[x+4].[x+1] =14
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HQ
Hà Quang Minh
Giáo viên
19 tháng 10 2023
a) Ta có (14 – 4) x 3 = 14 x 3 – 4 x 3
b) 29 x 2 – 9 x 2 = (29 – 9) x 2
= 20 x 2
= 40
7 x 214 – 7 x 14 = 7 x (214 – 14)
= 7 x 200
= 1 400
29 tháng 8 2015
A = 7/7.17 + 7/17.27 + 7/27.37 + ............ +7/1997.2007
A=7/10 ( 10/7.17 + 10/17.27 + 10/27.37 + ................+10/1997.2007)
A= 7/10 ( 1/7 -1/17 + 1/17 - 1/27 + 1/27 - 1/37 +...............+ 1/1997 - 1/2007)
A= 7/10 (1/7 - 1/2007)
A= 7/10 . 2000/14049
A=200/2007
bây h mk có vc rùi tích đúng nha tối mk lm típ cho
TT
1
15 tháng 1 2018
Mình làm mẫu câu a nha
a, pt <=> ( x-2/7 - 1 ) + ( x-1/8 - 1 ) = ( x-4/5 - 1 ) + ( x-3/6 - 1 )
<=> x-9/7 + x-9/8 = x-9/5 + x-9/6
<=> x-9/5 + x-9/6 - x-9/7 - x-9/8 = 0
<=> (x-9).(1/5+1/6-1/9-1/8) = 0
<=> x-9 = 0 ( vì 1/5+1/6-1/9-1/8 > 0 )
<=> x = 9
Vậy x = 9
Tk mk nha
MH
31 tháng 1 2021
1/ \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)
Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)
(1) \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\\ \Leftrightarrow\left(16+21+12-5\right)x=-5-28\\ \Leftrightarrow44x=-33\\ \Leftrightarrow x=-\dfrac{3}{4}\) (Thỏa mãn)
Vậy \(x=-\dfrac{3}{4}\).
2/ \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\) (2)
Điều kiện: \(x\ne\pm1\)
(2)\(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow x\left(x+1\right)-2x=0\\ \Leftrightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
So sánh với điều kiện \(\Rightarrow x=0\) là nghiệm của PT.
3/ \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\) (3)
Điều kiện: \(x\ne\pm3\)
(3)\(\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\\ \Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\\ \Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0\) (Vô nghiệm do \(x^2+x+8>0\qquad\forall x\)).
Vậy PT vô nghiệm.
4/ \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\) (4)
Điều kiện: \(x\ne\pm1\)
(4)\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1\) (loại)
Vậy PT vô nghiệm.
5/ \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (5)
Điều kiện: \(x\ne0\)
(5)\(\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2\)
Đặt \(t=x+\dfrac{1}{x}\), ta có: \(t=t^2-2\\ \Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-1\end{matrix}\right.\)
Với \(t=2\) ta có: \(x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thỏa mãn)
Với \(t=-1\) ta có: \(x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
6/ \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\) (6)
Điều kiện: \(x\ne-1\)
(6)\(\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.\)
\(x-1=0\Leftrightarrow x=1\) (Thỏa mãn)
\(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\\ \Leftrightarrow x^2-x+3=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
31 tháng 1 2021
1) ĐKXĐ: \(x\notin\left\{1;-\dfrac{4}{3}\right\}\)
Ta có: \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0\)
\(\Leftrightarrow20x+33=0\)
\(\Leftrightarrow20x=-33\)
\(\Leftrightarrow x=-\dfrac{33}{20}\)(nhận)
Vậy: \(S=\left\{-\dfrac{33}{20}\right\}\)
2) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
Suy ra: \(x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
3) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\)
\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(-x-3-14=x^2-9\)
\(\Leftrightarrow x^2-9=-x-17\)
\(\Leftrightarrow x^2-9+x+17=0\)
\(\Leftrightarrow x^2+x+8=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0\)(vô lý)
Vậy: \(S=\varnothing\)
4) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
5) ĐKXĐ: \(x\ne0\)
Ta có: \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}\)
\(\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)\)
\(\Leftrightarrow x^4+x^2=x^5+x\)
\(\Leftrightarrow x^5+x-x^4-x^2=0\)
\(\Leftrightarrow x\left(x^4-x^3-x+1\right)=0\)
\(\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\)
nên \(x\cdot\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy: S={1}
6) ĐKXĐ: \(x\in R\)
Ta có: \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2+x-3\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\)
nên x-1=0
hay x=1(nhận)
Vậy: S={1}
HH
5 tháng 2 2018
Giải:
a) \(4\left(x+1\right)-\left(3x+1\right)=14\)
\(\Leftrightarrow4x+4-3x-1=14\)
\(\Leftrightarrow x+3=14\)
\(\Leftrightarrow x=14-3=11\)
Vậy ...
