(x^2-2x+4)(x+2)-x(x-1)(x+1)+3=0 giải giúp mik vs mik cần gấp ạ cảm ơn mn
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1 tháng 7 2017
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
1 tháng 7 2017
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
KL
2
22 tháng 10 2023
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
XO
6 tháng 8 2020
Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)
=> \(2x-2=-\frac{1}{2}\)
=> \(2x=\frac{3}{2}\)
=> \(x=\frac{3}{4}\)
LB
0
14 tháng 7 2021
a, 1,5 +|2x - 2/3| = 3/2
|2x - 2/3| = 3/2 - 1,5
|2x - 2/3| = 0
<=> 2x - 2/3 = 0
<=> 2x = 0 + 2/3
<=> 2x = 2/3
<=> x = 2/3 : 2
<=> x = 1/3
Vậy x = 1/3
14 tháng 7 2021
b, 3/4 - |1/4 - x| = 5/8
|1/4 - x| = 3/4 - 5/8
|1/4 - x| = 1/8
<=> 1/4 - x = 1/8
1/4 - x = /1/8
<=> x = 1/4 - 1/8
x = 1/4 - ( -1/8)
<=> x = 1/8
x = 3/8
Vậy x thuộc { 1/8 ; 3/8 }
HT
1
4 tháng 4 2020
<=> x2 -4+3x2= 4x2+4x+1+2x
<=> 4x^2 - 4= 4x^2 +6x +1
<=> - 4=6x +1
<=> 6x= -5
<=> x= \(-\frac{5}{6}\)
11 tháng 7 2018
1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)\)
\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-8x^2+20x-8\)
\(=-2x^2+3x-3\)
11 tháng 7 2018
a) x(x+4)(x-4)-(x2+1)(x2-1)
=>x(x2-42)-(x4-12)
=>x3-16x-x4+1
=>-x4-x3-15x
b) 7x(4y-x)+4y(y-7x)-2(2y2-3.5x)
=>28xy-7x2+4y2-28xy-4y2+30x
=>-7x2+30x
c) (3x+1)(2x-5)-4(2x2-5x+2)
=>6x2-15x+2x-5-8x2+20x-8
=>-2x2+7x-13
GT
4
9 tháng 11 2018
\(a,x^2-4x+1=0.\)
\(\text{Áp dụng biệt thức }\Delta=b^2-4ac\text{, ta có:}\)(Lớp 9 kì 2 hok)
\(\Delta=-4^2-4.1.1=16-4=12\)
\(\Rightarrow\text{pt có 2 nghiệm }\orbr{\begin{cases}x_1=\frac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x_2=\frac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{cases}}\)
b,bn xem lại đề nếu đúng nói mk 1 tiếng mk làm tiếp cho
\(\left(x^2-2x+4\right)\left(x+2\right)-x\left(x-1\right)\left(x+1\right)+3=0\)\(\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)+4\left(x+2\right)-x\left(x-1\right)\left(x+1\right)+3=0\)\(\Leftrightarrow x^3+2x^2-2x^2-4x+4x+8-\left[x^2\left(x+1\right)-x\left(x+1\right)\right]+3=0\)
\(\Leftrightarrow x^3+\left(2x^2-2x^2\right)+\left(-4x+4x\right)+8-\left(x^3+x^2-x^2-x\right)+3=0\)
\(\Leftrightarrow x^3+8-\left(x^3-x\right)+3=0\)
\(\Leftrightarrow x^3+\left(8+3\right)-x^3+x=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+11+x=0\)
\(\Leftrightarrow11+x=0\)
\(\Leftrightarrow x=-11\)
Vậy x = -11