Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{3+2x}{2+x}-1=\frac{2-x}{2+x}\) (x khác -2)
\(\Leftrightarrow\frac{3+2x}{2+x}-\frac{2-x}{2+x}=1\)
\(\Leftrightarrow\frac{1+3x}{2+x}=1\)
\(\Leftrightarrow1+3x=2+x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b) \(\frac{5-2x}{3}+\frac{x^2-1}{3}x-1=\frac{\left(x-2\right)\left(1-3x\right)}{9x-3}\) (x khác 1/3)
\(\Leftrightarrow\frac{x^3-3x+5}{3}+\frac{\left(x-2\right)\left(3x-1\right)}{3\left(3x-1\right)}=1\)
\(\Leftrightarrow\frac{x^2-2x+3}{3}=1\)
\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\frac{1}{\left(3-2x\right)^2}-\frac{4}{\left(3+2x\right)^2}=\frac{3}{9-4x^2}\) (x khác +- 3/2)
\(\Leftrightarrow\frac{\left(3+2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}-\frac{4\left(3-2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}=\frac{9}{\left(3+2x\right)^2\left(3-2x\right)^2}\)
\(\Leftrightarrow9+12x+4x^2-4\left(9-12x+4x^2\right)-9=0\)
\(\Leftrightarrow-12x^2+60x-36=0\)
\(\Leftrightarrow-12\left(x^2-5x+3\right)=0\Leftrightarrow x^2-5x+3=0\)
\(\Rightarrow\Delta=b^2-4ac=25-12=13>0\)
\(x_1=\frac{-b+\sqrt{\Delta}}{2ac}=\frac{5+\sqrt{13}}{6}\)
\(x_2=\frac{5-\sqrt{13}}{6}\)
d) \(\frac{1}{x^2+2x+1}=\frac{4}{x+2x^2+x^3}=\frac{5}{2x+2x^2}\)
\(\Leftrightarrow\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}=\frac{x^2+2x+1-\left(x+2x^2+x^3\right)+2x+2x^2}{1-4+5}\)
(dấu bằng thứ nhất của câu d là dấu cộng à???)
a) P xác định khi và chỉ khi \(\hept{\begin{cases}2x+3\ne0\\2x+1\ne0\end{cases}}\Rightarrow x\ne\frac{-3}{2};x\ne\frac{-1}{2}\)
b) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{2\left(2x+1\right)+3\left(2x+3\right)-\left(6x+5\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{4x+6}{\left(2x+3\right)\left(2x+1\right)}=\frac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(=\frac{2}{2x+1}\)
Vậy \(P=\frac{2}{2x+1}\)
c) \(P=1\Leftrightarrow\frac{2}{2x+1}=1\Leftrightarrow2x+1=2\Leftrightarrow x=\frac{1}{2}\left(tmdkxđ\right)\)
\(P=-3\Leftrightarrow\frac{2}{2x+1}=-3\Leftrightarrow2x+1=\frac{-2}{3}\Leftrightarrow x=\frac{-5}{6}\left(tmđkđ\right)\)
Vậy \(x=\frac{1}{2}\)thì P = 1; \(x=\frac{-5}{6}\)thì P = -3
d) \(P>0\Leftrightarrow\frac{2}{2x+1}>0\Leftrightarrow2x+1>0\Leftrightarrow x>\frac{-1}{2}\)
Vậy \(x>\frac{-1}{2}\)thì P > 0
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)\(=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}+\frac{10\left(x-5\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{25\left(x+10\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+25\left(x+10\right)}{5x\left(x+5\right)}=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)\(=\frac{\left(x+5\right)^2}{5\left(x+5\right)}=\frac{x+5}{5}\)
b) \(x^2-3x=0\)\(\Leftrightarrow x\left(x-3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
So sánh với ĐKXĐ, ta thấy \(x=0\)không thoả mãn
Thay \(x=3\)vào biểu thức ta được: \(P=\frac{3+5}{5}=\frac{8}{5}\)
c) Để \(P=-4\)thì \(\frac{x+5}{5}=-4\)\(\Leftrightarrow x+5=-20\)\(\Leftrightarrow x=-25\)( thoả mãn ĐKXĐ )
Vậy \(P=-4\)\(\Leftrightarrow x=-25\)
d) Để \(P\ge0\)thì \(\frac{x+5}{5}\ge0\)\(\Leftrightarrow x+5\ge0\)( vì \(5>0\))\(\Leftrightarrow x\ge-5\)
So sánh với ĐKXĐ, ta thấy x phải thoả mãn \(x>-5\)và \(x\ne0\)
Vậy \(P\ge0\)\(\Leftrightarrow\)\(x>-5\)và \(x\ne0\)
a) ĐKXĐ : 9x2 - 16 # 0
=> ( 3x - 4)( 3x + 4) # 0
=> x # \(\dfrac{4}{3}\); x # \(-\dfrac{4}{3}\)
Vậy,...
