Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
ĐKXĐ của A:
\(x^2-25\ne0\Leftrightarrow x^2\ne25\Leftrightarrow x\ne\pm5\)
b)
Ta có:
\(A=\dfrac{2x}{x^2-25}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\left(\cdot\right)\)
c) Ta có:
\(x^2+5x=0\Rightarrow x\left(x+5\right)=0\Rightarrow x\in\left\{0;-5\right\}\)
Đối chiếu đkxđ có x=0( TM)
Thay vào (.) có:
\(A-\dfrac{4}{0-5}=\dfrac{4}{5}\)
a.
A xác định \(\Leftrightarrow x^2-25\ne0\Leftrightarrow x\ne5;x\ne-5\)
b.
\(A=\dfrac{2x}{x^2-25}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\\ =\dfrac{2x}{x^2-25}-\dfrac{5\left(x+5\right)}{x^2-25}-\dfrac{x-5}{x^2-25}\\ =\dfrac{2x-5x-25-x+5}{x^2-25}\\ =\dfrac{-4x-20}{x^2-25}\\ =\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)
c.
\(x^2+5x=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
x=0 => A = 4/5
x=-5 => A = 2/5
a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
2) \(\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\) MTC: \(\left(x-5\right)\left(x+5\right)\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5x+25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x+30}{\left(x-5\right)\left(x+5\right)}\)
3)
Thay \(x=\dfrac{4}{5}\) vào biểu thức A ta được:
\(\dfrac{-4.\left(\dfrac{4}{5}\right)+30}{\left[\left(\dfrac{4}{5}\right)-5\right]\left[\left(\dfrac{4}{5}\right)+5\right]}\)
\(=\dfrac{-3,2+30}{-4,2.5,8}\)
\(=\dfrac{26,8}{-24,36}\)
\(=\dfrac{-670}{609}\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{4}{5}\) là \(\dfrac{-670}{609}\)
a)
\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)
b)
x khác 1
c)
x khác 0; x khác 5
d) x khác 5 ; x khác -5
b: ĐKXĐ: x<>0; x<>-5
a: \(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(X+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
a: DKXĐ: x<>1; x<>-1
b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
Bài 1:
a, Ta có:
\(\dfrac{x.\dfrac{xy}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{y.\dfrac{xy}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\dfrac{x^2y}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{\dfrac{xy^2}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\left(\dfrac{x^2y}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)-\left(\dfrac{xy^2}{x-y}\right)\left(x+\dfrac{xy}{x-y}\right)}{\left(x+\dfrac{xy}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)}\)
\(=\dfrac{\dfrac{x^2y^2}{x-y}-\dfrac{x^3y^2}{\left(x-y\right)^2}-\dfrac{x^2y^2}{x-y}-\dfrac{x^2y^3}{\left(x-y\right)^2}}{xy-\dfrac{x^2y}{x-y}+\dfrac{xy^2}{x-y}-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{-\left(\dfrac{x^3y^2+x^2y^3}{\left(x-y\right)^2}\right)}{xy-\left(\dfrac{x^2y-xy^2}{x-y}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=-\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{xy-\left(\dfrac{xy\left(x-y\right)}{\left(x-y\right)}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{\dfrac{x^2y^2}{\left(x-y\right)^2}}=x+y\)
Chúc bạn học tốt!! Làm một câu mà toát cả mồ hôi!
điều kiện của x để gtrị của biểu thức đc xác định
=>\(2x+10\ne0;x\ne0:2x\left(x+5\right)\ne0\)
\(2x+5\ne0;x\ne0\)
=>\(x\ne-5;x\ne0\)
vậy đkxđ là \(x\ne-5;x\ne0\)
rút gon giống với bạn nguyen thuy hoa đến \(\dfrac{x-1}{2}\)
b,để bt =1=>\(\dfrac{x-1}{2}=1\)
=>x-1=2
=>x=3 thỏa mãn đkxđ
c,d giống như trên
1)
a) \(5x\left(x^2-3x+\dfrac{1}{5}\right)\)
\(=5x^3-15x^2+x\)
b) \(\left(x-3\right)\left(2x-1\right)\)
\(=2x^2-x-6x+3\)
\(=2x^2-7x+3\)
2)
a) \(3x^2-15xy\)
\(=3x\left(x-5y\right)\)
b) \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right)\left(x-3+y\right)\)
c) \(x^2+3x+2\)
\(=\left(x^2+x\right)+\left(2x+2\right)\)
\(=x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
bài 4
vì x2+1 >0 với mọi x , do đó GT của Q luôn xác định với mọi x
Q=\(\dfrac{2x^2-4x+5}{x^2+1}=\dfrac{\left(3x^2+3\right)+\left(2x^2-4x+2\right)}{x^2+1}\)=\(\dfrac{3\left(x^2+1\right)+2\left(x-1\right)^2}{x^2+1}=\dfrac{3\left(x^2+1\right)}{x^2+1}+\dfrac{2\left(x-1\right)^2}{x^2+1}\)=\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\)
Do (x-1)2 ≥ 0
=>2(x-1)2 ≥ 0
x2+1 ≥ 0
=>\(\dfrac{2\left(x-1\right)^2}{x^2+1}\ge0\)
=>\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\ge3\)
=> Q ≥ 3
=>GTNN của Q =3 khi
x-1=0
=>x=1
Vậy GTNN của Q =3 khi x=1
Bài 3:
a: DKDXĐ: x<>1
b: \(=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{2}{x-1}=\dfrac{x^2-2x+1}{\left(x-1\right)^2}\cdot\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+x+1}\)
c: Để C lớn nhất thì \(A=x^2+x+1_{MIN}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu = xảy ra khi x=-1/2
a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5
b: \(M=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)