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a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
\(A=\frac{x^2-10x+36}{x-5}=\frac{x^2-10x+25+9}{x-5}\) \(=\frac{\left(x-5\right)^2+9}{x-5}=x-5+\frac{9}{x-5}\)
để \(A\in Z\)
<=> \(\frac{9}{x-5}\in Z\)mà \(x\in Z\)
=> \(x-5\inƯ\left(9\right)\)
=> \(x-5\in\left(1;-1;3;-3;9;-9\right)\)
=> \(x\in\left(6;4;8;2;14;-4\right)\)
học tốt
cái đấy ko có GTNN và GTLN chỉ có giả trị của x để mấy cái trên nguyên thôi, đề bài sai rùi bạn ạ ko phải nghĩ nha
a. \(\frac{2x+3}{15}=\frac{7}{5}\)
\(\Leftrightarrow5\left(2x+3\right)=15.7\)
\(\Leftrightarrow10x+15=105\)
\(\Leftrightarrow10x=90\)
\(\Leftrightarrow x=9\)
b. \(\frac{x-2}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-2\right)=9.8\)
\(\Leftrightarrow3x-6=72\)
\(\Leftrightarrow3x=78\)
\(\Leftrightarrow x=26\)
c. \(\frac{-8}{x}=\frac{-x}{18}\)
\(\Leftrightarrow-x^2=-144\)
\(\Leftrightarrow x^2=12^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
Mấy câu kia tương tự
d, \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=6x-12\Leftrightarrow4x=-27\Leftrightarrow x=-\frac{27}{4}\)
e, \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x=132\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f, \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x=10\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
g, \(\left(2x-1\right)\left(2x+1\right)=63\Leftrightarrow4x^2+2x-2x-1=63\Leftrightarrow4x^2-64=0\)
\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)
h, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)
\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)
\(a)\) Ta có :
\(\frac{x}{18}=\frac{y}{9}\)\(\Leftrightarrow\)\(\frac{x}{2}=y\)
\(\Rightarrow\)\(x=2y\)
Thay \(x=2y\) vào \(A=\frac{2x-3y}{2x+3y}\) ta được :
\(A=\frac{2.2y-3y}{2.2y+3y}=\frac{4y-3y}{4y+3y}=\frac{y}{7y}=\frac{1}{7}\)
Vậy ... ( tự kết luận )
Chúc bạn học tốt ~
\(D=\frac{4x+1}{x+3}\inℤ\Leftrightarrow4x+1⋮x+3\)
\(\Rightarrow4x+12-11⋮x+3\)
\(\Rightarrow4\left(x+3\right)-11⋮x+3\)
\(\Rightarrow11⋮x+3\)
\(\Rightarrow x+3\in\left\{-1;1;-11;11\right\}\)
\(\Rightarrow x\in\left\{-4;-2;-14;8\right\}\)
a) \(D=\frac{4x+1}{x+3}\)
=> 4x + 1 \(⋮\)( x + 3 ) để D là số nguyên
Mà ( x + 3 ) \(⋮\)( x + 3 ) => 4( x + 3 ) \(⋮\)( x + 3 )
=> [ 4x + 1 - 4( x + 3 ) ] \(⋮\)( x + 3 )
=> [ 4x + 1 - 4x + 12 ] \(⋮\)( x + 3 )
=> 13 \(⋮\)( x + 3 )
=> \(x+3\inƯ\left(13\right)\)\(=\left\{\pm1;\pm13\right\}\)
| x + 3 | -1 | 1 | -13 | 13 |
| x | 2 | 4 | -10 | 16 |
Vậy \(x\in\left\{-10;2;4;16\right\}\)Để D là số nguyên
b) \(E=\frac{6x+2}{2x-3}\)
=> 6x + 2 \(⋮\)2x - 3 để E là số nguyên
Mà ( 2x - 3 ) \(⋮\)( 2x - 3 ) => 3( 2x - 3 ) \(⋮\)( 2x - 3 )
=> [ 6x + 2 - 3( 2x - 3 ) ] \(⋮\)( 2x - 3 )
=> [ 6x + 2 - 6x - 3 ] \(⋮\)( 2x - 3 )
=> -1 \(⋮\)( 2x - 3 )
=> ( 2x - 3 ) \(\inƯ\left(-1\right)=\left\{\pm1\right\}\)
| 2x - 3 | -1 | 1 |
| 2x | 2 | 4 |
| x | 1 | 2 |
Vậy x \(\in\left\{1;2\right\}\)để E là số nguyên
Còn phần còn lại cậu có thể làm tương tự.
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
:v ( cái đề bài )
Đề : Tìm x để biểu thức có giá trị là nguyên
\(d,\frac{x+4}{x-3}\)
ĐKXĐ : \(x\ne-3\)
Để biểu thức có giá trị là nguyên thì
\(x+4⋮x-3\)
=> \(\left(x-3\right)+7⋮x-3\)
=> \(7⋮x-3\)
=> \(x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\\\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\\\left[{}\begin{matrix}x=7+3=10\\x=-7+3=-4\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{4;2;10;-4\right\}\)
e, \(\frac{10x+9}{2x+3}\)
Để biểu thức có giá trị là nguyên thì
\(10x+9⋮2x+3\)
=>\(10x+30-21⋮2x+3\)
=> \(10\left(x+3\right)-21⋮2x+3\)
=> \(-21⋮2x+3\)
=> \(2x+3\inƯ\left(-21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x+3=1\\2x+3=-1\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=7\\2x+3=-7\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=21\\2x+3=-21\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=1-3=-2\\2x=-1-3=-4\end{matrix}\right.\\\left[{}\begin{matrix}2x=3-3=0\\2x=-3-3=-6\end{matrix}\right.\\\left[{}\begin{matrix}2x=7-3=4\\2x=-7-3=-10\end{matrix}\right.\\\left[{}\begin{matrix}2x=21-3=18\\2x=-21-3=-24\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-2:2=-1\\x=-4:2=-2\end{matrix}\right.\\\left[{}\begin{matrix}x=0:2=0\\x=-6:2=-3\end{matrix}\right.\\\left[{}\begin{matrix}x=4:2=2\\x=-10:2=-5\end{matrix}\right.\\\left[{}\begin{matrix}x=18:2=9\\x=-24:2=-12\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{-1;-2;0;-3;2;-5;9;12\right\}\)
f, \(\frac{4x+1}{2x-1}\)
ĐKXĐ : \(x\ne\frac{a}{b}\)( phân số tối giản )
Đẻ biểu thức trên có giá trị là nguyên thì
\(4x+1⋮2x-1\)
=> \(\left(4x-1\right)+2⋮2x-1\)
=> \(2⋮2x-1\)
=> \(2x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\\\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=1+1=2\\2x=-1+1=0\end{matrix}\right.\\\left[{}\begin{matrix}2x=2+1=3\\2x=-2+1=-1\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2:2=1\left(lấy\right)\\x=0:2=0\left(lấy\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=3:2=\frac{3}{2}\left(loại\right)\\x=-:2=-\frac{1}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{1;0\right\}\)