Ai đó vẽ giùm t cái hình

7) \(S_{ABCD}=12,5.6,72=84\left(cm^2\right)\)

Gọi độ dài đoạn BD là a, AC là b

\(\dfrac{a.b}{2}=84\Rightarrow ab=168\)

Áp dụng định lí Py-ta-go, ta có:

\(\left(\dfrac{1}{2}a\right)^2+\left(\dfrac{1}{2}b\right)^2=12,5^2\)

\(\dfrac{a^2+b^2}{4}=156,25\)

\(a^2+b^2=625\)

\(\left(a+b\right)^2=a^2+2ab+b^2=625+2.168=961\)

\(\Rightarrow a+b=31\)

\(a.b=168\)

\(a.\left(31-a\right)=168\)

\(31a-a^2-168=0\)

\(-\left(a^2-31a+168\right)=0\)

\(-\left[a^2-24a-7a+168\right]=0\)

\(-\left[a\left(a-7\right)-24\left(a-7\right)\right]=0\)

\(-\left(a-24\right)\left(a-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-24=0\Rightarrow a=24\Rightarrow b=7\\a-7=0\Rightarrow a=7\Rightarrow b=24\end{matrix}\right.\)

Ta có AC<BD hay b<a. Vậy chọn nghiệm đầu tiên

Độ dài đường chéo AC và BD lần lượt là 7 và 24

Gọi 4 số lần lượt là a,b,c,d

\(a+3=b-4=2c=\dfrac{d}{3}\)

\(\Rightarrow\left[{}\begin{matrix}a=2c-3\\b=2c+4\\c\\d=6c\end{matrix}\right.\)

\(a+b+c+d=67\)

\(\Leftrightarrow\left(2c-3\right)+c+\left(2c+4\right)+6c=67\)

\(11c+1=67\)

\(11c=66\)

\(c=6\)

\(\Rightarrow\left[{}\begin{matrix}a=2.6-3=9\\b=2.6+4=16\\c=6\\d=6.6=36\end{matrix}\right.\)

\(f\left(x\right)=x^3+5x^2+ax+b\)

\(f\left(-2\right)=0\Leftrightarrow12-2a+b=0\left(1\right)\)

\(f\left(3\right)=0\Leftrightarrow72+3a+b=0\left(2\right)\)

\(\left(2\right)-\left(1\right)=0\Leftrightarrow\left(72+3a+b\right)-\left(12-2a+b\right)=0\Leftrightarrow60+5a=0\Leftrightarrow5a=-60\Leftrightarrow a=-12\)

\(q\left(x\right)=x^2-2x-3=\left(x+1\right)\left(x-3\right)\)

\(f\left(x\right)⋮q\left(x\right)\) thì \(f\left(-1\right)=0\) và \(f\left(3\right)=0\)(Theo định lí Bezout)

\(f\left(-1\right)=0\Leftrightarrow1+p+q=0\left(1\right)\)

\(f\left(3\right)=0\Leftrightarrow81+9p+q=0\left(2\right)\)

\(\left(2\right)-\left(1\right)=0\Leftrightarrow\left(81+9p+q\right)-\left(1+p+q\right)=0\Leftrightarrow80+8p=0\)

\(80+8p=0\)

\(8\left(10+p\right)=0\)

\(10+p=0\Leftrightarrow p=-10\Rightarrow q=9\)

Vậy: p=-10; q=9

\(\dfrac{6x-5}{10x+1}=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{49.51}\)

\(\dfrac{6x-5}{10x+1}=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+......+\dfrac{2}{49.51}\right)\)

\(\dfrac{6x-5}{10+1}=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.....+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(\dfrac{6x-5}{10x+1}=\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)

\(\dfrac{6x-5}{10x+1}=\dfrac{25}{51}\)

\(\left(6x-5\right)51=\left(10x+1\right)25\)

\(306x-255=250x+25\)

\(306x-250x=255+25\)

\(56x=280\)

\(x=5\)

8) \(\dfrac{xy}{x^2+y^2}=\dfrac{3}{8}\Leftrightarrow8xy=3\left(x^2+y^2\right)\Leftrightarrow x^2+y^2=\dfrac{8}{3}xy\)

\(A=\dfrac{x^2+2xy+y^2}{x^2-2xy+y^2}=\dfrac{\dfrac{8}{3}xy+2xy}{\dfrac{8}{3}xy-2xy}=\dfrac{\dfrac{14}{3}xy}{\dfrac{2}{3}xy}=7\)

3) -__- \(3x^2+16y^2+12x-8xy+18=0\)

\(16y^2-8xy+x^2+2x^2+12x+18=0\)

\(\left(4y-x\right)^2+2\left(x+3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\4y-x=0\Leftrightarrow4y+3=0\Rightarrow y=\dfrac{-3}{4}\end{matrix}\right.\)

\(\Rightarrow x+y=-3,75\)

Hình ảnh chỉ mang tính chất minh họa, vẽ cho cốt hiểu

Vì ABCD là HCN nên AB=CD=8cm

Áp dụng định lí Py-ta-go vào tam giác ABX

\(BX^2=AX^2-AB^2\)

\(BX^2=17^2-8^2\)

\(BX^2=225\Rightarrow BX=15\left(cm\right)\)

Áp dụng định lí Py-ta-go vào tam giác CXD

\(CX^2=DX^2-DC^2\)

\(CX^2=10^2-8^2\)

\(CX^2=36\Rightarrow CX=6\left(cm\right)\)

\(BC=BX+CX=15+6=21\left(cm\right)\)

Vì ABCD là HCN nên BC=AD=21cm

\(S_{ADC}=\dfrac{AD.DC}{2}=\dfrac{21.8}{2}=84\left(cm^2\right)\)