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22 tháng 10 2021
\(14x\left(x+3\right)-7x^2\left(3+x\right)=\left(x+3\right)\left(14x-7x^2\right)=7x\left(2-x\right)\left(x+3\right)\\ \left(2x-5\right)\left(3+4x\right)-4x+10=\left(2x-5\right)\left(3+4x-2\right)=\left(2x-5\right)\left(4x+1\right)\)
NN
0
15 tháng 6 2022
\(=\dfrac{2}{x\left(x+2\right)}+\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{4}{\left(x+5\right)\left(x+9\right)}+\dfrac{1}{x+9}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+9}+\dfrac{1}{x+9}\)
=1/x
24 tháng 6 2021
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\)
\(=10x-6x^3-60-10x\)
\(=\) \(-6x^3-60\)
PT
24 tháng 6 2021
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)
b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)
c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\)
27 tháng 11 2016
bac hai thi bien doi ve tong binh phuong
\(A=\left(x^2-2.3x+9\right)+\left(y^2+2.\frac{5}{2}y+\frac{25}{4}\right)+\left(1-9-\frac{25}{4}\right)\)cu ep vao BP thua de ra ngoai
\(A=\left(x-3\right)^2+\left(y+\frac{5}{2}\right)^2+\left(1-9-\frac{25}{4}\right)\)
\(A\ge\left(1-9-\frac{25}{4}\right)\)co tinh de nguyen cac gia tri them bot de ban de hieu
dang thuc khi x=3; y=-5/2
NC
1
17 tháng 11 2022
a: \(=\dfrac{3x^3+4x^2+6x^2+8x+6x+8-5}{3x+4}\)
=x^2+2x+2-5/3x+4
b: \(\Leftrightarrow x^3-2x^2-x^2+2x+3x-6+m+4⋮x-2\)
=>m+4=0
=>m=-4
VN
3
18 tháng 6 2019
a) \(2x^2=x\Rightarrow2x^2-x=0\Rightarrow x\left(2x-1\right)\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
b) \(x^2-\frac{1}{36}=0\Rightarrow\left(x+\frac{1}{6}\right)\left(x-\frac{1}{6}\right)=0\Rightarrow\left[{}\begin{matrix}x+\frac{1}{6}=0\\x-\frac{1}{6}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=\frac{1}{6}\end{matrix}\right.\)
c) \(x^2-14x+49=0\Rightarrow\left(x-7\right)^2=0\Rightarrow x-7=0\Rightarrow x=7\)
d) \(x^3-3x^2+3x-1=0\Rightarrow\left(x^3-1\right)-\left(3x^2-3x\right)=0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x^2+x+1-3x\right)=0\Rightarrow x-1=0\Rightarrow x=-1\)
e) \(x\left(x-2\right)-5x+10=0\Rightarrow x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
TP
18 tháng 6 2019
a) \(2x^2=x\)
\(\Leftrightarrow2x^2-x=0\)
\(\Leftrightarrow2x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x^2-\frac{1}{36}=0\)
\(\Leftrightarrow x^2-\left(\frac{1}{6}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{6}\right)\left(x+\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{6}=0\\x+\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
c)\(\text{ x²-14x+49=0}\)
\(\Leftrightarrow x^2-2.7.x+7^2=0\)
\(\Leftrightarrow\left(x-7\right)^2=0\)
\(\Leftrightarrow x-7=0\)
\(\Leftrightarrow x=7\)
d)\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow(x-1)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(\text{e) x(x-2) -5x+10=0}\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
G
0
18 tháng 6 2019
a,2x^2=x
=>2x^2 -x=0
=>x(2x-1)=0
th1 x=0
th2 2x-1=0 => 2x=1 =>x=1/2
18 tháng 6 2019
b,x^2-1/36=0
<=> x^2-(1/6)^2=0
<=> (x-1/6)(x+1/6)=0
th1 x-1/6=0 =>x=1/6
th2 x+1/6=0 =>x=-1/6
dễ :>>>
\(-14x=10\)
\(x=10:\left(-14\right)\)
\(x=\frac{-5}{7}\)