(2x-1)^2+(x+3)^2-5(x+7)(x-7)
Tìm x???
Giú mik vs ạ mik đang cần gấp
bài 2:tìm x,biết:
a)x+3/9=7/6x2/3
b)x-2/3=1/8:5/4
giải ra giúp mik vs ạ! mik cảm ơn...mik đag cần gấp
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
1/2-5/12x=2/3
Giúp mik tìm x vs mik đang cần gấp ạ
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)
\(x=\dfrac{-2}{5}\)
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-2}{5}\)
Tìm x biết :
(x+2)^2-2(x+2)(x+3)+(x+5)^2=7
Mn giúp mik vs ạ mik cần gấp mai mik phải ik hok rùi ạ
ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)
\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)
Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé
\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)
\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)
\(\Leftrightarrow4x+10=0\)
\(\Leftrightarrow4x=-10\)
\(\Leftrightarrow x=\frac{-5}{2}\)
Trả lời:
( x + 2 )2 - 2 ( x + 2 ) ( x + 3 ) + ( x + 5 )2 = 7
<=> x2 + 4x + 4 - 2 ( x2 + 5x + 6 ) + x2 + 10x + 25 = 7
<=> x2 + 4x + 4 - 2x2 - 10x - 12 + x2 + 10x + 25 = 7
<=> 4x + 17 = 7
<=> 4x = - 10
<=> x = - 5/2
Vậy x = - 5/2 là nghiệm của pt.
bài 1 :tìm x
a ( 2x/5 -1 ) : (-5) = 1/7
b , x+25% .x = 1,25
c | 3,5 + 2/1/2 .x | - 2,5 = 7/2
d 3/1/2 .x -| 0,25 - 1/4 | = 5
mik tích cho mik đang cần gấp
a) \(\left(\frac{2x}{5}-1\right):\left(-5\right)=\frac{1}{7}\)
\(\frac{2x}{5}-1=\frac{1}{7}.\left(-5\right)\)
\(\frac{2x}{5}-1=\frac{-5}{7}\)
\(\frac{2x}{5}=\frac{-5}{7}+\frac{7}{7}\)
\(\frac{2x}{5}=\frac{2}{7}\)
\(=>2x.7=2.5\)
\(=>14x=10\)
\(=>x=\frac{5}{7}\)
c) \(\left|3,5+2,5x\right|-2,5=3,5\)
\(\left|3,5+2,5x\right|=3,5+2,5\)
\(\left|3,5+2,5x\right|=6\)
\(TH1\) \(3,5+2,5x=6\) \(TH2\) \(3,5+2,5x=-6\)
\(2,5x=6-3,5\) \(2,5x=-6-3,5\)
\(2,5x=2,5\) \(2,5x=-9.5\)
\(x=1\) \(x=-3,8\)
vậy \(x=1\) hoặc \(x=-3,8\)
câu d) làm tương tự như câu c)
Tìm \(x\):
a)\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\) b)\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
giúp mik vs ạ mik cần gấp
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
tìm x biết
a,(x-2)^2-(x+3)^2-4(x+1)=5
b, (2x-3)(2x+3)-(x-1)^2-3x(x-5)=-44
c, (5x+1)^2-(5x+3)(5x-3)=30
d,(x+3)^2+(x-2)(x+2)-2(x-1)^2=7
giúp mik vs mik cần gấp
a) |2x+1|=5
b) |2x+1|=0
c) |2x+1|=7
d) |2x+5|=|3x-7|
e) |2x+7|=x-1
g) |x-2|+|2x-3|=2
h) |x+2| + |1-x | =3x+2
Giúp mik với cần gấp ạ
`a)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
`b)|2x+1|=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
`c)|2x+1|=7`
`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`d)|2x+5|=|3x-7|`
`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\)
`e)|2x+7|=1`
`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
`g)|x-2|+|2x-3|=2`
Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`
`pt<=>x-2+2x-3=2`
`<=>3x-5=2`
`<=>3x=7`
`<=>x=7/3(tm)`
Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`
`pt<=>2-x+3-2x=2`
`<=>5-3x=2`
`<=>3x=3`
`<=>x=1(tm)`
Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`
`pt<=>2-x+2x-3=2`
`<=>x-1=2`
`<=>x=3(l)`
`h)|x+2|+|1-x|=3x+2`
Vì `VT>=0=>3x+2>=0=>x>=-2/3`
`=>|x+2|=x+2`
`pt<=>x+2+|1-x|=3x+2`
`<=>|1-x|=2x(x>=0)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
a.
