tìm x
( 2x + 8) : 4 = ( x+1)
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8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
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
17 tháng 8 2021
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20
Q
2
2
28 tháng 5 2022
`2//(5x-8)-3(4x-5)=4(3x-4)`
`<=>5x-8-12x+15=12x-16`
`<=>-19x=-23`
`<=>x=23/19` Vậy `x=23/19`
`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`
`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`
`<=>4x=8`
`<=>x=2` Vậy `x=2`
5 tháng 11 2023
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
22 tháng 6 2019
\(2x+\left(1+2+3+...+100\right)=15150\)
\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)
\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101
\(2x+\left[101\times50\right]=15150\)
\(2x=15150:5050\)
\(2x=3\)
\(x=3:2\)
\(x=1.5\)
SS
22 tháng 6 2019
a, 2x + (1+2+3+4+...+100) = 15150
=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150
=> 2x + \(\frac{101.100}{2}\)= 15150
=> 2x + 5050 = 15150
=> 2x = 15150 - 5050
=> 2x = 10100
=> x = 10100 : 2
=> x = 5050
Vậy x = 5050
b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36
=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36
=> 8x + 36 = 36
=> 8x = 0
=> x = 0
Vậy x = 0
c, 0+0+4+6+8+...+2x=110
Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110
SSH : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)
Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)
\(\Leftrightarrow\left(x+1\right)x=110\)
\(\Leftrightarrow\left(10+1\right).10=110\)
=> x = 10
Vậy x = 10
VM
0
CP
0
HN
3
BA
2 tháng 9 2019
\(a,\frac{8}{x}=\frac{x}{4}\)
\(=>x\cdot x=8\cdot4\)
\(=>x^2=32\)
\(=>x=\sqrt{32}\)
BA
2 tháng 9 2019
\(c,\frac{2x+3}{6}=\frac{x+1}{-8}\)
\(=>-8\cdot\left(2x+3\right)=6\cdot\left(x+1\right)\)
\(=>-16x-24=6x+6\)
\(=>-16x-6x=6+24\)
\(=>-22x=30\)
\(=>x=\frac{30}{-22}=-\frac{15}{11}\)
24 tháng 6 2021
`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`
`đk:x ne 0,x ne -2`
`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`
`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`
`=-4/(x+2)^2*(x+2)/4`
`=-1/(x+2)`
`b)P<0`
`<=>-1/(x+2)<0`
Vì `-1<0`
`<=>x+2>0`
`<=>x> -2`
`c)P=1/x+1(x ne 0)`
`<=>-1/(x+2)=1/x+1`
`<=>1/x+1+1/(x+2)=0``
`<=>x+2+x(x+2)+x=0`
`<=>x^2+4x+2=0`
`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\)
`d)|2x-1|=3`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\)
`x=-1=>P=-1/(-1+2)=-1`
`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?
24 tháng 6 2021
a) đk: \(x\ne-2;2\)
\(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)
= \(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)
= \(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}\) = \(\dfrac{-4}{4\left(x+2\right)}=\dfrac{-1}{x+2}\)
b) Để P < 0
<=> \(\dfrac{-1}{x+2}< 0\)
<=> x +2 > 0
<=> x > -2 ( x khác 2)
c) Để P= \(\dfrac{1}{x}+1\)
<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)
<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)
<=> x2 + 4x + 2 = 0
<=> (x+2)2 = 2
<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)
d) Để \(\left|2x-1\right|=3\)
<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)
Thay x = -1, ta có:
P = \(\dfrac{-1}{-1+2}=-1\)