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Ta có 3x( y + 1 ) + y + 1 = 7
3x( y + 1 ) + ( y + 1 ) = 7
( 3x + 1 )( y + 1 ) = 7
Vì x; y nguyên nên 3x + 1 và y + 1 nguyên
Vậy ( 3x + 1 ); ( y + 1 ) ϵ Ư( 7 ) = { 1; -1; 7; -7 }
Lập bảng giá trị
| 3x + 1 | 1 | -1 | -7 | 7 |
| y + 1 | 7 | -7 | -1 | 1 |
| x | 0 | \(\dfrac{-2}{3}\) ( loại ) | \(\dfrac{-8}{3}\) | 2 |
| y | 6 | 0 |
Vậy các cặp số nguyên ( x; y ) cần tìm để 3x( y + 1 ) + y + 1 = 7 là ( 0; 6 ); ( 2; 0 )
a) \(\left|x\right|< 1\Rightarrow-1< x< 1\Rightarrow x=0\)
b) \(\left|x+3\right|=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
c) \(\left|x+2\right|=\left|12-10\right|\)
\(\Leftrightarrow\left|x+2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=-2\\x+2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-2\right)-2\\x=2-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=0\end{matrix}\right.\)
d) \(\left|x+3\right|=2x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ge0\\\left[{}\begin{matrix}x+3=2x-2\\x+3=\left(-2x\right)+2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge2\\\left[{}\begin{matrix}x-2x=-2-3\\x-\left(-2x\right)=2-3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}-x=-5\\3x=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{-1}{3}\end{matrix}\right.\end{matrix}\right.\)
Vì \(\dfrac{-1}{3}< 1\) nên \(x=5\) thỏa mãn đề bài.
e) \(\left|x+1\right|>4\)
\(\Rightarrow\left[{}\begin{matrix}x+1>4\\x+1< 4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>3\\x< 3\end{matrix}\right.\)
f) \(\left|x-3\right|=\left|2x-1\right|\)
(cho thời gian suy nghĩ, mình chưa làm dạng này bao giờ)
g) \(\left|2x-1\right|-1+2x=0\)
\(\Rightarrow\left|2x-1\right|=-2x+1\)
Mà \(\left|2x-1\right|=\left|-2x+1\right|\)
\(\Rightarrow\left|-2x+1\right|=-2x+1\)
\(\Rightarrow-2x+1\ge0\)
\(\Rightarrow-2x\ge-1\)
\(\Rightarrow x\ge\dfrac{-1}{-2}=\dfrac{1}{2}\)
h) \(\left|3-2x\right|=2x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\\left[{}\begin{matrix}3-2x=2x-3\\3-2x=-2x+3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge3\\\left[{}\begin{matrix}3+3=2x+2x\\3-3=-2x+2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}6=4x\\0=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\end{matrix}\right.\end{matrix}\right.\)
Vì \(0=0\) luôn đúng nên ta có \(x=\dfrac{3}{2}\)
j) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\)
(đầu hàng)
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3 k lun
x/2-2/y=1/5
2/y=x/2-1/5
2/y=5x/10-2/10
2/y=5x-2/10
=>y(5x-2)=20
BẠN LÀM NỐT NHA!!
ĐƯA VỀ ƯỚC CỦA 1 SỐ NHA
CHÚC BN HOK TỐT
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{25}{28}\)
\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)
\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)
\(\Rightarrow x=-\dfrac{37}{30}\)
\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)
\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)
\(\Rightarrow35x=-70\)
\(\Rightarrow x=-2\)
\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)
\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)
\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)
\(\Rightarrow x=-\dfrac{8}{9}\)
b.(a+b)-(b-a)+c=2a+c
Xét VT: (a+b)-(b-a)+c = a + b - b + a + c = 2a+c
Mà VP = 2a+c
=> VT = VP
c.-(a+b-c)+(a-b-c)=-2b
Xét VT: -(a+b-c)+(a-b-c) = -a - b + c + a - b - c = -2b
Mà VP = -2b
=> VT = VP
d.a(b+c)-a(b+d)=a(c-d)
Xét VT: a(b+c)-a(b+d) = ab + ac - ab - ad = ac - ad = a(c-d)
Mà VP = a(c-d)
=> VT = VP
e.a(b-c)+a(d+c)=a(b+d)
Xét VT: a(b-c)+a(d+c)= ab -ac + ad + ac = ab + ad = a(b+d)
Mà VP = a(b+d)
=> VT = VP