Rút gọn biểu thức: \(A=\left|x+0,8\right|-\left|x-25\right|+1,9\)với \(x< -0,8\)
Rút gọn biểu thức A = |x + 0,8| - |x - 2,5| + 1,9. khi x < -0,8
A. -1,4
B. 3,6
C. 0,2
D. 5,2
Rút gọn biểu thức A=|x+0,8|-|x-2,5|+1,9 khi x< -0,8
\(x< -0,8\Leftrightarrow\left\{{}\begin{matrix}x+0,8< 0\\x-2,5< -3,3< 0\end{matrix}\right.\\ \Leftrightarrow A=-x-0,8+x-2,5+1,9=1,4\)
Rút gọn biểu thức
B= |x+0,8| - |x-2,5| + 1,9 với x <-0,8
Rút gọn biểu thức
\(A=\dfrac{x^2}{5x+25}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
\(A=\dfrac{x^2}{5x+25}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{5x+50}{x\left(x+5\right)}\)
\(=\dfrac{x^3+2\cdot5\left(x-5\right)\left(x+5\right)+5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)^2}{5\left(x+5\right)}=\dfrac{x+5}{5}\)
\(A=\dfrac{x^2}{5x+25}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\left(ĐKXĐ:x\ne0;x\ne-5\right)\)
\(A=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(A=\dfrac{x^2.x}{5x\left(x+5\right)}+\dfrac{2.5\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{5\left(50+5x\right)}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10.\left(x^2-25\right)}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10x^2-250}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3+10x^2-250+250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(A=\dfrac{\left(x+5\right)^2}{5\left(x+5\right)}\)
\(A=\dfrac{x+5}{5}\)
Rút gọn biểu thức:
\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)
a)rút gọn biểu thức \(\left(x+2\right)^2-\left(x-1\right).\left(x+1\right)\)
Rút gọn biểu thức: \(A = {\log _2}\left( {{x^3} - x} \right) - {\log _2}\left( {x + 1} \right) - {\log _2}\left( {x - 1} \right)\,\,\,\,\left( {x > 1} \right).\)
\(A=log_2\left(x^3-x\right)-log_2\left(x+1\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x^3-x}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)=log_2x\)
Rút gọn biểu thức sau:
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^2+3\left(x-1\right)\left(x+1\right)\)
b, \(\left(x^4-5x^2+25\right)\left(x^2+5\right)-\left(2+x^2\right)^2+3\left(1+x^2\right)^2\)
Rút gọn biểu thức Q với x =5; và x =– 5
Q=\(\left(\dfrac{1}{x+5}+\dfrac{1}{x-5}\right):\dfrac{2x}{x^2-25}\)
Ta có: \(Q=\left(\dfrac{1}{x+5}+\dfrac{1}{x-5}\right):\dfrac{2x}{x^2-25}\)
\(=\left(\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}+\dfrac{x+5}{\left(x-5\right)\left(x+5\right)}\right):\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x-5+x+5}{\left(x+5\right)\left(x-5\right)}:\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x}{\left(x+5\right)\left(x-5\right)}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{2x}\)
\(=1\)
Có: \(x^2-25=\left(x-5\right)\left(x+5\right)\)
ĐKXĐ của Q là x ≠ 5; x ≠ -5
Mà theo đề: x = 5; x = -5
=> Ko có giá trị của Q tìm đc