47837894237

ko

76276712

g) \(\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x^2-2x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)^2\left(x-2\right)=0\)

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

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

h) \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

i) \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)

\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow x=0\)hoặc \(x=\frac{1}{2}\)hoặc x = -3

\(h.x^2-5x+6=0\)

\(x^2-1x-6x+6=0\)

\(x\left(x-1\right)-6\left(x-1\right)=0\)

\(\left(x-6\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

3123123

a) \(ax-bx+ab-x^2=\left(ax+ab\right)-\left(bx+x^2\right)=a\left(x+b\right)-x\left(x+b\right)=\left(x+b\right)\left(a-x\right).\)

b) \(x^2-2xy+y^2-9=\left(x^2-2xy+y^2\right)-3^2=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x+y+3\right).\)

Học tốt nhé ^3^

\(a.=a\left(x+b\right)-x\left(b+x\right)\)

\(=a\left(x+b\right)-x\left(x+b\right)\)

\(=\left(a-x\right)\left(x+b\right)\)

\(b.=\left(x-y\right)^2-3^2\)

\(=\left(x-y-3\right)\left(x-y+3\right)\)

_Moon_

Ta có: x + y = 1 => x = 1 - y

Khi đó: x³ + y³ = (x + y)(x² - xy + y²)

= 1.(x² - xy + y²) = x² + 2xy + y² - 3xy

= (x + y)² - 3xy = 1 - 3xy

=  1 - 3y(1 - y)

= 1 - 3y + 3y²

= 3(y² - y + 1/4) + 1/4

= 3(y - 1/2)² + 1/4 \(\ge\)1/4 \(\forall\)y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}y-\frac{1}{2}=0\\x=1-y\end{cases}}\) <=> \(\hept{\begin{cases}y=\frac{1}{2}\\x=1-\frac{1}{2}=\frac{1}{2}\end{cases}}\)

Vậy Min của x³ + y³ = 1/4 <=> x = y = 1/2

Ta có : 

\(x+y=1\)

\(\Leftrightarrow\left(x+y\right)^2=1\)

\(\Leftrightarrow x^2+2xy+y^2=1\)

\(\Leftrightarrow x^2+y^2=1-2xy\)

Ta lại có :

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=1-2xy-xy=1-3xy=-3xy+1\ge1\)

Dấu '' = '' xảy ra

\(\Leftrightarrow-3xy=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\y=0\end{cases}}\)

Vậy...............

Cho x+y=a ; x.y=b tính giá trị nhé ạ 

+) Ta có : 

\(x+y=a\)

\(\Leftrightarrow\left(x+y\right)^2=a^2\)

\(\Leftrightarrow x^2+2xy+y^2=a^2\)

\(\Leftrightarrow x^2+y^2=a^2-2b\)

Ta có \(x-y=4\)   =>  \(\left(x-y\right)^2=4^2\)

                                  => \(x^2-2xy+y^2=16\)

                                 => \(x^2-2xy+y^2+4xy=16+4xy\)

                                => \(x^2+2xy+y^2=16+4.3\)

                                 => \(\left(x+y\right)^2=28\)