b, \(\Delta=\left(-8\right)^2-4.1.7=64-28=36>0\)

\(x_1=\dfrac{8+6}{2}=7,x_2=\dfrac{8-6}{2}=1\)

\(c,\Delta=\left(-6\right)^2-4.1.\left(-3\right)=36+12=48>0\)

\(x_1=\dfrac{6+4\sqrt{3}}{2}=3+2\sqrt{3},x_2=\dfrac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)

\(d,\Delta=\left(-6\right)^2-4.1.\left(-7\right)=36+28=64>0\)

\(x_1=\dfrac{6+8}{2}=7,x_2=\dfrac{6-8}{2}=-1\)

\(3x+2\left(1-2x\right)=5x+1\\ \Leftrightarrow3x+2-4x-5x-1=0\\ \Leftrightarrow-6x+1=0\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{6}\right\}\)

Tham khảo:

Var i,S:integer;

Begin

i:=0;

While i<=100 do If i mod 2 = 0 then S:= S+i;

Writeln(' tong cac so chan :',S);

Readln;

end.

C

 Cửa hàng bán được số hộp kẹo là:

\(56:\left(3+4\right)\times3=24\left(hộp\right)\)

 Cửa hàng bán được số hộp bánh là:

\(56-24=32\left(hộp\right)\)

Số thóc của kho thứ nhất là:
\(1350:\left(4+5\right)\times4=600\left(tấn\right)\)

Số thóc của kho thứ 2 là:
\(1350-600=750\left(tấn\right)\)​

C

C

\(\left\{{}\begin{matrix}x+2y=5\\3x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+4y=10\\3x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-5+2y=5\\x=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y=10\\x=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-5\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne-2\\y\ne3\end{matrix}\right.\\ \left\{{}\begin{matrix}\dfrac{3}{x+2}+\dfrac{1}{y-3}=4\\\dfrac{2}{x+2}-\dfrac{3}{y-3}=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+2}+\dfrac{3}{y-3}=12\\\dfrac{2}{x+2}-\dfrac{3}{y-3}=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+2}+\dfrac{1}{y-3}=4\\\dfrac{11}{x+2}=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{1}+\dfrac{1}{y-3}=4\\x+2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3+\dfrac{1}{y-3}=4\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-3}=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-3=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\x=-1\left(tm\right)\end{matrix}\right.\)