Ta có : \(ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}=k\)

          \(\Rightarrow k^2=\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\)

        \(\Rightarrow k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

        \(\Rightarrow k^2=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

Và \(k.k=\frac{a}{c}.\frac{b}{d}\)

 \(\Rightarrow k^2=\frac{ab}{cd}\left(2\right)\)

Từ (1) và (2) , ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

a +b+c+d=0

0 nha bn

\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)

                                     \(=\left(2x-1\right)^2-6^2\)

                                     \(=\left(2x-7\right)\left(2x+5\right)\)

\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)

\(8x^3-26x^2+13x+5=\)  \(8x^3-8x^2-18x^2+18x-5x+5\)

                                                \(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)

                                                \(=\)  \(\left(8x^2-18x-5\right)\left(x-1\right)\)

                                                \(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)

a)  \(x^2+6x+8\)

\(=\left(x^2-2x\right)-4x+8\)

\(=x\left(x-2\right)-4\left(x-2\right)\)

\(\left(x-2\right)\left(x-4\right)\)

b) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5y\right)\)

a) x² - 6x + 8

= x² -2x - 4x +8

= x( x-2) -4( x-2)

= ( x-2)(x-4)

Áp dụng hằng đẳng thức 1:

= (x+y)²+1

Vì (x+y)² > hoặc=0 với mọi x thuộc R

=> (x+y)²+1 >=1 với mọi x thuộc R

=> Amax=1

Theo đề bài có :

\(a^2+b^2+c^2=ab+bc+ac\)

Ta lại có :

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

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

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

\(\Leftrightarrow a-b=b-c=a-c=0\)

\(\Rightarrow a=b=c\)(đpcm)

Ta có :

\(a+b+c=0\)

\(\Rightarrow a=-\left(b+c\right)\)

\(\Rightarrow a^2-b^2-c^2=2bc\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+a^2c^2+b^2c^2\right)\)

Cộng \(a^4+b^4+c^4\)vào \(2\left(a^2b^2 +a^2c^2+b^2c^2\right)\)

=> Đpcm