cho hình vuông abcd và 2018 đường thẳng cùng có tính chất chia hình vuông này thành h tứ giác có tỉ số diện tích = 2/3. chứng minh rằng có ít nhất 505 đường thẳng đồng quy.
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\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ca+x^2\)
\(=4x^2-2ax-2bc-2cx+ab+bc+ca\)
\(=4x^2-2\left(a+b+c\right)x+ab+bc+ca\)
với \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\Rightarrow2x=a+b+c\)
\(\Rightarrow M=\left(a+b+c\right)^2-\left(a+b+c\right)^2+ab+bc+ca\)
\(=ab+bc+ca\)
\(x^2+\frac{81x^2}{\left(x+9\right)^2}=40^{^{\left(1\right)}}\)
\(ĐK:x\ne-9\)
\(\left(1\right)\Leftrightarrow x^2-2.x.\frac{9x}{x+9}+\frac{81x^2}{\left(x+9\right)^2}+\frac{18x^2}{x+9}=40\)
\(\Leftrightarrow\left(x-\frac{9x}{x+9}\right)^2+\frac{18x^2}{x+9}=40\)
\(\Leftrightarrow\left(\frac{x^2}{x+9}\right)^2+18.\frac{x^2}{x+9}=0\)
Đặt \(\frac{x^2}{x+9}=t\)ta có:
\(t^2-18t-40=0\)
\(\Leftrightarrow\left(t+2\right)\left(t-20\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+2=0\\t-20=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}t=-2\\t=20\end{cases}}\)
................
rồi tự thay vào nha
ĐKXĐ : \(x\ne\pm\frac{1}{2}\)
\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)
\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)
\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)
\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)
\(E=\frac{8x^3+1}{1+4x^2}\)
Study well
E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)
E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{4x^3+1}{1+4x^2}\)
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha )
\(\left(2x+5\right)^2=\left(3x-1\right)^2\)
\(2x+5=3x-1\)
\(2x-3x=-1-5\)
\(-1x=-6\)
\(x=6\)