D=\(\frac{x^2+4x+7}{x^2+4}\)
C=\(\frac{x^2+7}{x+4}\)
tìm x để d,c nguyên với x\(\varepsilon\)Q
K
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16 tháng 9 2018
1 Giải :
\(\frac{3x+7}{x-1}\)là phân số <=> x - 1 \(\ne\)0 => x \(\ne\)1
Ta có : \(\frac{3x+7}{x-1}=\frac{3\left(x-1\right)+8}{x-1}=3+\frac{8}{x-1}\)
Để \(\frac{3x+7}{x-1}\)là số nguyên thì 8 \(⋮\)x - 1 => x - 1 \(\in\)Ư(1; -1; 2; -2; 4; -4; 8; -8}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
| x | 2 | 0 | 3 | -1 | 5 | -3 | 9 | -7 |
Vậy x \(\in\){2; 0; 3; -1; 5; -3; 9; -7} thì \(\frac{3x+7}{x-1}\)là số nguyên
D
16 tháng 9 2018
Đặt \(A=\frac{3x+7}{x-1}\)
Ta có: \(A=\frac{3x+7}{x-1}=\frac{3x-3+10}{x-1}=\frac{3x-3}{x-1}+\frac{10}{x-1}=3+\frac{10}{x-1}\)
Để \(A\in Z\)thì \(\frac{10}{x-1}\in Z\Rightarrow10⋮x-1\Leftrightarrow x-1\in U\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(5\) | \(-5\) | \(10\) | \(-10\) |
| \(x\) | \(2\) | \(0\) | \(3\) | \(-1\) | \(6\) | \(-4\) | \(11\) | \(-9\) |
Vậy, với \(x\in\left\{-9;-4;-1;0;2;3;6;11\right\}\)thì \(A=\frac{3x+7}{x-1}\in Z\)
10 tháng 2 2021
a, \(A=\frac{\left(x+2\right)^2}{x}\left(1-\frac{x^2}{x+2}\right)=\frac{\left(x+2\right)^2}{x}\left(\frac{x+2-x^2}{x+2}\right)\)
\(=\frac{-\left(x+2\right)^2\left(x-2\right)\left(x+1\right)}{x\left(x+2\right)}=\frac{-\left(x\pm2\right)\left(x+1\right)}{x}\)
c, Theo bài ra ta có : \(C=\frac{A}{B}\)hay \(\frac{\frac{-\left(x\pm2\right)\left(x+1\right)}{x}}{\frac{4}{\left(x-2\right)^2}}=\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}\)
d, Theo bài ra ta có :
\(C>0\)hay \(\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}>0\)
\(\Leftrightarrow\frac{-\left(x+2\right)\left(x+1\right)}{x}.\frac{x-2}{4}>0\)
\(\Leftrightarrow-\left(x+2\right)\left(x+1\right)>0\Leftrightarrow\left(x+2\right)\left(x+1\right)>0\)
\(\Leftrightarrow x>-2;x>-1\Rightarrow x>-1\)
MN
3 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x\left(x+2\right)+x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}:\frac{6}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{-6x}{6x\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{-1}{x+2}\)
b) Khi \(\left|x\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)
c) Để P = 7
\(\Leftrightarrow-\frac{1}{x+2}=7\)
\(\Leftrightarrow7\left(x+2\right)=-1\)
\(\Leftrightarrow7x+14=-1\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\frac{15}{7}\)
Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)
d) Để \(P\inℤ\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)
PN
21 tháng 5 2021
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
PN
21 tháng 5 2021
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
16 tháng 12 2022
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
16 tháng 8 2019
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
16 tháng 8 2019
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự