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Có sai đề hông má :??
Tooy sửa đề :\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\left(1\right)\)
Đặt \(t=x^2+3x+1\)ta được:
\(\left(1\right)=t\left(t+1\right)-6\)
\(=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)\)
\(=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)
\(\rightarrow[\left(x+2\right).\left(x+5\right)].[\left(x+3\right).\left(x+4\right)]-24\)
\(\rightarrow\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
\(\text{Đặt}:\)\(x^2+7x+11=a\)
\(a.\left(a+2\right)-24\)
\(\rightarrow a^2+2a-24\)
\(\rightarrow a^2+6a-4a-24\)
\(\rightarrow a.\left(a+6\right)-4.\left(a+6\right)\)
\(\rightarrow\left(a-4\right).\left(a-6\right)\)
\(\rightarrow\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\left(1\right)\)
Đặt \(t=x^2+7x+10\)ta được:
\(\left(1\right)=t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=t^2-4t+6t-24\)
\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t-4\right)\left(t+6\right)\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
hì em mới hc lp 5 anh ạ nên em ko biết xin anh thông cảm
x3 - 11x2 + 30x = x(x2 - 11x + 30) = x(x2 - 6x - 5x + 30) = x[x(x - 6) - 5(x - 6)] = x(x - 5)(x - 6)
Khó quá..............................................................................
Với \(x=0\Rightarrow y=\ne2\)
Với \(x>1\Rightarrow\)VT lẻ \(\Rightarrow y=2x+1\)
\(2^x+2=\left(2x+1\right)^2-1=4x\left(x+1\right)\)
\(\Leftrightarrow2^{x-1}+1=2x\left(x+1\right)\)
do \(x>1\Rightarrow2^{x-1}\)chẵn \(\Rightarrow\)VT lẻ , mà VP chẵn
\(\Rightarrow\)P/t vô nghiệm
Vậy p/t có nghiệm là \(\hept{\begin{cases}x=0\\y=\ne2\end{cases}}\)
a.Xét \(\Delta ADB\)và \(\Delta CAB\)có:
\(\widehat{ADB}=\widehat{CAB}=90^o\)
\(\widehat{ABC}\)chung
\(\Rightarrow\Delta ADB~\Delta CAB\left(g.g\right)\)
b.Kí hiệu: \(\widehat{ABE}=\widehat{B_1};\widehat{EBC}=\widehat{B_2}\)
Ta có:\(\widehat{B}=2\widehat{C}\)
\(\Rightarrow\widehat{B_1}=\widehat{B_2}=\widehat{C}\)
Vì \(\Delta ADB~\Delta CAB\left(g.g\right)\)
\(\Rightarrow\frac{AB}{AE}=\frac{AC}{AB}\)
\(\Rightarrow AB^2=AE.AC\)
c.Ta có:\(\Delta ABB~\Delta CAB\left(g.g\right)\)(cm câu a)
\(\Rightarrow\frac{BA}{BC}=\frac{BD}{AB}\)
Theo t/c đường p/g ta có: \(\frac{BA}{BC}=\frac{EA}{EC}\)và \(\frac{BD}{BA}=\frac{FD}{FA}\)
\(\Rightarrow\frac{FD}{FA}=\frac{EA}{EC}\left(đpcm\right)\)
d.Ta có:\(AB=2BD\left(gt\right)\)
\(\Rightarrow\frac{BD}{AB}=\frac{1}{2}\)
Mà \(\frac{BD}{AB}=\frac{FD}{FA}\)(câu c)
\(\Rightarrow\frac{BD}{AB}=\frac{FD}{FA}=\frac{1}{2}\)
\(\Rightarrow FA=2FD\)
Mà \(S_{ABC}=\frac{1}{2}BC.AD\)
và \(S_{BFC}=\frac{1}{2}BC.FD\)
\(\Rightarrow S_{ABC}=3S_{BFC}\left(đpcm\right)\)