x+1/x+2.x+2/x+3.x+3/x+4
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13 tháng 8 2017
a, 2x-3-x+5=x+2-x+1
2x-x-x+x=2+1+3-5
0x=1
=> x thuộc rỗng (vì số nào nhân với 0 cũng bằng 0)
b, 2x-2-5x+10=-10
2x-5x=-10+2-10
-3x=2
x=-2/3
c, 2x-10-3x+21=14
2x-3x=14+10-21
-x=3
x=-3
d, 5x-6-2x+6=12
5x-2x=12+6-6
3x=12
x=4
e, -35+7x-2x+10=15
7x-2x=15+35-10
5x=40
x=8
18 tháng 2 2023
a: =>1+3x-6=7-x
=>3x-5=7-x
=>4x=12
=>x=3(nhận)
b: \(\Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)
=>\(x^3-3x^2-x^2+3x-x^3-3x^2=-7x^2+3x\)
=>\(-7x^2+3x=-7x^2+3x\)
=>0x=0(luôn đúng)
Vậy: S=R\{3;-3}
c: =>x(x+2)+(2x-1)(x+1)=0
=>2x^2+2x-x-1+x^2+2x=0
=>3x^2+3x-1=0
\(x=\dfrac{-3\pm\sqrt{21}}{6}\)
d: =>2(x-2)-x-1=3x-11
=>3x-11=2x-4-x-1=x-5
=>2x=6
=>x=3(nhận)
TS
Bài 1: Tìm x
a) (x-5) (x-3)+ 2(x-5)=0
b) (x-2)(x^2+2x+4)-(x+2)(x^2-2x+4)=2(x+2)
giúp e với ạ, e cảm ơn
2
KV
30 tháng 8 2023
a) (x - 5)(x - 3) + 2(x - 5) = 0
(x - 5)(x - 3 + 2) = 0
(x - 5)(x - 1) = 0
x - 5 = 0 hoặc x - 1 = 0
*) x - 5 = 0
x = 5
*) x - 1 = 0
x = 1
Vậy x = 1; x = 5
b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)
x³ - 8 - x³ - 8 = 2x + 4
2x = -8 - 8 - 4
2x = -20
x = -20 : 2
x = -10
30 tháng 8 2023
a)
\(\left(x-5\right)\left(x-3\right)+2\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-3+2\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(x-5=0\) hoặc \(x-1=0\)
+) \(x-5=0\\ \Rightarrow x=5\)
+) \(x-1=0\\ \Rightarrow x=1\)
Vậy \(x=1\) hoặc \(x=5\)
b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)
\(x^3-8-x^3-8=2x+4\)
\(2x=-8-8-4\)
\(2x=-20\)
\(x=-20:2\)
\(x=-10\)
Vậy \(x=-10\)
NV
1
7 tháng 9 2017
(x2 _ 1)3 _ (x4 + x2 + 1) (x2 _ 1)
= x6 _ 1 _ ( x6 + x4 + x2 _ x4 _ x2 _ 1)
= x6 _ 1 _ x6 _ x4 _ x2 + x4 + x2 + 1
= 0
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
21 tháng 6 2020
a) 2x - 3 = -12
=> 2x = -12 + 3 = -9
=> x = \(-\frac{9}{2}\)
b) \(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
=> \(\frac{1}{2}+2x=-\frac{5}{6}\cdot\frac{3}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\cdot\frac{1}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\)
=> \(2x=-\frac{5}{2}-\frac{1}{2}=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
c) \(1< \frac{x}{5}< 2\)
=> \(\frac{5}{5}< \frac{x}{5}< \frac{10}{5}\)
=> 5 < x < 10
=> x \(\in\){6,7,8,9}
Dù bạn có cho âm vào nx thì nó vẫn sai nhá
d) Đặt \(A=\frac{x+5}{x-2}=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)
=> \(x-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
+) x - 2 = 1 => x = 3(T/M)
x - 2 = -1 => x = -1 +2 = 1(t/m)
x - 2 = 7 => x = 9 (t/m)
x - 2 = -7 => x = -7 + 2 = -5(t/m)
e) làm nốt ...
PN
21 tháng 6 2020
a,\(2x-3=-12\)
\(< =>2x=-12+3=-9\)
\(< =>x=-\frac{9}{2}\)
b,\(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
\(< =>\frac{1}{2}+\frac{4x}{2}=-\frac{5}{6}.\frac{3}{2}\)\(< =>\frac{4x+1}{2}=-\frac{5}{4}\)
\(< =>\frac{8x+2}{4}=-\frac{5}{4}\)\(< =>8x+2=-5\)
\(< =>8x=-5-2=-7\)\(< =>x=-\frac{7}{8}\)
31 tháng 8 2023
x^2+y^2=(x+y)^2-2xy
=5^2-2*3
=25-6
=19
x^3+y^3=(x+y)^3-3xy(x+y)
=5^3-3*3*5
=125-9*5
=80
(x-y)^2=(x+y)^2-4xy=5^2-4*3=13
=>\(x-y=\sqrt{13}\)
16 tháng 11 2021
ý bạn là tìm x hay sao?
\(a,\Leftrightarrow\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=\dfrac{x^2-3x-2}{x-1}\left(x\ne\pm1\right)\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=x^2-3x-2\\ \Leftrightarrow x^2-3x+2=x^2-3x-2\\ \Leftrightarrow2=-2\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\\ \Leftrightarrow x+2=x+2\\ \Leftrightarrow x\in R\)
16 tháng 11 2021
alo cho tui hỏi bạn có phải Dương Ngọc Lan Hương Trường THCS Minh Thuận 3 k dọ
14 tháng 6 2023
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
14 tháng 6 2023
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
Rút gọn đa thức ?
Với \(x\ne\left\{-2;-3;-4\right\}\) thì:
\(\frac{x+1}{x+2}\cdot\frac{x+2}{x+3}\cdot\frac{x+3}{x+4}=\frac{x+1}{x+4}\)