Cho x< 0. Rút gọn P=\(\left|2x-\sqrt{\left(5x-1\right)^2}\right|\)
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a) Ta có: \(B=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
=1-a
b) Ta có: \(2x^2-5x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};2\right\}\)
\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)
Nhưng \(2\sqrt{x}+1\ge1\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(x\in\left\{0;4\right\}\)
a: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b: \(=3x^2-6x-5x+5x^2-8x^2+24\)
=-11x+24
7:
a: =>0,5x-5=2 hoặc 0,5x-5=-2
=>0,5x=3 hoặc 0,5x=7
=>x=6 hoặc x=14
b: |5x-2|=-3
mà |5x-2|>=0
nên ptvn
c: =>1/4x+3=0
=>1/4x=-3
=>x=-12
Bài 1:
a: Ta có: \(x^2-2\sqrt{5}x+5=0\)
\(\Leftrightarrow x-\sqrt{5}=0\)
hay \(x=\sqrt{5}\)
b: Ta có: \(\sqrt{x+3}=1\)
\(\Leftrightarrow x+3=1\)
hay x=-2