tim x biet
(2x+4)(3x+1)>0
Tim x biet
(2x-5)(4+3x).x<0
tim x biet /2x-1/-3x=0
=> |2x-1|=3x
=> 2x-1=3x hoặc 2x-1=-3x
=> x = -1 hoặc x = 1/5
k mk nha
tim x biet
3x(2x-1)-1-2x=0
\(3x\left(2x-1\right)-1-2x=0\)
\(\Leftrightarrow6x^2-3x-1-2x=0\)
\(\Leftrightarrow6x^2-5x-1=0\)
chắc bài này sai đề. sửa lại:
\(3x\left(2x-1\right)+1-2x=0\)
\(\Leftrightarrow3x\left(2x-1\right)-\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)
Tim xthuoc Z biet:
1,|2x-5|-|2x+9|=0
2,|x+1|-|x+2|-|3-x|=7
3,|2x+3|+|3x+2|-|4-x|=10
Tim x biet rang:
a) 4/4^x + 3.4^2-x = 832
b) (3x - 2)^2002 = (3x - 2)^2004
c) /1 - 2x/ + x + 2 = 0
tim x biet;(2x-1)(3x+1)+(3x-4)(3-2x)=5
ta co (2x-1)(3x+1)+(3x+4)(3-2x)=5
(=)6x2-3x+2x-1+6x-6x2+12-8x=5
(=)-4x+11=5
(=)-4x=-6
(=)x=3/2
(2x-1)(3x+1)+(3x-4)(3-2x)=5
<=> 6x2+2x-3x-1+9x-6x2-12+8x=5
<=> 16x-13=5
<=> 16x = 18
<=> x=9/8
bai1.tim x biet:
a,(x+2).(x+3)-(x-2).(x+5)=0
b,(2x+3).(x-4)+(x-5).(x-2)=(3x-5).(x-4)
c,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)=33
,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn
Tim x biet
\(x^4-2x^3-2x^2+3x+2=0\)
\(x^4-2x^3-2x^2+3x+2=0\)
\(\Leftrightarrow x^4-2x^3-2x^2+4x-x+2=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)-\left(2x^2-4x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x^3\left(x-2\right)-2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^3-x\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^2-x\right)\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x-1\right)=0\)
Đến đây ez r
Tim x biet:
A. (x-3/4).(3x+1/2)lon hon hoac bang 0
B. (2x+1).(4x+3)be hon hoac bang 0