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a) \(x^3-0,25x=0\\ < =>x\left(x^2-0,25\right)=0\\ =>\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\sqrt{0,25}\end{matrix}\right.\)
b) \(x^2-10x=-25\\ < =>x^2-10x+25=0\\ < =>\left(x-5\right)^2=0\\ < =>x-5=0\\=>x=5\)
a) \(x^3-0,25x=0\)
\(x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x^2-0,25=0\)
\(\Leftrightarrow x=0\) hoặc \(x=0,25\) hoặc \(x=-0,25\)
b) \(x^2-10x=-25\)
\(\Leftrightarrow x\left(x-10\right)=-25\)
\(\Leftrightarrow x=-25\) hoặc \(\Leftrightarrow x-10=-25\)
\(\Leftrightarrow x=-25\) hoặc x=-15
![](https://rs.olm.vn/images/avt/0.png?1311)
d) x-0,25x=0
=>(1-0,25)x=0
=>0,75x=0
=>x=0
e)x2-10x=-25
=>x2-10x+25=0
=>x2-2x.5+52=0
=>(x-5)2=0
=>x-5=0
=>x=5
Chúc bn học giỏi nhoa!!!
Ta có : x - 0,25x = 0
=> x (1 - 0,25) = 0
Mà 1 - 0,25 = 0
Nên x = 0
Ta có : x2 - 10x = - 25
=> x2 - 10x + 25 = 0
=> x2 - 2.x.5 + 55 = 0
=> (x - 5)2 = 0
=> x - 5 = 0
=> x = 5
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a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
a)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
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a)\(x\left(x^2-0,25\right)=0\)
TH1:\(x=0\) TH2:\(x^2-0,25=0\)
\(x^2=0,25=>x=0,5\)
Vậy x E \(\hept{0,5;0}\)
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Bài 1 :
Theo giả thiết đã ra ta có :
\(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=\left(n+1\right)\left(n^2+2n\right)=n\left(n+1\right)\left(n+2\right)\) .
\(n\left(n+1\right)\left(n+2\right)\) là tích của 3 số nguyên liên tiếp nên luôn chia hết cho 6 .
Vì vậy \(n^2\left(n+1\right)+2n\left(n+1\right)\) luôn chia hết cho 6 với mọi số nguyên n ( đpcm )
Bài 2 :
Câu a : \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10.40=400\)
Câu b : \(87^2+73^2-27^2-13^2=\left(87^2-13^2\right)+\left(73^2-27^2\right)\)
\(=\left(87+13\right)\left(87-13\right)+\left(73+27\right)\left(73-27\right)=100.74+100.46\)
\(=100\left(74+46\right)=100.120=12000\)
Bài 3 :
Câu a :
\(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=\dfrac{1}{2}\)
Câu b :
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
a) \(x^3-0,25x=0\)
\(\Rightarrow x^3=\dfrac{1}{4}x\)
\(\Rightarrow x^2=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b) \(x^2-10x=-25\)
\(\Rightarrow x^2=-25+10x\)
\(\Rightarrow\left[{}\begin{matrix}x=-25+10x\\x=-\left(-25+10x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}10x-x=-25\\-10x-x=25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}9x=-25\\-11x=25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{9}\\x=-\dfrac{25}{11}\end{matrix}\right.\)