Bài 1: Tính nhanh
a) 15.64 + 25.100 + 36.15 + 60.1000
b) 37,5. 6,5 – 7,5 .3,4 -6,6 .7,5 +3,5. 37,5
Bài 2: Tìm x
a) x.(x-2) + x - 2 = 0
b) 5x.(x-3) - x+3 =0
Bài 3 Phân tích các đa thức sau thành nhân tử
a) x3 - 2x + x
b) x2 + 2x +1
c) x2 -16
d) (2x-1)2 –(x+3)2
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\(a,x^2-xy+9x-9y\)
\(=x\left(x-y\right)+9\left(x-y\right)\)
\(=\left(x+9\right)\left(x-y\right)\)
a)\(37.x6.5-7.5x3.4-6.6x7.5+3.5x37.5\)
=\(7.5x5x6.5-7.5x3.4-6.6x7.5+3.5x7.5x5\)
=\(7.5x\left(5x6.5-3.4-6.6+3.5x5\right)\)
=\(7.5x\left(2x6.5-10\right)\)
=\(7.5x\left(13-10\right)\)
=\(7.5x3\)
=\(22.5\)
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(2x\left(x-3\right)-x+3=0\)
=>\(2x\left(x-3\right)-\left(x-3\right)=0\)
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>-x=2
=>x=-2
a) x\(^2\)- xy + 9x -9y = x( x + 9 ) - y ( x + 9 ) = (x - y ) ( x + 9 )
b) x\(^2\)- 2xy - 5x + 10y = x ( x - 5 ) - 2y ( x - 5 ) = ( x - 2y ) ( x - 5 )
d) x\(^2\)+ 4x - y\(^2\)+ 4y = x\(^2\) - y\(^2\) + 4 ( x + y ) = ( x+ y ) ( x- y ) + 4 ( x+ y )
= (x+ y ) ( x - y + 4 )
Bài 1:
a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)
b: \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)
\(=7xy\left(2x-3y+4xy\right)\)
c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)
\(=10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10x+8y\right)\)
\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)
\(=2\left(5x+4y\right)\left(x-y\right)\)
bài 2:
a: Đề thiếu vế phải rồi bạn
b: \(x^3-13x=0\)
=>\(x\left(x^2-13\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)
Bài 1:
a, $3x-6y$
$=3(x-2y)$
b, $14x^2y-21xy^2+28x^2y^2$
$=7xy(2x-3y+4xy)$
c, $10x(x-y)-8y(y-x)$
$=10x(x-y)-8y[-(x-y)]$
$=10x(x-y)+8y(x-y)$
$=(x-y)(10x+8y)$
$=2(x-y)(5x+4y)$
Bài 2:
a, Đề thiếu rồi bạn nhé.
b, \(x^3-13x=0\)
\(\Rightarrow x\left(x^2-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)