Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
a)\(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
b)\(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)\)
c)\(y\left(x-z\right)+7\left(z-x\right)\)
\(=7\left(z-x\right)-y\left(z-x\right)\)
\(=\left(7-y\right)\left(z-x\right)\)
\(=\left(x-5\right)\left(x+3\right)\)
d)\(36-12x+x^2\)
\(=x^2-12x+36\)
\(=\left(x-6\right)^2\)
Bài 1:
a)\(2x\left(x^2-7x-3\right)=2x^3-14x^2-6x\)
b)\(\left(-2x^3+34y^2-7xy\right)\cdot4xy^2=136xy^4-28x^2y^3-8x^4y^2\)
c)\(\left(x^2-2x+3\right)\left(x-4\right)\)
\(=x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)\)
\(=x^3-4x^2-2x^2+8x+3x-12\)
\(=x^3-6x^2+11x-12\)
d)\(\left(2x^3-3x-1\right)\left(5x+2\right)\)
\(=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)\)
\(=10x^4-15x^2-5x+4x^3-6x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
Sửa đề : \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2^2}{4x^2-2x}\)
\(=\frac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{2x\left(3x-2\right)}{2x\left(2x-1\right)}+\frac{3x-4}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-6x+3x+6x^2-4x+3x-4}{2x\left(2x-1\right)}\)
\(=\frac{-2x+6x^2-5}{2x\left(2x-1\right)}\)
Thay x = 1/234 vào tính là ra giá trị biểu thức nhé !!!
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Rightarrow\left(x^3+2^3\right)-x^3-2x=15\)
\(\Rightarrow x^3+8-x^3-2x=15\)
\(\Rightarrow8-2x=15\)
=>2x=8-15=-7
=>x=\(\frac{-7}{2}\)
\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]=0\)
\(\Rightarrow\left(x^2-1\right)\left[\left(x^4-2x^2+1\right)-\left(x^4+x^2+1\right)\right]=0\)
\(\Rightarrow\left(x^2-1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left(-3x^2\right)=0\)
=>x2-1=0 hoặc -3x2=0
+)Nếu x2-1=0
=>x2=1
=>x=-1 hoặc x=1
+)Nếu -3x2=0
=>3x2=0
=>x2=0
=>x=0
Vậy x=-1 hoặc x=1 hoặc x=0
a, ta có :
( x + 3 ) . ( \(x^2\) - 3x + 9 )
= \(\left(x+3\right)\) ( \(\left(x^2-3.x+3^2\right)\)
= \(x^3\) + \(3^3\) = \(x^3\) + 27
x2 - 2x - 15 = 0
x2 - 25 - 2x + 10 =0
( x2 - 25) - ( 2x -10) =0
(x-5)(x+5) - 2( x-5) =0
(x-5) ( x+5-2) =0
(x-5)(x+3)
\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
kết luận x \(\in\) { -3; 5}