giải phương trình, dấu lớn, bé là ngoặc nha\(\frac{1-x}{2x^2-4x}-\frac{1}{4x-4}=\frac{x-1}{2x< x-2>}-\frac{1}{2x}\)
giải phương trình sau
(2x-1)2+(2-x) (1-2x)=0
[(3-4x)(x+2)] =x2+4x+4
\(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1-\frac{x^2+x-3}{1-x}\)
giải phương trình hộ minh nha mấy bạn <3
a) \(\frac{3x-1}{x-1}-\frac{2x+5}{3}+\frac{4}{x^2-2x-3}=1\)
b) \(\frac{5}{x^2+x-6}+\frac{2}{x^2+4x+3}=\frac{-3}{2x-1}\)
c) \(\frac{4x^2+16}{x^2+16}=\frac{3}{x^2+1}+\frac{5}{x^2+3}+\frac{7}{x^2+5}\)
Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà
Giải phương trình sau: \(\frac{x^2-4x}{x^2+4x}+\frac{27}{2x^2+7x-4}=\frac{7-2x}{2x-1}-1\)
\(\frac{x^2-4x}{x^2+4x}+\frac{27}{2x^2+7x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4x\right)}{x\left(x+4x\right)}+\frac{27}{2x^2+7x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{2x^2+8x-x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{2x\left(x+4\right)-\left(x+4\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x-4}{x+4}+\frac{27}{\left(x+4\right)\left(2x-1\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\left(x-4\right)\left(2x-1\right)+27=\left(7-2x\right)\left(x+4\right)-\left(x+4\right)\left(2x-1\right)\)
\(\Leftrightarrow2x^4-9x+31=-8x+32-4x^2\)
\(\Leftrightarrow2x^2-9x+31+8x-32+4x^2=0\)
\(\Leftrightarrow6x^2-x-1=0\)
\(\Leftrightarrow6x^2+2x-3x-1=0\)
\(\Leftrightarrow2x\left(3x+1\right)-\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\left(\text{nhận}\right)\\x=\frac{1}{2}\left(\text{loại}\right)\end{cases}}\)
\(\Rightarrow x=-\frac{1}{3}\)
Vậy: nghiệm phương trình là \(-\frac{1}{3}\)
Giải các phương trình sau :
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(b,\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(c,\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)
Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)
\(\Leftrightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)
\(b,ĐKXĐ:x\ne\pm1;-3\)
Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)
\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)
\(\Leftrightarrow9x^2+14x+13=0\)
\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)
Pt vô nghiệm
\(c,ĐKXĐ:x\ne1\)
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)
Kết hợp vs ĐKXĐ được x = -1
Vậy pt có nghiệm duy nhất x = -1
làm lần lượt nha(bài nào k bt bỏ qua)
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow-2x-5=4\)
\(\Rightarrow-2x=9\)
\(\Rightarrow x=\frac{9}{-2}\)
Phần mẫu số là 2x-1 nha m.n. Còn đề bài là giải phương trình nha... m.n giúp mik nha.. mik cảm mơn nhìu
\(\frac{5}{^{x^2+x-6}}-\frac{2}{x^2+4x+3}=\frac{-3}{2x-1}\)
ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
Giải phương trình:
\(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
Phân tích : x2-3x +2=(x-1)(x-2) , x2-4x +3 = (x-1 )(x-3) , điều kiện : x # 1, x # 2 ,x # 3
pt tương đương với : \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{2x+5+x+1}{\left(x-1\right)\left(x-3\right)}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{3\left(x+2\right)}{\left(x-1\right)\left(x-3\right)}\)
<=> \(\frac{\left(x+4\right)\left(x-3\right)-3\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(1-2x\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
<=> x=0 hoặc x=1/2
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
Giải các phương trình sau:
a. \(\frac{4}{2x+3}-\frac{7}{3x-5}=0\)
b. \(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\)
c. \(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\)
d. \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)
\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(< =>4x-2+x=14x+7\)
\(< =>14x-5x=-2-7\)
\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)
Giải phương trình sau
\(\frac{2x-1}{4x^2+2x+1}\) \(-\frac{2}{2x-1}=\frac{8x+2}{1-8x^3}\)
\(\frac{2x+9}{x^2+9x+8}-\frac{2x+15}{x^2+15x+56}+\frac{2x+10}{x^2+10x+21}=\frac{4}{3}\)