( x - 1) : \(\frac{1}{2}\) = \(\frac{1}{2}\) + \(1\frac{1}{2}\)
Tìm x biết :
\(\frac{1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}}-x=\frac{\frac{\frac{\frac{2}{3}+1}{3}+1}{3}+1}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}\)
\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)
\(\frac{1+\frac{1+\frac{5}{4}}{2}}{1+\frac{2}{1+\frac{6}{5}}}-x=\frac{\frac{\frac{\frac{5}{3}}{3}+1}{3}+1}{\frac{3}{\frac{\frac{3}{2}+1}{\frac{5}{2}}}+1}\)
\(\frac{1+\frac{9}{\frac{4}{2}}}{1+\frac{2}{\frac{11}{5}}}-x=\frac{\frac{\frac{14}{9}}{3}+1}{\frac{3}{1}+1}\)
Thực hiện các phép tính sau:
\(a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x + 1}} - \frac{3}{{x + 2}}\)
\(b)\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\)
\(\begin{array}{l}a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x - 1}} - \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} - \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} - \frac{2}{{x - 1}}} \right) + \left( {\frac{3}{{x + 2}} - \frac{3}{{x + 2}}} \right)\\ = 0 + \frac{2}{{x + 1}} - \frac{2}{{x - 1}} + 0\\ = \frac{{2\left( {x - 1} \right) - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{2{\rm{x}} - 2 - 2{\rm{x}} - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\end{array}\)
\(\begin{array}{l}b)\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{{x - 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} - 9}} - \frac{3}{{x + 3}}} \right)\\ = 0 + 0 + \frac{3}{{\left( {x + 3} \right)\left( {x - 3} \right)}} - \frac{3}{{x + 3}}\\ = \frac{{3 - 3\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{12 - 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\end{array}\)
Rút gọn biểu thức sau:
a) \(\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\)
b) \(\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\)
c) \(\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\)
d) \(1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\)
a)
\(\begin{array}{l}\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{2}{{3{\rm{x}}}} - \frac{x}{{1 - x}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4\left( {1 - x} \right) - 6{{\rm{x}}^2} + 3\left( {6{{\rm{x}}^2} - 4} \right)}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4 - 4{\rm{x}} - 6{{\rm{x}}^2} + 18{{\rm{x}}^2} - 12}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{12{{\rm{x}}^2} - 4{\rm{x}} - 8}}{{6{\rm{x}}\left( {1 - x} \right)}}\end{array}\)
b)
\(\begin{array}{l}\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{x}{{{x^3} - 1}}\end{array}\)
c)
\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{ - 4{\rm{x - 2}}}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{\left( { - 4{\rm{x}} - 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 8{\rm{x}} - 2{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 6{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {4{{\rm{x}}^2} - 1} \right)}}\end{array}\)
d)
\(\begin{array}{l}1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}\\ = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}\\ = 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}\\ = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\\ = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\\ = \frac{1}{{{x^2} + 1}}\end{array}\)
Tính giá trị của :
D=\(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\right)\)
Đặt \(a=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}\)
\(b=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\)
Khi đó : \(D=ab-\left(b+1\right)\left(a-1\right)\)
\(\Rightarrow D=ab-\left(ab+a-b-1\right)\)
\(\Rightarrow D=b-a+1=\frac{1}{2020^2}-1+1=\frac{1}{2020^2}\)
help me
1) \(\frac{3-7x}{1+x}=\frac{1}{2}\)
2)\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\)
3) \(\frac{y-1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)
4) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2+1}\)
5) \(1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
6) \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
7) \(\frac{2x}{x+2}-\frac{x}{x-2}=\frac{-4x}{x^2-4}\)
8) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
9) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
10) \(\frac{2x-3}{x+2}-\frac{x+2}{x-2}=\frac{2}{x^2-4}\)
11)\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
12)\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
bài 1
\(ĐKXĐ:1+x\ne0\Rightarrow x\ne-1\)
\(\frac{3-7x}{1+x}=\frac{1}{2}\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\\
\Leftrightarrow-14x-x=1-6\\
\Leftrightarrow-15x=-5\\
\Leftrightarrow x=\frac{1}{3}\left(N\right)\)
giải pt
1,\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
2,\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
3,\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
4,\(\frac{2x}{x-1}+\frac{4}{x^2+2x-3=}=\frac{2x-5}{x+3}\)
5,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
6,\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
7,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
Bài 1:
ĐKXĐ: x≠1
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 2:
ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)
Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(tm)
Vậy: x=-4
Bài 3:
ĐKXĐ: x≠1; x≠-1
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x-\frac{3x\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)
\(\Leftrightarrow-6x^2+10x=0\)
\(\Leftrightarrow2x\left(-3x+5\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)
Bài 4:
ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)
\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)
\(\Leftrightarrow13x-1=0\)
\(\Leftrightarrow13x=1\)
hay \(x=\frac{1}{13}\)(tm)
Vậy: \(x=\frac{1}{13}\)
Bài 5:
ĐKXĐ: x≠1; x≠-2
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)
\(\Leftrightarrow x+2-7x+7-3=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Leftrightarrow-6\left(x-1\right)=0\)
Vì -6≠0
nên x-1=0
hay x=1(ktm)
Vậy: x∈∅
Bài 6:
ĐKXĐ: x≠4; x≠2
Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 7:
ĐKXĐ: x≠1; x≠-2; x≠-1
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)
\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)
\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)
\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)
\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)
\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)
Còn lại tương tự mà làm nhé!
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
bài 1: tính
\(A=\frac{4^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(B=\frac{3x+1}{x^2-2x+1}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)
\(C=\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x2}\right)\)
\(D=\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right)\)
\(E=\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(F=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)
Mấy thánh ơi giúp em với mai phải nộp rồi!!!!!!!!!!!!!!!
cho x,y,z là các số thực dương thỏa mãn\(xy+yz+zx=1\). Chứng minh rằng \(\text{x/căn(1+x^2)+y/căn(1+y^2)+z/căn(1+z^2)+1/x^2+1/y^2+1/z^2>=21/2}\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
\(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)