2x^5 + 2 = 4
Những câu hỏi liên quan
1. 2x*(3x^2+4)-6x*(x-7)
2. (x+1)*(2x-2)-(2x-1)^2
3. (2x-3)^2-2*(2x-3)*(2x+5)+(2x+5)^2
4. (x-2)^4+(x+2)^4
5. (x+1)^5-(x-1)^5
tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho
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Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
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tìm x, biết
1. -11/2x + 1= 1/3x - 1/4
2. 2x- 2/3 - 7x = 3/2 - 1
3. 3/2x - 2/5 = 1/3x - 1/4
4. 2/3 - 5/3x= 7/10x + 5/6
5. 2x -1/4 = 5/6 - 1/2x
6. 3x - 5/3 = x - 1/4
7. - 5/6 + 3x = 2/3 - 1/2x
8. 1/2 ( x + 2 ) - 4( x - 1/4 ) = 1/2x
9. 5/2( x - 3/5 ) - 1/10 = x-3
10. -4/3( x - 1/4 ) = 3/2( 2x - 1 )
Giúp mk với !!!
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
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Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
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Bài 3: \(\dfrac{3}{2}\)\(x\) - \(\dfrac{2}{5}\) = \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)
\(\dfrac{3}{2}\)\(x\) - \(\dfrac{1}{3}x\) = \(-\dfrac{1}{4}\) + \(\dfrac{2}{5}\)
(\(\dfrac{9}{6}\) - \(\dfrac{2}{6}\))\(x\) = \(\dfrac{-5}{20}\) + \(\dfrac{8}{20}\)
\(\dfrac{7}{6}x\) = \(\dfrac{3}{20}\)
\(x\) = \(\dfrac{3}{20}\) : \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{9}{70}\)
Vậy \(x=\dfrac{9}{70}\)
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Xem thêm câu trả lời
8 tháng 9 2021 lúc 14:22
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
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a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
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1) (x + 1/2).(2/3 - 2x) = 0 2) 2/3x + 1/2x = 5/2 : 3 và 3/4 3) (2x - 3)(6 - 2x)= 0 4) -5(x + 1/5) - 1/2(x - 2/3) = 3/2x - 5/6
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
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cho A=1/101+1/102+1/103+...+1/199+1/200 chứng minh 1/2 <A<1
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1. (x-6)^2 = 2(x-6) ||||| 2. 2(x-3)^2 = (x-3)(x+5) ||||| 3. 4(x-3)=2x-5(2x+3) | 4. x2 +4 -2 (x-1) = (x-2)^2 |||| 4. x-3/5 = 6 - 1-2x/3 ||||| 5. x+2 = 6-5x/2 ||||| 6. x+2/5 - x+3 = x-2/2 ||||| 7. 2x-5/x-4 = 2x+1/x+2 ||||| 8. X+3/x-3 - x-1/x+3 = x2 + 4x + 6/x2 -9 |||| 9. 3x-3/x2-9 -1/x-3 = x+1/x+3 |||| 10. X+1/x-1 - x-1/x+1 = 4/x2 -1
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(x + 2)2-2x-4 = 0
(2x+5)2-2(2x +1)(2x +5 ) + (2x + 1)2
(x + 2)2 - 2x - 4 = 0
<=> (x + 2)2 - 2(x + 2) = 0
<=> (x + 2)(x + 2 - 2) = 0
<=> x(x + 2) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
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1.(x+2)3+(x-3)2-x2(x+5)
2.(2x+3).(x-5)+2x(3-x)+x-10
3.(x+5).(x2-5x+25)-x(x-4)2+16x
4.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
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a,-2\3x+1\5=1\10
b,2|1\2x-3\8|-3\2=1\4
c,-5(x+1\5)-1\2(x-2\3)=3\2x-5\6
d,3(x-1\2)- 5(x+3\5)=-x+1\5
e,3\4-2.|2x-0,125|=2
f,2|1\2x-1\3|-3\2=1\4
g,4\5-1\2x=1\10
a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{1}{10}-\frac{1}{5}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)
\(\Leftrightarrow x=\frac{3}{20}\)
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
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