K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a) Ta có: \(2x+x^2=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b) Ta có: \(\left(2x+1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

2 tháng 8 2021

đặt cái chung ra ngoài

 

 

a) \(5x+10y=5\left(x+2y\right)\)

b) \(3x^2y+9xy^2z=3xy\left(x+3yz\right)\)

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

h) \(x^2+9x+8=\left(x+8\right)\left(x+1\right)\)

l) \(x^2-10x+9=\left(x-1\right)\left(x-9\right)\)

k) \(x^2+x-12=\left(x+4\right)\left(x-3\right)\)

l) \(3x^2+8x+4=\left(3x+2\right)\left(x+2\right)\)

Em cần giúp câu nào hả em? Em nên chụp 1-2 ý cho 1 lần hỏi nhá, như thế mọi người sẽ dễ dàng giúp em hơn

12 tháng 5 2021

13

a, \(3x-4=-x+8\)

\(< =>3x+x=8+4\)

\(< =>4x=12\)

\(< =>x=\frac{12}{4}=3\)

b, \(\frac{2x+1}{6}+\frac{x-7}{12}=10\)

\(< =>\frac{2\left(2x+1\right)}{12}+\frac{x-7}{12}=\frac{120}{12}\)

\(< =>4x+2+x-7=120\)

\(< =>5x=120+5=125\)

\(< =>x=\frac{125}{5}=\frac{5^3}{5}=5^2=25\)

9 tháng 12 2021

\(a,=\dfrac{4xy-1-2xy+1}{5x^2y}=\dfrac{6xy}{5x^2y}=\dfrac{6}{5x}\\ b,=\dfrac{x^2+8x-2x+8}{x\left(x-4\right)\left(x+4\right)}=\dfrac{\left(x+2\right)\left(x+4\right)}{x\left(x-4\right)\left(x+4\right)}=\dfrac{x+2}{x\left(x-4\right)}\\ c,=\dfrac{x^2+3x-x+1}{x\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\\ d,=\dfrac{x-3-x-3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{3-x}\\ e,=\dfrac{x+1-1}{x+1}=\dfrac{x}{x+1}\\ f,=\dfrac{3x+5-5+9x}{6x^2y}=\dfrac{12x}{6xy}=\dfrac{2}{y}\)

9 tháng 12 2021

\(g,=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ h,=\dfrac{3x+1-3x+1+2x-3}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{2x-1}{\left(3x-1\right)\left(3x+1\right)}\\ j,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x^2+5x}{x^2+6x}\\ k,=\dfrac{\left(x-7\right)\left(x+7\right)}{2x+1}\cdot\dfrac{-3}{x-7}=\dfrac{-3\left(x+7\right)}{2x+1}\\ l,=\dfrac{x\left(3x-2\right)}{x^2-1}\cdot\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{\left(3x-2\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(3x-2\right)^2}\)

23 tháng 11 2023

a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

\(Q=\dfrac{x\left(-x^2-2x-6\right)}{x}\)

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

\(Q=-\left(x^2+2x+6\right)\)

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)