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20 tháng 7 2018
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
30 tháng 8 2016
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
DQ
3
5 tháng 8 2016
\(2x:15=0,4:0,8\)
\(=2x:15=0,5\)
\(\Rightarrow2x=7,5:2\)
\(\Rightarrow x=3,75\)
Mk pt nhiu đóa à!
5 tháng 8 2016
4.(3x-2)-10=4(x-2)
=>4.(3x-2)-10-4.(x-2)=0
=> 4.[ (3x-2)-(x-2)] -10=0
=> 4.2x-10=0
=>8x-10=0
=>8x=10
=>x= 5/4
29 tháng 12 2019
\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)
\(=\frac{\left(x^2-36\right).3}{\left(2x+10\right)\left(6-x\right)}\)
\(=\frac{3\left(x+6\right)\left(x-6\right)}{\left(2x+10\right)\left(6-x\right)}\)
\(=-\frac{3\left(x+6\right)\left(x-6\right)}{2\left(x+5\right)\left(x-6\right)}\)
\(=-\frac{3\left(x+6\right)}{2\left(x+5\right)}\)
ZD
0
M
1 tháng 4 2020
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
HH
7 tháng 8 2017
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
17 tháng 7 2021
a,thay P(1),P(2),P(3),P(4) vào P(x(=) rồi giải hệ pt
câu b thì thay x=567 vào P(x) tính đc ở trên nhờ có các hệ số a,b,c,d
Ko biết có đúng ko hì hì
HN
16 tháng 12 2018
\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)
\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)
\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)
\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)
\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)
\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)
C1 : \(\frac{2x+4}{10}+\frac{2-x}{15}=30x+60=20-10x\)
C2 : \(\frac{2x-4}{10}+\frac{2-x}{15}=\frac{6x-12}{30}+\frac{4-2x}{30}=\frac{4x-8}{30}\)