$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix}
2x+1=5\\
2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=2\\
x=-3\end{matrix}\right.\)
b.
$|2x+1|=0$
$\Leftrightarrow 2x+1=0$
$\Leftrightarrow x=-\frac{1}{2}$
c.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
d.
$|2x+5|=|3x-7|$
\(\Leftrightarrow \left[\begin{matrix} 2x+5=3x-7\\ 2x+5=7-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=12\\ x=0,4\end{matrix}\right.\)
e.
$|2x+7|=x-1\Rightarrow x-1\geq 0\Leftrightarrow x\geq 1$
Với $x\geq 1$ thì $|2x+7|=2x+7$
Khi đó pt trở thành:
$2x+7=x-1$
$\Leftrightarrow x=-8< 1$ (vô lý)
Vậy pt vô nghiệm.
g.
$|x-2|+|2x-3|=2$
Nếu $x\geq 2$ thì pt trở thành:
$x-2+2x-3=2$
$\Leftrightarrow 3x-5=2$
$\Leftrightarrow x=\frac{7}{3}$ (thỏa mãn)
Nếu $\frac{3}{2}\leq x< 2$ thì pt trở thành:
$2-x+2x-3=2$
$\Leftrightarrow x=3$ (không thỏa mãn)
Nếu $x< \frac{3}{2}$ thì pt trở thành:
$2-x+3-2x=2$
$\Leftrightarrow 5-3x=2$
$\Leftrightarrow x=1$ (thỏa mãn)
Vậy..........
h.
Từ đề suy ra $x\geq \frac{-2}{3}$
$\Rightarrow |x+2|=x+2$
Nếu $x\geq 1$ thì $|1-x|=x-1$. PT trở thành:
$x+2+x-1=3x+2$
$\Leftrightarrow 2x+1=3x+2$
$\Leftrightarrow x=-1$ (vô lý)
Nếu $\frac{-2}{3}\leq x< 1$ thì $|1-x|=1-x$. PT trở thành:
$x+2+1-x=3x+2$
$\Leftrightarrow 3=3x+2$
$\Leftrightarrow x=\frac{1}{3}$ (thỏa mãn)
tìm x biết:
a)2/3x-3/2x=5/12
b)2/5+3/5.(3x-3,7)=-53/10
c)7/9:(2+3/4x)+5/9=23/27
d)-2/3.x+1/5=3/10
e)|x|-3/4=5/3
f)|2x-1/3|+5/6=1
giúp mik vs mik đang cần gấp bn nào giải mik cũng tick nha
giải hết giúp mik các bn nhé
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)
\(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)
\(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{21}{8}\)
\(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2
\(\dfrac{3}{4}\)\(x\) = \(\dfrac{5}{8}\)
\(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{5}{6}\)
Vậy \(x=\dfrac{5}{6}\)
tìm số nguyên x để phân thức sau có giá trị là số nguyên:mọi người làm đk thì giúp mik vs nhé vì mik đang cần gấp ạ cảm ơn mọi người nhiều nhé
a,\(\frac{3}{2x-1}\)b,\(\frac{5}{x^2+1}\)c,\(\frac{7}{x^2-x+1}\)d,\(\frac{x^2-59}{x+8}\)e,\(\frac{x+2}{x^2+4}\)
a, 2x-1 thuộc ước của 2,rồi giải ra
b,c tương tự
d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d