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a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)=\left(\dfrac{1-2x+x^2}{x\left(x+1\right)}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\left(\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\right)\cdot\left(\dfrac{x}{\left(x-1\right)^2}\right)=\dfrac{\left(x-1\right)^2\cdot x}{\left(x-1\right)^2\cdot x\cdot\left(x+1\right)}=\dfrac{1}{x+1}\)
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???)
bạn ơi tại sao lại bằng 2x +6, bạn có thể giải đáp cho mình đc ko
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
a.\(\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) = \(\frac{5x+1-1+3x-2x^2+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\) =\(\frac{10x+2}{x^3-1}\)
b.\(\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{x^3+1}\)( đến đây dễ r đúng ko)
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 )
Ta có: \(\dfrac{2x+3}{1-x^2}+\dfrac{2x+1}{x^2-2x+1}\)
\(=\dfrac{-2x-3}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x+1}{\left(x-1\right)^2}\)
\(=\dfrac{\left(-2x-3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}+\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x+1\right)\cdot\left(x-1\right)^2}\)
\(=\dfrac{-2x^2+2x-3x+3+2x^2+2x+x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{2x+4}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
thực ra mình cũng cố rồi nhưng mà IQ có hạn nên nghĩ mãi ko ra, thế nên mới phải cầu cứu mấy bạn giỏi hơn đấy =)