K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
16 tháng 8 2021
\(a,4x^2-6x=2x\left(2x-3\right)\\ b,9x^4y^3+3x^2y^4=3x^2y^3\left(2x^2+y\right)\\ c,x^3-2x^2+5x=x\left(x^2-2x+5\right)\\ d,3x\left(x-1\right)+5\left(x-1\right)=\left(3x+5\right)\left(x-1\right)\\ e,2x^2\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(2x^2+4\right)=2\left(x+1\right)\left(x^2+2\right)\\ f,2x^2y-4xy^2+6xy=2xy\left(x-y+3\right)\\ g,4x^3+4x^2+4x=4x\left(x^2+x+1\right)\\ h,x^3+x^2-3x-27=x^3-3x^2+4x^2-12x+9x-27=x^2\left(x-3\right)+4x\left(x-3\right)+9\left(x-3\right)=\left(x^2+4x+9\right)\left(x-3\right)\\ i,4x^2-12x+9=\left(2x-3\right)^2\\ k,8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\\ l,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
Tick nha 😘
16 tháng 8 2021
a: Ta có: \(2x+3=x+1\)
\(\Leftrightarrow2x-x=1-3\)
hay x=-2
b: Ta có: \(2x\left(2x-1\right)-\left(2x+3\right)^2=5\)
\(\Leftrightarrow4x^2-2x-4x^2-12x-9=5\)
\(\Leftrightarrow-14x=14\)
hay x=-1
c: Ta có: \(4x^2-25\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5x-10\right)\left(2x+5x+10\right)=0\)
\(\Leftrightarrow\left(-3x-10\right)\left(7x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=-\dfrac{10}{7}\end{matrix}\right.\)
16 tháng 8 2021
d: Ta có: \(2x^2+7x+5=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e: Ta có: \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
f: Ta có: \(\dfrac{1}{9}x^3-x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{9}x^2-1\right)=0\)
\(\Leftrightarrow x\left(\dfrac{1}{3}x-1\right)\left(\dfrac{1}{3}x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
g: Ta có: \(x^3+3x^2+3x=7\)
\(\Leftrightarrow\left(x+1\right)^3=8\)
\(\Leftrightarrow x+1=2\)
hay x=1
18 tháng 10 2021
b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x-4\right)\left(x-1+2x+4\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
QM
2
QM
1
QM
1
QM
1
1 tháng 7 2023
a: (x-4)(x+5)>0
=>x-4>0 hoặc x+5<0
=>x>4 hoặc x<-5
b: (2x+1)(x-3)<0
=>2x+1>0 và x-3<0
=>-1/2<x<3
c: (x-7)(3-x)<0
=>(x-7)(x-3)>0
=>x>7 hoặc x<3
d: x^2+6x-16<0
=>(x+8)(x-2)<0
=>-8<x<2
e: 3x^2+7x+4<0
=>3x^2+3x+4x+4<0
=>(x+1)(3x+4)<0
=>3x+4>0 và x+1<0
=>-4/3<x<-1
f: 5x^2-9x+4>0
=>(x-1)(5x-4)>0
=>x>1 hoặc x<4/5
g: x^2+6x-16<0
=>(x+8)(x-2)<0
=>-8<x<2
h: x^2+4x-21>0
=>(x+7)(x-3)>0
=>x>3 hoặc x<-7
i: x^2-9x-22<0
=>(x-11)(x+2)<0
=>-2<x<11
l: 16x^2+40x+25<0
=>(2x+5)^2<0(loại)
m: 3x^2-4x-4>=0
=>3x^2-6x+2x-4>=0
=>(x-2)(3x+2)>=0
=>x>=2 hoặc x<=-2/3
18 tháng 10 2021
a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
b: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)
d: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
NT
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\) (đk \(x\)≠ -1; 1)
\(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)
\(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);
\(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{1}{x^2-1}\) = \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)
2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\) - 2)
\(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\) đk \(x\) ≠ 2; -2
\(\dfrac{x}{2x-4}\) = \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\)
\(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)
\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)
1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)
2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)
3) \(3x^3-4x+1=\left(x-1\right)\left(3x^2+3x-1\right)\)
4) \(x^4-3x^2+3x-1=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
1: Ta có: \(x^3-8x+7\)
\(=x^3-x-7x+7\)
\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-7\right)\)
2: Ta có: \(x^3+8x^2-9\)
\(=x^3-x^2+9x^2-9\)
\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+9x+9\right)\)
3: Ta có: \(3x^3-4x+1\)
\(=3x^3-3x-x+1\)
\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^2+3x-1\right)\)
4: Ta có: \(x^4-3x^2+3x-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-3x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x^3+x+x^2+1-3x\right)\)
\(=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)