a)  \(A=x^2+2xy+y^2-4x-4y+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

b)  \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37=100\)

c)  \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2.5+10=25\)

a) \(A=x^2+2xy+y^2-4x-4v+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

\(\left(\frac{3}{2}x+y\right)^3=\left(\frac{3}{2}x\right)^3+3.\left(\frac{3}{2}\right)^2.y+3.\frac{3}{2}.y^2+y^3\)

\(\left(\frac{3}{2}x+y\right)^3\)

\(=\frac{27}{8}x^3+\frac{27}{4}x^2y+3xy^2+y^3\)

\(\frac{-x^6}{125}-\frac{y^3}{64}\)

\(=\frac{-\left(x^2\right)^3}{5^3}-\frac{y^3}{4^3}\)

\(=\left(\frac{-x^2}{5}\right)^3-\left(\frac{y}{4}\right)^3\)

\(=\left(\frac{-x^2}{5}-\frac{y}{4}\right)\cdot\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)

Tham khảo nhé~

\(-\frac{x^6}{125}-\frac{y^3}{64}\)

\(=-\left(\frac{x^6}{125}+\frac{y^3}{64}\right)\)

\(=-\left(\frac{x^2}{5}+\frac{y}{4}\right)\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)

             \(x^3+16x=6x^2+9\)

\(\Leftrightarrow\)\(\left(x-2\right)^3+4\left(x-2\right)=-7\)   (1)

           \(9y^2+32=y^3+31y\)

\(\Leftrightarrow\)\(\left(3-y\right)^3+4\left(3-y\right)=7\)   (2)

Đặt  \(a=x-2;\) \(b=3-y\) từ (1) và (2) suy ra:

     \(\hept{\begin{cases}a^3+4a=-7\\b^3+4b=7\end{cases}}\)

nên     \(\left(a^3+b^3\right)+4\left(a+b\right)=0\)

\(\Leftrightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2+4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}a+b=0\\a^2-ab+b^2+4=0\end{cases}}\)

+)  \(a+b=0\) \(\Rightarrow\)\(x-2+3-y=0\)\(\Rightarrow\)\(x-y=-1\)\(\Rightarrow\)\(B=-1\)

+)  \(a^2-ab+b^2+4=0\)\(\Leftrightarrow\)\(\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}+4=0\)  (vô lí)

Vậy  \(B=-1\)

p/s: tham khảo nhé

cảm ơn bạn nha %%%
 

mk bổ sung AH là đường cao:

(hình hơi xấu, thông cảm nhé)

\(\Delta ABC\)vuông tại A nên ta có:  \(AB^2+AC^2=BC^2\)

  \(\frac{AB}{AC}=\frac{5}{12}\) \(\Rightarrow\)\(\frac{AB}{5}=\frac{AC}{12}\)\(\Rightarrow\)\(\frac{AB^2}{25}=\frac{AC^2}{144}=\frac{AB^2+AC^2}{25+144}=\frac{26^2}{169}=4\)

suy ra:  \(\frac{AB^2}{25}=4\)\(\Rightarrow\)\(AB=10\)

             \(\frac{AC^2}{144}=4\)\(\Rightarrow\)\(AC=24\)

Áp dụng hệ thức lượng ta có:

        \(AB.AC=AH.BC\)\(\Rightarrow\)\(AH=\frac{AB.AC}{BC}=\frac{10.24}{26}=9\frac{3}{13}\)

        \(AB^2=BH.BC\)\(\Rightarrow\)\(BH=\frac{AB^2}{BC}=\frac{10^2}{26}=3\frac{11}{13}\)

       \(HC=BC-BH=26-3\frac{11}{13}=22\frac{2}{13}\)

\(A=\left(\frac{a^2-a}{a-1}-1\right)\left(\frac{a^2+a}{a+1}+1\right)\)

\(\Leftrightarrow A=\left(\frac{a\left(a-1\right)}{a-1}-1\right)\left(\frac{a\left(a+1\right)}{a+1}+1\right)\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)\)

\(\Leftrightarrow A=a^2-1\)

Vậy \(A=a^2-1\)

Xin 1 slot xíu nữa làm giờ đang bận 

đừng bắt trc t hiếu à , m càng ngày càng giống t rồi đấy , đờ mờ