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Hai bài bị trùng nhau nên các bạn nhìn ảnh hay văn bản đều như nhau ạ
c: =>x+2>0
hay x>-2
d: =>-4<=x<=3
e: =>\(x\in\varnothing\)
f: \(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -6\end{matrix}\right.\)
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a) \(\left(x-5\right)\left(2x-3^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)
b) \(2\left(3x-15\right)\left(5-x\right)=0\)
\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)
(x - 5)(2x - 32) = 0
=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)
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Vì 1/x(x+1) = 1/x - 1/(x+1) nên
vế trái = 2(1/2.3 + 1/3.4 + ... + 1/x(x+1))
=2( 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/x - 1/(x+1)
=2( 1/2 - 1/(x+1)
= x/(x+1)
Đến đây thì bạn tự thay vào rồi giải tiếp
nha
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\(\left(x^2+1\right)\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x^2+1=0\left(vô.lí.vì.x^2\ge0,1>0\right)\\x-5=0\end{matrix}\right.\\ \Rightarrow x=5\)
\(\left(x^2+1\right)\left(x-5\right)=0\)
TH1 : x^2 + 1 = 0 ( vô lí vì x^2 + 1 > 0 )
TH2 : x - 5 = 0 <=> x = 5
Vậy x = 5
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\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)
\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)
\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)
hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)
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\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
\(\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)