tìm số nguyên x
5/x+1+4/x+1=3/-13
-x+2+2x+3+x+1/4+2x+1/6=8/3
3/2x+1+10/4x+2-6/6x+2=12/26
giúp mình mik đang vội]
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Leftrightarrow2x+1=13\)
hay x=6
tìm x biết
a) (6x-3) (2x+4) + (4x-1) (5-3x) = -21
b) 6x (3x+5) - 2x (9x-2) + (17-x) (x-1) + x (x-18) =0
c) (15-2x) (4x+1) - (13-4x) (2x-3) - (x-1) (x+2) + x2=52
d) (8x-3) (3x+2) - (4x+7) (x+4) = (2x+1) (5x-1) - 33
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21
<=> 12x2 + 24x - 6x - 12 + 20x - 12x2 - 5 + 3x = -21
<=> 41x = -21 + 12 + 5
<=> 41x = -4
<=> x = -4/41
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 \(x=\dfrac{5}{6}\)
. Help Me ! Please :)) Mik đang gấp lắm nhé nên nếu các bạn biết thì giải giúp mik nhé :3 Cảm ơn nhiều nhiều lắm nek ~~~ Bạn nào làm đúng mik sẽ tik nhé =)) Phần nào các bạn làm đk thì giúp mik, bài dài nên mik ko mún làm mất quá nhìu thời gian các bạn đâu ^^
~ ĐỀ BÀI: Phân tích các đa thức sau thành nhân tử
1) 6x^2 - 11x + 3
2) 2x^2 + 3x - 27
3) 2x^2 - 5xy - 3y^2
4) x^3 + 2x - 3
5) x^3 - 7x + 6
6) x^3 - 9x^2 + 6x + 16
7) x^3 + x^2 - x + 2
8) x^3 - 6x^2 - x + 30
9) 4x^4 - 32x^2 + 1
10) x^6 + 27
11) 4x^4 + 1
12) (x^2-2x).(x^2 - 2x - 1) - 6
13) x^2 - 2xy + y^2 + 3x - 3y - 4
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
K MIK NHA BẠN !!!!!!!!!!
bÀI 1
bÀI 2 :
Bài 3 :
Bài 4:
5,
6,
7,
8,
9,
10,
11,
12,
13,
K MIK NHA BẠN !!!!!!!!!!
Nguyễn Quang Trung Linh Nhi : Mik cảm ơm 2 bạn nhiều nhóe ;)
a) 2x+3/5x-3 - 3/4x-6 = 2/5
b) 2x-1/x-1 + x/(x-1)(x-2) = 6x-2/x-2
Mình dốt toán lắm, giúp mik với nha!!! Gấp lắm!!!
b)\(\frac{2x-1}{x-1}+\frac{x}{\left(x-1\right)\left(x-2\right)}=\frac{6x-2}{x-2}\)
\(< =>\left(2x-1\right)\left(x-2\right)+x-\left(6x-2\right)\left(x-1\right)=0\)
\(< =>2x^2-x-4x+2+x-\left(6x^2-2x-6x+2\right)=0\)
\(< =>2x^2-4x+2-6x^2+8x-2=0\)
\(< =>-4x^2+4x=0\)
\(< =>-4x\left(x+1\right)=0\)
\(TH1:-4x=0< =>x=0\) \(TH2:x+1=0< =>x=-1\)
\(S=\left\{-1;0\right\}\)
ai giúp mik với mik nhiều bt lắm
a, 1/x-1 - 7/x+2 = 3/x2+x-2
b,x+3/x-4 + x-1/x-2 = 2/6x-8-x2
c,1/x+1 + 2/x3-x2x+1 = 3/1-x2
d, 3x-1/x-1 - 2x+5/x+3 = 1 - 4/x2+2x-3
e, x2+2x+1/x2+2x+2 + x2+2x+2/x2+2x+3 = 7/6
f, 1/4x2-12x+9 - 3/9-4x2=4/4x2+12x+9
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
a, 3y-2y=2y-3
3y-2y-2y=3
-y=3
y=-3
b, 3-4x+24+6x=x+27+3x
-4x+6x-x-3x =27-3-24
-2x =0
x =0
c, 5-(6-x)=4.(3-2x)
5-6+x =12-8x
x+8x =12+6-5
9x =13
x =13/9
d, 4.(x+3)=-7x+17
4x+12 =-7x+17
4x+7x =17-12
11x =5
x =5/11
A) \(3y-2y=2y-3\) \(\Leftrightarrow y-2y=-3\) \(\Leftrightarrow y=3\)
B) \(3-4x+24+6x=x+27-3x\)\(\Leftrightarrow3-4x+24+6x-x-27-3x=0\)
\(\Leftrightarrow-2x=0\)\(\Leftrightarrow x=0\)
C) \(5-\left(6-x\right)=4\left(3-2x\right)\)\(\Leftrightarrow5-6x+x-12+8x=0\)\(\Leftrightarrow9x-13=0\)\(\Leftrightarrow x=\frac{13}{9}\)
D) \(4\left(x+3\right)=-7x+17\) \(\Leftrightarrow4x+12+7x-17=0\)\(\Leftrightarrow11x-5=0\)\(\Leftrightarrow x=\frac{5}{11}\)
E) \(11x+42-2x=100-9x-22\)\(\Leftrightarrow11x+42-2x-100+9x+22=0\)
\(\Leftrightarrow18x-36=0\)\(\Leftrightarrow x=2\)
F) \(5x+\frac{3}{12}=1+\frac{2x}{9}\)\(\Leftrightarrow180x+9-36-8x=0\)\(\Leftrightarrow172x-27=0\)\(\Leftrightarrow x=\frac{27}{172}\)
TK nka !!! Mk giải tiếp !!
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
A(x) = 2x - 1
B(x) = 3 - 6/5x
C(x) = 4x^2 - 25
D(x) = ( x + 1/4 )^2 - 16/9
E(x) = 8x^2 + 27
F(x) = x^2 + 3x
Ai giúp mik nhanh với nha, mik đang gấp lắm
`A(x) =2x-1`
`2x-1=0`
`=> 2x=0+1`
`=>2x=1`
`=>x=1/2`
__
`B(x) =3 - 6/5x`
`3-6/5x=0`
`=> 6/5x=3-0`
`=> 6/5x=3`
`=> x= 3 : 6/5`
`=> x= 3 xx 5/6`
`=> x=15/6`
__
`C(x) = 4x^2 - 25`
`4x^2 - 25=0`
`=> 4x^2 = 0+25`
`=> 4x^2 =25`
`=> 4x^2 = (+-5)^2`
`=> x= 5/4` hoặc `x=-5/4`
__
`D(x) = ( x + 1/4 )^2 - 16/9`
` ( x + 1/4 )^2 - 16/9=0`
`=> ( x + 1/4 )^2 = 16/9`
`=>( x + 1/4 )^2 =(+-4/3)^2`
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{4}{3}\\x+\dfrac{1}{4}=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
__
`E(x) = 8x^2 + 27`
`8x^2 +27=0`
`=>8x^2=0-27`
`=> 8x^2 =-27`
`->` đề hơi sai;-;.
__
`F(x) = x^2 + 3x`
`x^2 +3x=0`
`=>x(x+3)=0`
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
`@ yl`
