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Câu a:
\(\frac{x+5}{x+9}\) = \(\frac{x+4}{x+3}\)
(\(x\) + 5)(\(x+3\)) = (\(x+4\))(\(x+9\))
\(x^2\) + 3\(x\) + 5\(x\) + 15 = \(x^2\) + 9\(x\) + 4\(x\) + 36
\(x^2+3x+5x\) - \(x^2-9x\) - 4\(x\) = 36 - 15
(\(x^2-x^2\)) + (\(3x+5x\) - 9\(x-4x\)) = 21
0 + (8\(x\) - 9\(x\) - 4\(x\)) = 21
-\(x-4x\) = 21
-5\(x\) = 21
\(x\) = 21 : (-5)
\(x\) = - \(\frac{21}{5}\)
Vậy \(x=-\frac{21}{5}\)
a, 3x - 2 ⋮ x + 3
=> 3x + 9 - 11 ⋮ x + 3
=> 3(x + 3) - 11 ⋮ x + 3
=> 11 ⋮ x + 3
b, x ⋮ 2x + 1
=> 2x ⋮ 2x + 1
=> 2x + 1 - 1 ⋮ 2x + 1
=> 1 ⋮ 2x + 1
c, 3x + 6 ⋮ x + 1
=> 3x + 3 + 3 ⋮ x + 1
=> 3(x + 1) + 3 ⋮ x + 1
=> 3 ⋮ x + 1
d, em không biết làm
câu a,b,c bn Cả Út lm r
mik làm câu d
\(x^2⋮x-2\)
\(\Rightarrow x\left(x-2\right)+2x⋮x-2\)
\(\Rightarrow2x⋮x-2\)
\(\Rightarrow2\left(x-2\right)+4⋮x-2\)
\(\Rightarrow4⋮x-2\)
\(\Rightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{3;1;4;0;6;-2\right\}\)
Vậy..............................
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
hok tốt!!
b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
a) \(|x+1|=3\)
\(\Rightarrow x+1=\pm3\)
+) \(x+1=3\) +) \(x+1=-3\)
\(\Rightarrow x=2\) \(\Rightarrow x=-4\)
Vậy \(x\in\left\{2;-4\right\}\)
b) \(3^2x+2^4=5^2\)
\(9x+16=25\)
\(9x=25-16\)
\(9x=9\)
\(x=1\)
c) \(\frac{4+x}{7+y}=\frac{4}{7}\)
\(\Rightarrow\left(4+x\right).7=\left(7+y\right).4\)
\(\Rightarrow28+7x=28+4y\)
\(\Rightarrow7x=4y\)
Mà \(\left(7,4\right)=1\) và \(x+y=11\)
Vậy \(x=4;y=7\)
a) Ta có: \(\left|x+1\right|=3\)
\(\Rightarrow x+1=\pm3\)
Nếu x + 1 = 3 => x = 2
Nếu x + 1 = -3 => x = -4
Vậy x = {2;-4}
b) \(3^2x+2^4=5^2\)
\(\Rightarrow9x+16=25\)
\(\Rightarrow9x=9\)
\(\Rightarrow x=1\)
Vậy x = 1
c) \(\frac{4+x}{7+x}=\frac{4}{7}\)
\(\Rightarrow7\left(4+x\right)=4\left(7+x\right)\)
\(\Rightarrow28+7x=28+4x\)
\(\Rightarrow7x-4x=0\)
\(\Rightarrow x=0\)
Vậy x = 0
\(a,2x-138=2^3:\left(-3\right)^2\)
\(\Rightarrow2x-138=8:9\)
\(\Rightarrow2x=\frac{8}{9}+138\)
\(\Rightarrow2x=\frac{1250}{9}\)
\(\Rightarrow x=\frac{626}{9}\)
\(10+2x=\left(-4\right)^5:\left(-4\right)^3\)
\(10+2x=-1024:\left(-64\right)\)
\(10+2x=16\)
\(2x=16-10\)
\(2x=6\)
\(x=6:2=3\)
a) 6(x+1)-3=2(x+2)+11
6x + 6 - 3= 2x+ 4+11
6x-2x +6-3= 4+11
4x +6-3= 15
4x+6= 15+3
4x+6= 18
4x= 18-6
4x= 12
x= 12:4
x=3
Vậy...
b) x-7\(⋮\)x+6
\(\Rightarrow\)x+6-13\(⋮\)x+6
\(\Rightarrow\)13\(⋮\)x+6
\(\Rightarrow\)x+6 \(\in\) Ư(13)
Ta có Ư(13)={\(\pm1;\pm13\)}
nếu x+6=1 thì x=-5
nếu x+6=-1 thì x=-7
nếu x+6=13 thì x=7
nếu x+6=-13 thì x=-19
Vậy x \(\in\) {-5;-7;7;-19}
a) Ta có: \(-2x-11⋮3x+2\)
\(\Rightarrow-3\left(-2x-11\right)⋮3x+2\)
\(\Rightarrow6x+33⋮3x+2\)
\(\Rightarrow\left(6x+4\right)+29⋮3x+2\)
\(\Rightarrow2\left(3x+4\right)+29⋮3x+2\)
\(\Rightarrow29⋮3x+2\)
\(\Rightarrow3x+2\in\left\{1;-1;29;-29\right\}\)
\(\left[\begin{matrix}3x+2=1\\3x+2=-1\\3x+2=29\\3x+2=-29\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=-1\\x=9\\x=-\frac{31}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{3};-1;9;\frac{-31}{3}\right\}\)
b) \(x-7⋮x+6\)
\(\Rightarrow\left(x+6\right)-13⋮x+6\)
\(\Rightarrow13⋮x+6\)
\(\Rightarrow x+6\in\left\{1;-1;13;-13\right\}\)
\(\left[\begin{matrix}x+6=1\\x+6=-1\\x+6=13\\x+6=-13\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-5\\x=-7\\x=7\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{-5;-7;7;-19\right\}\)
a)-2x-11\(⋮\)3x+2
\(\Rightarrow\)-3(-2x-11)\(⋮\)3x+2
\(\Rightarrow\)6x+33\(⋮\)3x+2
\(\Rightarrow\)(6x+4)+29\(⋮\)3x+2
\(\Rightarrow\)2(3x+4)+29\(⋮\)3x+2
\(\Rightarrow\)29\(⋮\)3x+2
\(\Rightarrow\)3x+2 \(\in\) Ư(29)={\(\pm1;\pm29\)}
nếu 3x+2=1 thì x=\(\frac{-1}{3}\)
nếu 3x+2=-1 thì x=-1
nếu 3x+2=29 thì x=9
nếu 3x+2=-29 thì x=\(\frac{-31}{3}\)
Vậy x \(\in\) {\(\frac{-1}{3};-1;9;\frac{-31}{3}\)}