b) \(-2\left(x-1\right)-\left(2-3x\right)=2\left(x+1\right)-4\)
\(\Leftrightarrow-2x+2-2+3x=2x+2-4\)
\(\Leftrightarrow x=2x-2\)
\(\Leftrightarrow x-2x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)
Vậy ...
c) \(-\left(x-3\right)+\left(4x-1\right)-14=2\left(x-2\right)-5\)
\(\Leftrightarrow-x+3+4x-1-14=2x-4-5\)
\(\Leftrightarrow3x-15=2x-9\)
\(\Leftrightarrow3x-2x=-9+15\)
\(\Leftrightarrow x=6\)
Vậy ...
VK
4
13 tháng 7 2018
1, \(14:x+6=8\)
\(14:x=8-6=2\)
\(x=14:2=6\)
2, \(x:\frac{1}{4}\cdot5=50\)
\(x:\frac{1}{4}=50:5=10\)
\(x=10\cdot\frac{1}{4}=2,5\)
13 tháng 7 2018
3, \(x:\frac{7}{3}+4=16\)
\(x:\frac{7}{3}=16-4=12\)
\(x=12\cdot\frac{7}{3}=28\)
4, \(24-x\cdot\frac{1}{3}=14\)
\(x\cdot\frac{1}{3}=24-14=10\)
\(x=10:\frac{1}{3}=0,3\)
TK MK NHA. CÂU CUỐI MK KO LÀM ĐÂU. ^_^
NA
12 tháng 3 2022
\(a.=\dfrac{2019}{2020}\times\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\\ =\dfrac{2019}{2020}\times1=\dfrac{2019}{2020}\\ b.=\dfrac{25}{27}\times\left(\dfrac{17}{14}-\dfrac{1}{14}-\dfrac{2}{14}\right)\\ =\dfrac{25}{27}\times1=\dfrac{25}{27}\)
LP
3
H9
HT.Phong (9A5)
CTVHS
23 tháng 3 2023
a) \(x-\dfrac{5}{8}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{5}{8}\)
\(\Rightarrow x=\dfrac{2}{8}+\dfrac{5}{8}=\dfrac{7}{8}\)
b) \(\dfrac{4}{7}:x=-\dfrac{5}{14}\)
\(\Rightarrow x=\dfrac{4}{7}:-\dfrac{5}{14}\)
\(\Rightarrow x=-\dfrac{8}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 7 2023
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 7 2023
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
HM
1
2 tháng 5 2022
\(x:\dfrac{3}{4}\) = \(\dfrac{9}{14}-\dfrac{1}{7}\)
\(x:\dfrac{3}{4}\) = \(\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\) x \(\dfrac{3}{4}\)
\(x=\dfrac{3}{8}\)
\(x\)x \(\dfrac{1}{2}-\dfrac{1}{4}\)=\(\dfrac{2}{3}\)
\(x\) x \(\dfrac{1}{4}\) =\(\dfrac{2}{3}\)
\(x\) = \(\dfrac{2}{3}:\dfrac{1}{4}\)
\(x\) \(\dfrac{8}{3}\)
(x+4).(x+1)=14
\(\Rightarrow\)x+4=14 hoặc x+1=14
x= 10 x= 13
Vậy x \(\in\){ 10; 13 }
(x+4).(x+1)=14
\(\Rightarrow\)x+4=14 hoặc x+1=14
x=10 x=13
Vậy x\(\in\){ 10; 13 }