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Em chuyển 9x = 8y - 31 thành 8b - 9b = 31 cho dễ làm ạ
Từ \(8b-9a=31\Rightarrow b=\frac{31+9a}{8}=\frac{32-1+8a+a}{8}\in N\)
\(\Rightarrow a-1⋮8\Rightarrow a=8k+1\left(k\in N\right)\Rightarrow b=\frac{31+72k+9}{8}=9k+5\)
\(\Rightarrow\frac{a}{b}=\frac{8k+1}{9k+5}\)Mà \(\frac{11}{17}< \frac{a}{b}< \frac{2329\Rightarrow11}{17}< \frac{8k+1}{9k+5}< \frac{23}{29} \)
+ Với \(\frac{11}{17}< \frac{8k+1}{9k+5}\Rightarrow11.\left(9k+5\right)< 17.\left(8k+1\right)\Rightarrow99k+55< 136k+17\Rightarrow37k>38\)
\(\Rightarrow k>\frac{38}{37}\Rightarrow k>1\) (1)
Với \(\frac{8k+1}{9k+5}< \frac{23}{29}\Rightarrow29.\left(8k+1\right)< 23.\left(9k+5\right)\Rightarrow232k+29< 207k+115\Rightarrow25k< 86\)
\(\Rightarrow k< \frac{86}{25}\Rightarrow k< 4\) (2)
Từ (1) và (2) suy ra \(1< k< 4\)mà \(k\in N\)nên \(k\in\left\{2;3\right\}\)
Với \(k=2\)thì \(\frac{a}{b}=\frac{17}{25}\)
Với \(k=3\)thì \(\frac{a}{b}=\frac{25}{32}\)
Vậy............
d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7
\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>x^2+11x+28=54
=>x^2+11x-26=0
=>(x+13)(x-2)=0
=>x=2 hoặc x=-13
e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)
=>x-258=0
=>x=258
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
\(a.C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}=\dfrac{x^{28}+x^{24}+...+x^8+x^4+1}{x^3\left(x^{28}+x^{24}+...+x^8+x^4+1\right)}=\dfrac{1}{x^3}\) Tại x = 2015 thì : \(C=\dfrac{1}{x^3}=\dfrac{1}{2015^3}\)
\(b.F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
\(3F=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{2011.2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{2011.2012.2013}-\dfrac{1}{2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2012.2013.2014}\)
Tới đây dễ rồi , bạn tự tính nốt .
Vì làm vậy để triệt tiêu dần mà ( dang bài kiểu ... này thường là phải triệt tiêu ) Triệu Tử Dương
2) \(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{2}-\dfrac{1}{3}\)
<=>\(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{6}\)
=>15x-3x<5
<=>12x<5
<=>x<\(\dfrac{5}{12}\)
=> S={x|x<\(\dfrac{5}{12}\)}
a: \(\Leftrightarrow\dfrac{2\left(2x+3\right)}{4x-6}-\dfrac{3}{4x-6}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5\left(4x+6-3\right)}{5\left(4x-6\right)}=\dfrac{2\left(4x-6\right)}{5\left(4x-6\right)}\)
=>5(4x+3)=2(4x-6)
=>20x+15=8x-12
=>12x=-27
hay x=-9/4
b: \(\Leftrightarrow\dfrac{x+29}{31}+1-\dfrac{x+27}{33}-1=\dfrac{x+17}{43}+1-\dfrac{x+15}{45}-1\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\)
=>x+60=0
hay x=-60
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)
b) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)
Đính chính
Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)
\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}-10=0\)
\(\Leftrightarrow(\dfrac{x-241}{17}-1)+(\dfrac{x-220}{19}-2)+(\dfrac{x-195}{21}-3)+(\dfrac{x-166}{23}-4)=0\)
\(\Leftrightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)
\(Do\) \(\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)\ne0\) \(nên\) \(để\) \(gt=0\)
\(\Leftrightarrow x-258=0\)
\(\Leftrightarrow x=258\)
\(Vậy...\)