Tìm X biết:3-(Xx2+1/2):1/2=2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)
\(\Leftrightarrow\left(x-4\right)4=0\)
\(\Leftrightarrow x=4\)
2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)
3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)
Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)
1.
x2 - 16 - x(x - 4) = 0
<=> (x2 - 42) - x(x - 4) = 0
<=> (x - 4)(x + 4) - x(x - 4) = 0
<=> (x + 4 - x)(x + 4) = 0
<=> 4(x + 4) = 0
<=> x + 4 = 0
<=> x = -4
2.
(x + 3)2 - (x - 3)(x + 5)
= x2 + 6x + 9 - (x2 + 5x - 3x - 15)
= x2 + 6x + 9 - x2 + 5x - 3x - 15
= x2 - x2 + 6x + 5x - 3x + 9 - 15
= 8x - 6
a: \(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x^2-1+x+2-x^2}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x-1}\)
b: P<1
=>P-1<0
=>\(\frac{x^2-x+1}{x-1}<0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được: x<1 và x∉{0;-1}
2/5 x 1/x + 1/x x 2 + 2/5 = 1/2
1/x x ( 2/5 + 2 ) + 2/5 = 1/2
1/x x 6/5 + 2/5 = 1/2
1/x x 6/5 = 1/2 - 2/3
( 1 : x) x 6/5 = -1/6
1:x = -1/6 : 6/5
1: x = -5/36
⇔ x = -36/5.
Vậy x = -36/5.
b: =>x/23=1+3/4+4/7=65/28
=>x=23*65/28=1495/28
c: =>3/5:x=3/5-1/4-1/2=9/40
=>x=3/5:9/40=8/3




\(3-\left(2x+\frac{1}{2}\right)\div\frac{1}{2}=2\)
\(\Leftrightarrow\left(2x+\frac{1}{2}\right)\div\frac{1}{2}=3-2\)
\(\Leftrightarrow\left(2x+\frac{1}{2}\right)\times2=1\)
\(\Leftrightarrow2x+\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow2x=\frac{1}{2}-\frac{1}{2}\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\div2\)
\(\Leftrightarrow x=0\)
Vậy x = 0
\(3-\left(x\times2+\frac{1}{2}\right):\frac{1}{2}=2\)
\(\left(x\times2+\frac{1}{2}\right):\frac{1}{2}=3-2\)
\(\left(x\times2+\frac{1}{2}\right):\frac{1}{2}=1\)
\(x\times2+\frac{1}{2}=1\times\frac{1}{2}\)
\(x\times2+\frac{1}{2}=\frac{1}{2}\)
\(x\times2=\frac{1}{2}-\frac{1}{2}\)
\(x\times2=0\)
x = 0 : 2
x = 0
Vậy x = 0