b) ĐKXĐ : x2 - 4x + 4 # 0
=> ( x - 2)2 # 0
=> x # 2
Vậy,...
c) ĐKXĐ : x2 - 1# 0
=> x # 1 ; x # -1
vậy,..
d) ĐKXĐ : 2x2 - x # 0
=> x( 2x - 1) # 0
=> x # 0 ; x # \(\dfrac{1}{2}\)
Vậy,...
a,\(\dfrac{x^2-4}{9x^2-16}\)
Phân thức trên được xác định \(\Leftrightarrow9x^2-16\ne0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4\ne0\\3x+4\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{4}{3}\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)
Vậy...
b,\(\dfrac{2x-1}{x^2-4x+4}\)
Phân thức trên được xác định \(\Leftrightarrow x^2-4x+4\ne0\)
\(\Leftrightarrow\left(x-2\right)^2\ne0\)
\(\Leftrightarrow x-2\ne0\)
\(\Leftrightarrow x\ne2\)
c,\(\dfrac{x^2-4}{x^2-1}\)
Phân thức trên được xác định \(\Leftrightarrow x^2-1\ne0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Vậy...
d,\(\dfrac{5x-3}{2x^2-x}\)
Phân thức trên được xác định \(\Leftrightarrow2x^2-x\ne0\)
\(\Leftrightarrow x\left(2x-1\right)\ne0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\2x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
Vậy...
a) \(\frac{2x+1}{x-1}\)=\(\frac{5\left(x-1\right)}{x+1}\):dkxd x\(\ne\)\(\pm\)1
\(\Rightarrow\)(2x+1)(x+1)=5(x-1)2
\(\Leftrightarrow\)2x2+2x+x+1=5(x2-2x+1)
\(\Leftrightarrow\)2x2+2x+x+1=5x2-10x+5
\(\Leftrightarrow\)2x2+2x+x+1-5x2+10x-5=0
\(\Leftrightarrow\)-3x2+13x-4=0
\(\Leftrightarrow\)-3x2+12x+1x-4=0
\(\Leftrightarrow\)-4x(x-4)+(x-4)=0
\(\Leftrightarrow\)(x-4)(-4x+1)=0
\(\Leftrightarrow\)x-4=0 hoac -4x+1=0
\(\Leftrightarrow\)x=4(tmdkxd) \(\Leftrightarrow\)x=1/4(tmdkxd)
vay s={4;1/4}
b)\(\frac{x}{x-1}\)-\(\frac{2x}{x^{ }2^{ }-1}\)=0 dkxd x\(\ne\)\(\pm\)1
\(\Leftrightarrow\)\(\frac{x\left(X+1\right)-2x^{ }}{\left(x-1\right)\left(x+1\right)}\)=0
\(\Rightarrow\)x2+x-2x=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)x=0 hoac x-1=0
\(\Leftrightarrow\)x=0(tmdkxd)\(\Leftrightarrow\)x=1(ktmdkxd)
vay s={0}
c.\(\frac{1}{x-2}\)+3=\(\frac{x-3}{2-x}\) dkxd x\(\ne\)2
\(\Leftrightarrow\)\(\frac{1}{x-2}\)+3=\(\frac{-\left(x-3\right)}{x-2}\)
\(\Leftrightarrow\)\(\frac{1+3\left(x-2\right)}{x-2}\)=\(\frac{-x+3}{x-2}\)
\(\Rightarrow\)1+3x-6=-x+3
\(\Leftrightarrow\)4x=8
\(\Leftrightarrow\)x=2(ktmdkxd)
vay s=\(\varnothing\)
