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`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`
a) -(3x - 5) - (-5x - 7 ) = 6
-3x + 5 + 5x + 7 = 6
( -3 + 5 )x + 5 + 7 = 6
2x + 12 = 6
2x = -6
x = -3
c) -5x - 17 = -2x + 4
-5x + 2x = 4 + 17
-3x = 21
x = -7
a) -(3x-5)-(-5x-7) =6
-3x+5 +5x+7 =6
(-3x+5x)+(5+7)=6
[(-3+5)x]+ 12 =6
2x + 12 =6
2x =6-12
2x =-6
x=-6:2
x=-3
b) (-5x-12)-(x+3) =1
-5x-12 - x-3 =1
(-5x-x) -(12-3)=1
(-5x-1x) - 9 =1
[(-5-1)x] - 9 =1
-6x-9=1
-6x=1+9
-6x=10
Ta có :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(=>35x^2+17x+2=35x^2+44x-7\)
\(=>17x+2=44x-7\)
\(=>44x-17x=2+7\)
\(=>27x=9\)
\(=>x=\frac{9}{27}=\frac{1}{3}\)
** Bổ sung điều kiện $x,y$ là số nguyên.
a/
$(5x-1)(y+1)=4$
Với $x,y$ nguyên thì $5x-1, y+1$ nguyên. Mà tích của chúng bằng 4 nên ta có các trường hợp sau:
TH1: $5x-1=1, y+1=4\Rightarrow x=\frac{2}{5}$ (loại)
TH2: $5x-1=-1, y+1=-4\Rightarrow x=0; y=-5$
TH3: $5x-1=2, y+1=2\Rightarrow x=\frac{3}{5}$ (loại)
TH4: $5x-1=-2, y+1=-2\Rightarrow x=\frac{-1}{5}$ (loại)
TH5: $5x-1=4, y+1=1\Rightarrow x=1; y=0$
TH6: $5x-1=-4; y+1=-1\Rightarrow x=\frac{-3}{5}$ (loại)
Vậy......
b/
$xy-7y+5x=0$
$y(x-7)+5(x-7)=-35$
$(x-7)(y+5)=-35$
Vì $x,y$ nguyên nên $x-7, y+5$ nguyên. $(x-7)(y+5)=-35\Rightarrow x-7$ là ước của $-35$.
Mà $x\geq 3\Rightarrow x-7\geq -4$
$\Rightarrow x-7\in \left\{-1; 1; 5; 7; 35\right\}$
Nếu $x-7=-1\Rightarrow y+5=35$
$\Rightarrow x=6; y=30$
Nếu $x-7=1\Rightarrow y+5=-35$
$\Rightarrow x=8; y=-40$
Nếu $x-7=5\Rightarrow y+5=-7$
$\Rightarrow x=12; y=-12$
Nếu $x-7=7\Rightarrow y+5=-5$
$\Rightarrow x=14; y=-10$
Nếu $x-7=35; y+5=-1$
$\Rightarrow x=42; y=-6$
Ta có 5x+2=y(5x+2)+7
suy ra y=0 Vì nếu y thuộc N* thì y(5x+2).7>5x+2
Ta được 5x+2=0(5x+2)+7
5x+2=0+7
5x+2=7
5x=7-2
5x=5
x=5:5
x=1
Vậy y=0;x=1
15 - 2 | 4 + 5x | = 59 <=> 2 | 4 + 5x | = -44 <=> | 4 + 5x | = -22
Vì \(\left|4+5x\right|\ge0\) nên không có x thỏa mãn | 4 + 5x | = -22
-------------------------------
TH1: \(x\le\frac{1}{5}\)=>3-x+1-5x=7=>4-6x=7=>-3=6x=>x=-1/2(nhận)
TH2:\(\frac{1}{5}< x\le3\)=>3-x+5x-1=7=>2+4x=7=>4x=5=>x=5/4(nhận)
TH3:x>3=>x-3+5x-1=7=>6x-4=7=>6x=11=>x=11/6(loại)
Vậy x=-1/2 hoặc x=5/4
\(\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{3}{7}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{3}{7}+\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{13}{14}\)
\(\Leftrightarrow x=\dfrac{13}{14}:\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{13}{14}\cdot\dfrac{5}{3}\)
\(\Leftrightarrow x=\dfrac{65}{42}\)
Vậy \(x=\dfrac{65}{42}\)
\(\dfrac{3}{5x}-\dfrac{1}{2}=\dfrac{3}{7}\) (ĐKXĐ: \(x\ne0\))
\(\Leftrightarrow\dfrac{3}{5x}=\dfrac{13}{14}\)
\(\Rightarrow65x=42\)
\(\Leftrightarrow x=\dfrac{42}{65}\)
Vậy \(x=\dfrac{42}{65}\)