Ta có : \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+ab+a+b\right)\left(b^2+ab+a+b\right)\)

\(=\left(a+1\right)\left(a+b\right)\left(b+1\right)\left(a+b\right)=\left(ab+a+b+1\right)\left(a+b\right)^2\)

\(=\left(1+1\right)\left(a+b\right)^2=2\left(a+b\right)^2\)(đpcm)

chịu ai bt đc 90% là 2k10 mà

Cho (a-b)(b-c)(c-a) = (a+b)(b+c)(c+a) .Chứng minh a^2b + b^2c+ c^2a+ abc=0 - H

Ta có:\(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left(a^2c-ac^2+bc^2-b^2c+ab^2-a^2b\right)-\left(2abc+ac^2+a^2c+b^2c+bc^2+a^2b+ab^2\right)=0\)

\(\Leftrightarrow a^2c-ac^2+bc^2-b^2c+ab^2-a^2b-2abc-ac^2-a^2c-b^2c-bc^2-a^2b-ab^2=0\)

\(\Leftrightarrow-2a^2b-2b^2c-2ac^2-2abc=0\)

\(\Leftrightarrow-2\left(a^2b+b^2c+c^2a+abc\right)=0\)

\(\Leftrightarrow a^2b+b^2c+c^2a+abc=0\left(đpcm\right)\)

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 \(S_{BFC}=\frac{1}{2}BC.FD\)

\(\Rightarrow S_{ABC}=3S_{BFC}\left(đpcm\right)\)

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

x⁵ + x + 1 = x⁵ - x² + x² + x + 1 = x²(x³ - 1) + x² + x + 1 = x²(x - 1)(x² + x + 1) + x² + x + 1 

= (x² + x + 1)[x²(x - 1) + 1] = (x² + x + 1)(x³ - x² + 1)

x³ - 11x² + 30x = x(x² - 11x + 30) = x(x² - 6x - 5x + 30) = x[x(x - 6) - 5(x - 6)] = x(x - 5)(x - 6)

Trả lời:

x³ - 11x² + 30x

= x ( x² - 11x + 30 )

= x ( x² - 5x - 6x + 30 ) 

= x [ ( x² - 5x ) - ( 6x - 30 ) ] 

= x [ x ( x - 5 ) - 6 ( x - 5 ) ]

= x ( x - 6 ) ( x - 5 )

49 LÀ SỐ ĐÓ NHA

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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 \(\hept{\begin{cases}x=0\\y=\ne2\end{cases}}\)