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1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
Câu 1:
a: ĐKXĐ: \(x\notin\left\{0;1;\dfrac{1}{2}\right\}\)
\(B=\dfrac{x^2+x}{x^2+x+1}-\dfrac{2x^3+x^2-x-2x^3+2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}-\dfrac{-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}+\dfrac{2x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}+\dfrac{-x}{x^2+x+1}=\dfrac{x^2}{x^2+x+1}\)
b: Để \(B=\dfrac{4}{3}\) thì \(\dfrac{x^2}{x^2+x+1}=\dfrac{4}{3}\)
\(\Leftrightarrow4x^2+4x+4-3x^2=0\)
=>x=-2(nhận)
a)
2x-3=0 => x=3/2
b)
2x^2 +1 =0 => vô nghiệm
c) x^2 -25 =0 => x=5 loiaj
x=-5 nhân
d)
x^2 -25 =0 => x=5 loại
x=-5 loại
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
a: \(B=\left(\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{x-1}\right)\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3x-3-2}{x-1}\)
\(=\dfrac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-5}=\dfrac{-1}{2x-3}\)
b: Để B>0 thì 2x-3<0
hay x<3/2
a: \(=\left(\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{2x-8}{\left(2x-5\right)\left(x-4\right)}-\dfrac{3}{2x-1}\right)\cdot\dfrac{4x^2+4x-3}{-8x^2+2x+21}+1\)
\(=\dfrac{2x-3-2\left(2x-1\right)-3\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)}\cdot\dfrac{\left(2x+3\right)\left(2x-1\right)}{\left(4x-7\right)\cdot\left(-2x-3\right)}+1\)
\(=\dfrac{2x-3-4x+2-3x+12}{1}\cdot\dfrac{-1}{4x-7}+1\)
\(=\dfrac{5x-11}{4x-7}+1=\dfrac{9x-4}{4x-7}\)
b: |x|=1/2
=>x=1/2(loại) hoặc x=-1/2(nhận)
Khi x=-1/2 thì \(P=\dfrac{\dfrac{-9}{2}-4}{-2-7}=-\dfrac{17}{2}:\left(-9\right)=\dfrac{17}{18}\)
c: Để P là số nguyên thì \(36x-16⋮4x-7\)
\(\Leftrightarrow4x-7\in\left\{1;-1;47;-47\right\}\)
hay \(x\in\left\{2;\dfrac{3}{2};\dfrac{27}{2};-10\right\}\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a: DKXĐ: x<>1; x<>-1
b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
5A+4B=0
=>10/2x+1+12/2x-1=0
=>10(2x-1)+12(2x+1)=0
=>20x-10+24x+12=0
=>44x+2=0
=>x=-1/22(nhận)