^2+1}=\dfrac{14\left(y^2+1\right)}{y\left(y^2+1\right)}=\dfrac{14}{y}\)

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

\(\left(d\right):y=ax+b//y=2x-3\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b\ne-3\end{matrix}\right.\)

\(B\left(2;2\right)\in\left(d\right)\Leftrightarrow y_B=2x_B+b\Leftrightarrow2=2.2+b\Leftrightarrow b=-2\)

Vậy \(\left(d\right):y=2x-2\)