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a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
b) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
a)xy(x2+2y)=xy.x2+xy.2y
=x3y+2xy2
b)-4(6x2-xy)=-4.6x2+4.xy
=-24x2+4xy
c)4x[x2+6x-1/2]
=4x.x2+4x.6x-4x.1/2
=4x3+24x2-2x
Bạn chú thích hơi quá lố :)
Ta có :( 5x - 3y + 4z ) . ( 5x - 3y - 4z ) \(=\left(5x-3y\right)^2-16z^2\)
\(=25x^2-30xy+9y^2-16z^2\)
Mà x^2=y^2 + z^2 nên ( 5x - 3y + 4z ) . ( 5x - 3y - 4z )\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)
Học tốt !
a) \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xz\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
b) \(x^2-6x+9-9y^2\)
\(=\left(x^2-2\cdot x\cdot3+3^2\right)-\left(3y\right)^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
c) \(x^2+9x+20\)
\(=x^2+5x+4x+20\)
\(=x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(x+5\right)\left(x+4\right)\)
d) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+4-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a/\(x^3+x^2y-x^2z-xyz\)
\(=\left(x^3-x^2y\right)+\left(x^2y-xyz\right)\)
\(=x^2\left(x-z\right)+xy\left(x-z\right)\)
\(=\left(x-z\right)\left(x^2+xy\right)\)
b/\(x^2-6x+9-9y^2\)
\(=\left(x^2-6x+9\right)-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3+3y\right)\left(x-3-3y\right)\)
c/\(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=\left(x^2+4x\right)+\left(5x+20\right)\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+5\right)\left(x+4\right)\)
d/\(x^4+4\)
\(=x^4+4x^2-4x^2+4\)
\(=\left(x^2+4x^2+4\right)-4x^2\)
\(=\left(x+2\right)^2-\left(2x\right)^2\)
\(=\left(x+2-2x\right)\left(x+2+2x\right)\)
e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
e)x3-4x+14x(x-2)=0
⇔ x(x2-4)+14x(x-2)=0
⇔ x(x-2)(x+2)+14x(x-2)=0
⇔ (x-2)(x2+2x+14x)=0
⇔ x(x-2)(x+16)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)
g)x2(x+1)-x(x+1)+x(x-1)=0
⇔ (x+1)(x2-x)+x(x-1)=0
⇔ x(x+1)(x-1)+x(x-1)=0
⇔ x(x-1)(x+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
X[X - Y] + Y[X + Y]
= x2-xy+xy+y2
= x2+y2
X[X2 - Y] - X2 [X + Y] + Y[X2 - Y]
=x3-xy-x3 -x2y+ x2y-y2
= -xy-y2
~ chúc bạn học tốt ~
(2+x).(2-x) = (4-x2)
* hằng đẵng thức đáng nhớ
ý mk là nhân hai cái với nhau chứ ai cũng bt là = (4-x^2) rồi.
giống kiểu (2+x).(2+x)=4+2x+2x+x^2