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c) Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{1-x}{x+1}+1>0\)
\(\Leftrightarrow\dfrac{1-x+x+1}{x+1}>0\)
\(\Leftrightarrow\dfrac{2}{x+1}>0\)
mà 2>0
nên x+1>0
hay x>-1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)
a) Ta có: \(A=\left(\dfrac{x+1}{x-1}-\dfrac{1-x}{x+1}+\dfrac{4x^2}{1-x^2}\right):\dfrac{2x^2-2}{x^2-2x+1}\)
\(=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{x^2+2x+1+x^2-2x+1-4x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{-2x^2+2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2\left(x+1\right)}\)
\(=\dfrac{-2\cdot\left(x-1\right)}{2\left(x+1\right)}\)
\(=\dfrac{1-x}{x+1}\)
a: ĐKXĐ: \(x\notin\left\{10;-10;\sqrt{10};-\sqrt{10}\right\}\)
b: \(A=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)
\(=\dfrac{10x^3+40}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)
a: Ta có: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2-9-x^2-3x+10=6\)
\(\Leftrightarrow-3x=5\)
hay \(x=-\dfrac{5}{3}\)
c: \(4x^2-9=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2-9-x^2-3x+10=6\\ \Leftrightarrow-3x=5\Leftrightarrow x=-\dfrac{5}{3}\\ b,\Leftrightarrow2x^2+3x^2-3=5x^2+5x\\ \Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\\ c,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(5-2x\right)^2-4=0\\ \Leftrightarrow\left(5-2x-2\right)\left(5-2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\\ e,\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(f,\Leftrightarrow\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{9}{2}\end{matrix}\right.\\ g,\Leftrightarrow\left(x^2-4\right)\left(3x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\\ h,\Leftrightarrow\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^4+2x^2+1-x^2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\)
Cko mình sửa đề 1 chút. Số trừ là = ((2x+3)/x^2-1)
Và biểu thức trên = 0 nữa ngken các bạn. Do mình lần đầu làm nên không rành, moq các bạn giúp đỡ mình nhiều hơn. ^.^
\(x^2-5\)
\(=x^2-\left(\sqrt{5}\right)^2\)
\(=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
dap an ra \(\left(x-3+\sqrt{6}\right).\left(x-3-\sqrt{6}\right)\)
Ta có : x = 99
=> 100 = x + 1
Thay vào A ta có : A = x2018 - 100x2017 + 100x2016 - ...... + 100x2 - 100x + 2019
=> A = x2018 - (x + 1)x2017 + (x + 1)x2016 - ...... + (x + 1)x2 - (x + 1)x + 2019
=> A = x2018 - x2018 - x2017 + x2017 + x2016 -.......+ x3 + x2 - x2 + x + 2019
=> A = x + 2019
=> A = 99 + 2019
=> A = 2118
P/s : ko cần ! :D
Theo đề bài ra ta có :
x = 99
Thay vào A ta có :
A = x2018 - 100x2017 + 100x2016 - ... + 100x2 - 100x + 2019
\(\Rightarrow\) A = x2018 - ( x + 1 ) x2017 + ( x + 1 ) x2016 - ... + ( x + 1 ) x2 - ( x + 1 ) x + 2019
\(\Rightarrow\) A = x2018 - x2018 - x2017 + x2017 + x2016- ... + x3 + x2 - x2 + x + 2019
\(\Rightarrow\) A = x + 2019
\(\Rightarrow\) A = 99 + 2019
\(\Rightarrow\) A = 2118