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\(\left(3x-1\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(3x+3-4\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(-4\right)⋮\left(x+1\right)\)
\(\Rightarrow x+1\inƯ\left(-4\right)=\left\{-4;-1;1;4\right\}\)
\(\Rightarrow x\in\left\{-5;-2;0;3\right\}\)
b)Ta có : (x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 7450
<=> ( x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 7450
<=> 100 .x + 5050 = 7450
<=> 100.x = 7450 - 5050
<=> 100. x = 2400
<=> x = 2400 : 100
<=> x = 24
Vậy x = 24
c) Có số số hạng là :
( x - 1 ) + 1 ( số hạng )
Tổng của dãy số là :
(x + 1 ) . x : 2 = 78
=> ( x + 1 ) . x = 156
=> (x + 1 ) . x =13 . 12 = 156
=> x = 12
Vậy x = 12
d) 12.x + 13.x = 2000
<=> x . ( 12 + 13 ) = 2000
<=> x . 25 = 2000
<=> x =2000 : 25
<=> x = 80
Vậy x = 80
e) 6.x + 4.x = 2010
<=> x . ( 6 + 4 ) = 2010
<=> x . 10 =2010
<=> x = 2010 : 10
<=> x = 201
Vậy x = 201
f) 5.x - 3.x - x = 20
<=> x . ( 5 - 3 - 1 ) = 20
<=> x . 1 = 20
<=> x = 20
Vậy x = 20
Còn câu a thì đợi mình tí ,lười nghĩ
x+y+xy+1=-1
x+y+xy=-1-1
x+y+xy=-2
ma x+y=xy
suy ra x+y+xy
=xy+xy
ta co
2xy=-2
xy=-2:2
xy=-1
Vay neu x=1 thi y=-1 (va nguoc lai )
tik nha ^_^
Bài giải
a, Ta có : \(\frac{2x+5}{x+2}=\frac{2\left(x+2\right)+1}{x+2}=\frac{2\left(x+2\right)}{x+2}+\frac{1}{x+2}=2+\frac{1}{x+2}\)
\(2x+5\text{ }⋮\text{ }x+2\text{ khi }1\text{ }⋮\text{ }x+2\text{ }\Rightarrow\text{ }x+2\inƯ\left(1\right)\)
\(\Rightarrow\orbr{\begin{cases}x+2=-1\\x+2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }-1\right\}\)
a) \(2\left(x+2\right)+1⋮x+2\)
\(\Leftrightarrow1⋮x+2\)
b) \(3x+5⋮x-2\)
\(\Leftrightarrow3\left(x-2\right)+11⋮x-2\)
\(\Leftrightarrow11⋮x-2\)
c) \(x^2+3⋮x+4\)
\(\Leftrightarrow\left(x^2-16\right)+19⋮x+4\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)+19⋮x+4\)
\(\Leftrightarrow19⋮x+4\)
P/s : Mình chỉ làm đến bước này thôi, các bước tiếp theo bạn tự làm nhé. Chúc bạn học tốt !
\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)
\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)
\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)
\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)
\(\Rightarrow x=-\frac{25}{32}\)
\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)
\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)
\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)
\(\Rightarrow x=10\)
\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)
\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)
\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)
\(\Rightarrow|x|=\frac{33}{20}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)
\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)
bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
a) \(-6⋮\left(2x-1\right)\)
\(\Leftrightarrow2x-1\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Ta có bảng sau :
Vậy \(x\in\left\{-1;0;1;2\right\}\)
b) \(\left(3x-2\right)⋮\left(x-3\right)\)
\(\Leftrightarrow\left(3x+9-7\right)⋮\left(x+3\right)\)
Vì \(\left(3x+9\right)⋮\left(x+3\right)\)nên \(7⋮\left(x+3\right)\)
\(\Leftrightarrow x+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng sau :
Vậy \(x\in\left\{-10;-4;-2;4\right\}\)
\(\left(-6\right)⋮\left(2x-1\right)\Rightarrow2x-1\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Vì 2x-1chia 2 dư 1
\(\Rightarrow2x-1\in\left\{\pm1;\pm3\right\}\)
\(\Rightarrow x\in\left\{1;0;2;-1\right\}\)
Vậy......................................
\(\left(3x+2\right)⋮\left(x+3\right)\)
\(\Rightarrow3\left(x+3\right)-7⋮x-3\)
\(\Rightarrow7⋮x-3\Rightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{4;2;10;-4\right\}\)
Vậy....................................