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a) \(\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{\left(x-7\right)\left(x+7\right)}{2x+1}.\frac{-3}{x-7}=\frac{-3\left(x-7\right)\left(x+7\right)}{\left(2x+1\right)\left(x-7\right)}=\frac{-3\left(x+7\right)}{2x+1}\)
\(=\frac{-3x-21}{2x+1}\)
b) \(\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{\left(2-3x\right)^3}=\frac{x\left(3x-2\right)}{x^2-1}.\frac{x^4-1}{\left(3x-2\right)^3}=\frac{x\left(3x-2\right)}{x^2-1}.\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(3x-2\right)^3}\)
\(=\frac{x\left(3x-2\right)\left(x^2-1\right)\left(x^2+1\right)}{\left(x^2-1\right)\left(3x-2\right)^3}=\frac{x\left(x^2+1\right)}{\left(3x-2\right)^2}=\frac{x^3+x}{\left(3x-2\right)^2}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{x^2-4}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{x^2-4}{-x}=\frac{2\left(x-2\right)}{-\left(x+2\right)}=\frac{-2\left(x-2\right)}{x+2}\)
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
1. \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{x^2-1}\)
= \(-\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\)
= \(\frac{-x-1+x-1+2}{\left(x-1\right)\left(x+1\right)}=0\)
c) \(\left(\frac{x^2-16}{x^2+8x+16}+\frac{6}{x+4}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16}{\left(x+4\right)^2}+\frac{6\left(x+4\right)}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16+6x+24}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\frac{x^2+6x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x-2}\)
= \(\frac{x^2+4x+2x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}\)
= \(\frac{\left(x+4\right)\left(x+2\right)}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}=\frac{2x}{x+4}\)
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 \(S=\left\{\frac{3}{2}\right\}\)
Tại vì nó được đề bài cho nên có nghĩa,k có nghĩa thì lm kiểu đếch j?
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
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, \(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
\(\frac{12x}{12}-\frac{2\left(5x+2\right)}{12}=\frac{3\left(7-3x\right)}{12}\)
\(12x-10x-4=21-9x\)
\(11x=25\)
\(x=\frac{24}{11}\)
\(b,\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\frac{10x+3}{12}=\frac{15+8x}{9}\)
\(9\left(10x+3\right)=12\left(15+8x\right)\)
\(3\left(10x+3\right)=4\left(8x+15\right)\)
\(30x+9=32x+60\)
\(-2x=51\)
\(x=-\frac{51}{2}\)
\(c,\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\frac{2x}{6}-\frac{3\left(2x+1\right)}{6}=\frac{x-6x}{6}\)
\(2x-6x-3=x-6x\)
\(x=3\)
P/s: Bn xem lại đề bài phần d nha!
=.= hk tốt!!
Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
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0 7
ai trả lời đúng mk cho
\(B=\left(\frac{1}{2-x}+\frac{3x}{x^2-4}-\frac{2}{2+x}\right):\left(\frac{x+4}{4-x^2}+4\right)\)\(=\left(\frac{1}{2-x}+\frac{3x}{x^2-4}-\frac{2}{2+x}\right):\left(\frac{-x-4}{x^2-4}+4\right)\)
\(=\left(\frac{-1}{x-2}+\frac{3x}{\left(x-2\right)\left(x+2\right)}+\frac{-2}{x+2}\right):\left(\frac{-x-4+4x^2-16}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\frac{\left(-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\)\(:\left(\frac{4x^2-x-20}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\frac{-x-2+3x-2x+4}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x-2\right)\left(x+2\right)}{4x^2-x-20}\)
\(=\frac{2}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x-2\right)\left(x+2\right)}{4x^2-x-20}\)
\(=\frac{2}{4x^2-x-20}\)