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a: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
b: \(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
c: \(\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0\)
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)
b: \(=\dfrac{2x-10}{x-5}=2\)
c: \(=\dfrac{x-2-3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2}{x+2}\)
\(b,\dfrac{2x}{x-5}-\dfrac{10}{x-5}=\dfrac{2\left(x-5\right)}{x-5}=2\)
\(h,=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{-4\left(x-2\right)}{x+2}=-5\)
\(m,\dfrac{5x-15}{4x+4}:\dfrac{x^2-9}{x^2+2x+1}=\dfrac{5\left(x-3\right)}{4\left(x+1\right)}.\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
1 A lot of money have won by the postman.
2 What time can the papers be handed by the people?
3 How did the lost girl be found by the police?
4 Chịu
5 Chịu
nha
Khanh ơi, mình k cho bạn rồi đấy. Giải cho mình bài toán đó đi. Nhanh lên nhé, mình cần gấp lắm😢
\(\text{ nhìn thì thiệt là rắc rối nhưng bạn chỉ để ý 1chút là được thui.}\)
\(\text{M=1.chi tiết cách giải nha: }\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)vì\left(a+b=1\right)\)
\(M=a^3+b^3+\left(3ab\left(a^2+b^2\right)+6a^2b^2\right)\)
\(M=a^3+b^3+3ab\left(a^2+b^2+2ab\right)\)
\(M=a^3+b^3+3ab\left(a+b\right)^2\)
\(M=\left(a^3+b^3\right)+3ab\)
\(M=\left(a+b\right)\left(a^2-2ab+b^2\right)+3ab\)
\(M=a^2-ab+b^2+3ab\)
\(M=a^2+b^2+2ab=\left(a+b\right)^2=1^2=1\)
\(x^2+2xy+y^2-xz-zy\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
a: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
f: =>(3x-5)(2x-5)=0
=>x=5/3 hoặc x=5/2
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(2x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(e,2x^3+6x^2=x^2+3x\\ \Rightarrow2x^3+6x^2-x^2-3x\\ \Rightarrow2x^3+5x^2-3x=0\\ \Rightarrow x\left(2x^2+5x-3\right)=0\\ \Rightarrow x\left(2x^2+6x-x-3\right)=0\\ \Rightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\\ \Rightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(f,\left(3x-5\right)^2-x\left(3x-5\right)=0\\ \Rightarrow\left(3x-5\right)\left(3x-5-x\right)=0\\ \Rightarrow\left(3x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\)
\(g,2x^2+5x=0\\ \Rightarrow x\left(2x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)