Tìm x sao cho:
(2x+3)(x-1)+(2x-3)(1-x)=0
Giúp mình với ạ😣
Tìm x
(x-3/2).(2x+1)>0
(2-x).(4/5-x)<0
Giúp mình với ạ
\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)
Th1:
\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)
\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)
( 1 )
Th2:
\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)
\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)
( 2 )
Từ ( 1 ) và ( 2 ), ta có:
\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)
\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
Th1:
\(2-x>0\Leftrightarrow x>2\)
\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)
( Loại )
Th2:
\(2-x< 0\Leftrightarrow x< 2\)
\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)
=> \(\dfrac{4}{5}< x< 2\)
Tìm x:
(x/3 + 1/2) (75%- 1 1/2x)=0
GIÚP MÌNH VỚI Ạ. CẢM ƠN MỌI NGƯỜI!
`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; 1/2}.`
Tìm x: d) (4x-3)(2x-5) +(3-4x)(x-1)=0 giúp mình với ạ.
(4x-3)(2x-5) +(3-4x)(x-1)=0
(4x-3)(2x-5)-(4x-3)(x-1)=0
(4x-3)(2x-5-x+1)=0
(4x-3)(x-4)=0
4x-3=0 hoặc x-4=0
x=\(\frac{3}{4}\)hoặc x=4
Bài 10: Tìm x biết:
1/ 5(x+43) – 2x(3+4x) = 0
Giúp mình với!! Mình cảm ơn trước ạ!! ^^
Ai đó giúp mình nha. Tìm x lớp 6
1) 1/3 + 2/3 : x = -7
2) 1/3x + 2/5(x-1) = 0
3) ( 2x-3)(6-2x)=0
4) 2|1/2x-1/3| - 2/3= 1/4
Mình xin cảm ơn trc ạ. Ai làm giúp mình cả 4 cau nhé . Cảm ơn ạ
Tìm x: (x-3)(x-1)-x(x-2)=0 (x+2y)²-(2x-y)² [Giúp mình ạ vs ạ]
a: (x-3)(x-1)-x(x-2)=0
=>\(x^2-4x+3-x^2+2x=0\)
=>\(-2x+3=0\)
=>-2x=-3
=>\(x=\dfrac{3}{2}\)
b: \(\left(x+2y\right)^2-\left(2x-y\right)^2\)
\(=\left(x+2y+2x-y\right)\left(x+2y-2x+y\right)\)
\(=\left(3x+y\right)\left(-x+3y\right)\)
Câu 1: Giải các phương trình sau:
a) 3x-2(x-3)=0
b) (x+1) (2x-3) = ( 2x -1) (x +5)
c) 2x/ x-1 -x/x+1 =1
d) (2x +3) (3x-5)=0
e) x-2/x+2-3/x-2 = 2(x-11)/ x2
giúp mình với ạ huhu\(^{ }\)
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
Mọi người ơi giúp mình bài này với ạ,mai mình học rồi , cho mình câu trả lời rõ ý nhất với nha, mik cảm ơn ạ.
bài tìm x:
a,2x2+3(x-1)(x+1)=5x(x+1)
b,(x-3)3+3-x=0
c,5x(x-2000)-x+2000=0
d, 3(2x - 3) + 2(2 - x) = -3
e,x+6x2=0
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
__________________________________________
`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
__________________________________________
`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
Tìm x biết:
b) 2x(x+3)-3(x^2+1)=x+1-x(x-2)
Giúp mình với ạ
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)
B=2x^2+6x-3x^2-3=x+1-x(x-2)=0
=-x^2+6x-3=x+1-x^2+x=0
=4x-3=0
x=3/4