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\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
a/ Do \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{3}{x}-1}-\frac{2}{x+\frac{3}{x}-3}=-1\)
Đặt \(x+\frac{3}{x}-3=a\) ta được:
\(\frac{3}{a+2}-\frac{2}{a}=-1\)
\(\Leftrightarrow3a-2\left(a+2\right)=-a\left(a+2\right)\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\)
b/ Đặt \(x^2+2x+\frac{5}{2}=a>0\)
Phương trình trở thành:
\(\frac{1}{\left(a-\frac{1}{2}\right)^2}+\frac{1}{\left(a+\frac{1}{2}\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow4\left(a+\frac{1}{2}\right)^2+4\left(a-\frac{1}{2}\right)^2=5\left(a^2-\frac{1}{4}\right)^2\)
\(\Leftrightarrow8a^2+2=5\left(a^4-\frac{1}{2}a^2+\frac{1}{16}\right)\)
\(\Leftrightarrow5a^4-\frac{21}{2}a^2-\frac{27}{16}=0\Rightarrow\left[{}\begin{matrix}a^2=\frac{9}{4}\\a^2=-\frac{3}{20}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x+\frac{5}{2}=\frac{3}{2}\\x^2+2x+\frac{5}{2}=-\frac{3}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2+\frac{2x^2}{x^2-1}-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
Đặt \(\frac{2x^2}{x^2-1}=a\)
\(\Rightarrow a^2-a-\frac{10}{9}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{5}{3}\\a=-\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2x^2}{x^2-1}=\frac{5}{3}\\\frac{2x^2}{x^2-1}=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=-5\left(l\right)\\x^2=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{1}{2}\)
d/ĐKXĐ: ...
\(\Leftrightarrow\left(x^2+\frac{36}{x^2}\right)-13\left(x-\frac{6}{x}\right)=0\)
Đặt \(x-\frac{6}{x}=a\Rightarrow x+\frac{36}{x^2}=a^2+12\)
\(\Rightarrow a^2-13a+12=0\Rightarrow\left[{}\begin{matrix}a=1\\a=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=1\\x-\frac{6}{x}=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-6=0\\x^2-12x-6=0\end{matrix}\right.\)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
b, lấy cái dễ nhất làm cho nhanh :PP làm tương tự thôi nhé !!!
\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^2+2\right)\left(x-1\right)}=\frac{\left(4-x\right)\left(x-1\right)}{x\left(x^2+2\right)\left(x-1\right)}-\frac{x\left(x+5\right)}{x\left(x^2+2\right)\left(x-1\right)}\)
\(=\frac{4x-4-x^2+x-x^2-5x}{MTC}=\frac{-4-2x^2}{MTC}\)
\(=\frac{-2\left(2+x^2\right)}{x\left(x^2+2\right)\left(x-1\right)}=\frac{-2}{x\left(x-1\right)}\)