chuc ban hoc tot
a.\(\frac{2x+1}{x-1}\) = \(\frac{5\left(x-1\right)}{x+1}\)
\(\leftrightarrow\) 2x+1 = 5x - 5
\(\leftrightarrow\) 2x - 5= -1-5
\(\leftrightarrow\) -3x = -6
\(\leftrightarrow\) x =2
Vậy S=\(\left\{2\right\}\)
b.\(\frac{x}{x-1}\) - \(\frac{2x}{x^2-1}\) =0
\(\leftrightarrow\) \(\frac{x}{x-1}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\)= 0 (ĐK : x\(_{\ne}\) -1 và 1)
\(\leftrightarrow\)\(\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\) =0
\(\leftrightarrow\) x2 + x -2x = 0
\(\leftrightarrow\)(x2 + x) -2x =0
\(\leftrightarrow\)x(x+1) -2x =0
\(\leftrightarrow\) x =0 -> x=0
x+1 =0 -> x = -1(Loại)
-2x = 0 -> x= 2(TM)
Vậy x =\(\left\{0,2\right\}\)
(BẠN NHỚ COI LẠI CÁI CÂU TRẢ LỜI Ở CUỐI MỖI BÀI NHA ,MÌNH KO CHẮC CÂU TRẢ LỜI ĐÓ )
a) ĐKXĐ: \(x\ne3;x\ne\pm2\)
\(C=\frac{2a-a^2}{a+3}\cdot\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(C=\frac{-a^2+2a}{a+3}\cdot\left(-\frac{4a}{a-2}\right)\)
\(C=-\frac{2a-a^2}{a+3}\cdot\frac{4a}{a-2}\)
\(C=-\frac{\left(2a-a^2\right)\cdot4a}{\left(a+3\right)\left(a-2\right)}\)
\(C=\frac{4a^2}{a+3}\)
b) \(C=\frac{4.4^2}{4+3}=\frac{46}{7}\)
c) \(\frac{4a^2}{a+3}=1\)
<=> 4a2 = a + 3
<=> 4a2 - a - 3 = 0
<=> 4a2 - 3a - 4a - 3 = 0
<=> a(4a + 3) - (4a + 3) = 0
<=> (4a + 3)(a - 1) = 0
<=> 4a + 3 = 0 hoặc a - 1 = 0
<=> a = -3/4 hoặc a = 1
a) đk: \(x\ne\pm3\)
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(-\frac{21}{9-x^2}-\frac{\left(x-4\right)\left(3+x\right)}{9-x^2}-\frac{\left(x-1\right)\left(3-x\right)}{9-x^2}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\frac{-6-3x}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3\left(x+2\right)}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3}{3-x}\)
b) \(\left|2x+1\right|=5\Leftrightarrow\left[\begin{matrix}2x+1=-5\\2x+1=5\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\left(loại\right)\\x=2\end{matrix}\right.\)
\(B\left(2\right)=-\frac{3}{3-2}=-3\)
c) \(B=-\frac{3}{5}\Leftrightarrow-\frac{3}{3-x}=-\frac{3}{5}\Leftrightarrow3-x=5\Leftrightarrow x=-2\)
d) \(B< 0\Leftrightarrow-\frac{3}{3-x}< 0\Leftrightarrow3-x>0\Leftrightarrow x< 3\)
a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)