ghi nhầm giá trị nha mấy chế

a) \(P=x^2-2x+5\)

\(P=x^2-2x+1+4\)

\(P=\left(x-1\right)^2+4\ge4\)

Dấu '' = '' xảy ra khi 

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Vậy ......................

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)

                           \(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2ac+2bc-3ac-3bc-3ab\right)+3abc\)

                           \(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ac\right)\right]+3abc\)

Ta có: \(a+b+c⋮3\Rightarrow a^3+b^3+c^3⋮3\)

\(\RightarrowĐPCM\)

$a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc$

                           $=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2ac+2bc-3ac-3bc-3ab\right)+3abc$

                           $=\left(a+b+c\right)\left[{\left(a+b+c\right)}^2-3\left(ab+bc+ac\right)\right]+3abc$

Ta có: $a+b+c⋮3\Rightarrow a^3+b^3+c^3⋮3$

\(\RightarrowĐPCM\)

a) \(36-4x^2+4xy-y^2\)

\(=36-\left(2x-y\right)^2\)

\(=\left(6+2x-y\right)\left(6-2x+y\right)\)

b) \(2x^4+3x^2-5\)

\(=2x^4-2x^2+5x^2-5\)

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

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

thank bn

1p=1n xấp xỉ=1 đvC

C nặng 12 đvC

C nặng 1,9926 nhân 10^-23

 tầm 1/3600 khoi luong ca nguyen tu

\(x^2-408x+2015=\left(x^2-5x\right)-\left(403x-2015\right)\)

\(=x\left(x-5\right)-403\left(x-5\right)=\left(x-5\right)\left(x-403\right)\)

a) x² - 408x + 2015

= x² - 403x - 5x + 2015

= ( x² - 403x ) - ( 5x - 2015 )

= x( x - 403 ) - 5( x - 403 )

=( x - 403 )( x - 5 )

b) x² - ( a² + b² )x + a²b²

= x² - a²x - b²x + a²b²

= ( x² - a²x ) - ( b²x - a²b² )

= x( x - a² ) - b²( x - a² )

= ( x - a² )( x - b² )

Ta có :

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\)

\(\Rightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b}{a+b}-\frac{b}{b+c}+\frac{d}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)+d\left(a+b\right)\left(b+c\right)=0\)( vì c khác a )

\(\Leftrightarrow abc-acd+bd^2-b^2d=0\)

\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow ac-bd=0\)

\(\Leftrightarrow ac=bd\)

\(\Rightarrow abcd=\left(ac\right)\left(bd\right)=\left(ac\right)^2\)

Vậy ......................................