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\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
\(ĐKXĐ:x\ne\pm1\)
\(\frac{4}{x^3-x^2-x+1}-\frac{3}{1-x^2}=\frac{1}{x+1}\)
\(\Rightarrow\frac{4}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x+1}\)
\(\Rightarrow\frac{4}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x+1}\)
Đặt\(x+1=u;x-1=v\)
Phương trình trở thành \(\frac{4}{uv^2}+\frac{3}{uv}=\frac{1}{u}\)
\(\Rightarrow\frac{4}{uv^2}+\frac{3v}{uv^2}=\frac{v^2}{uv^2}\)
\(\Rightarrow4+3v=v^2\Leftrightarrow v^2-3x-4=0\)
Ta có \(\Delta=\left(-3\right)^2+4.1.4=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}v=\frac{3+5}{2}=4\\v=\frac{3-5}{2}=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=0\end{cases}}\)
Vậy tập nghiệm S = {0;5}
17) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow-4x^2+3x+1=0\)
\(\Leftrightarrow-\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-\frac{1}{4}\left(tm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{4}\right\}\)
18) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
19) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\\x\ne\frac{1}{2}\end{cases}}\)
\(\frac{x+4}{2x^3-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(2x-1\right)\left(x-2\right)}+\frac{x+1}{\left(2x-1\right)\left(x-3\right)}-\frac{2x+5}{\left(2x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-12+x^2-x-2-2x^2-x+10}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow x=-4\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{-4\right\}\)
20) \(ĐKXĐ:x\ne0\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}-\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)-3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow x^4+x-x^4+x-3=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{3}{2}\right\}\)
Đặt \(x^2+x+10=u\)
Phương trình trở thành: \(\frac{u-6}{2}+\frac{u-3}{3}=\frac{u+3}{5}+\frac{u+6}{6}\)
\(\Rightarrow\frac{u}{2}-3+\frac{u}{3}-1=\frac{u}{5}+\frac{3}{5}+\frac{u}{6}+1\)
\(\Rightarrow\frac{u}{2}+\frac{u}{3}-\frac{u}{5}-\frac{u}{6}=3+1+1+\frac{3}{5}\)
\(\Rightarrow u\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=\frac{28}{5}\)
\(\Rightarrow u.\frac{7}{15}=\frac{28}{5}\Rightarrow u=12\)
Lúc đó \(x^2+x+10=12\)
\(x^2+x-2=0\)
Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\)
ĐKXĐ : \(x\ne2,x\ne4\)
Pt \(\Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\) (2)
Đặt \(\frac{x+1}{x-2}=a,\frac{x-2}{x-4}=b\Rightarrow ab=\frac{x+1}{x-4}\)
Khi đó pt (2) trở thành :
\(a^2+ab-12b=0\)
\(\Leftrightarrow a^2-3ab+4ab-12b=0\)
\(\Leftrightarrow a\left(a-3b\right)+4b\left(a-3b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=3b\\a=-4b\end{cases}}\)
Bạn thay vào tính, được nghiệm là \(S=\left\{3,\frac{4}{3}\right\}\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x=0
<=> x=0
Vậy x=0
+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)
\(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x.\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,6\right\}\)
+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)
\(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)