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Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
Làm ngắn gọn thôi nhé :v
\(A=\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)
\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)
\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x+2}{x-3}\)
\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)
\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)
\(B=\frac{\left(-x-2\right)\left(x+2\right)\left(x+2\right)\left(x-2\right)}{\left(-x-2\right)\left(x-2\right)\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{x+2}{x-2}\)
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{10x}{-x^2+9}\)
\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)
\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)
\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)
\(D=\frac{51x-15}{2x^3-18x}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)
\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)
\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(E=\frac{10x^2+10}{x^4-2x+1}\)
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
