giải pt sau:a,x.(x-3)=(2-x).(x-3)
b,x-1/2+x-1/3+x-1/2016=0
c,2x/3+2x-1/6=4
d,7+2x=4.(5-x)
e,x+2/x-2-1/x=2/x.(x-2)
Giải các pt sau:
a) (x-3)-(x-3)(2x-5)/6=(x-3)(3-x)/4
b) (2x-7)^2-x^2+8x-16=0
c) (3x+1)(x-3)=(3x+1)(2x-5)
\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)
Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)
\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)
\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)
-Giải pt :
A. x(x+3)-(2x-1)(x+3)=0
B. 3x-5(x+2)=3(4-2x)
C. (4x-3)(5x-6)=(4x-3)(2x-3)
\(a, x(x+3)-(2x-1)(x+3)=0\)
\(⇔(x+3)(1-x)=0\)
\(⇔\left[\begin{array}{} x+3=0\\ 1-x=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=-3\\ x=1 \end{array}\right.\)
Vậy phương trình có tập nghiệm là S={\(-3; 1\)}
\(b, 3x-5(x+2)=3(4-2x)\)
\(⇔3x-5x-10=12-6x\)
\(⇔3x-5x+6x=12+10\)
\(⇔4x=22\)
\(⇔x=\dfrac{22}{4}\)
Vậy pt có 1 nghiệm là \(x=\dfrac{22}{4}\)
\(c, (4x-3)(5x-6)=(4x-3)(2x-3)\)
\(⇔5x-6=2x-3\)
\(⇔5x-2x=-3+6\)
\(⇔3x=3\)
\(⇔x=1\)
Vậy pt có 1 nghiệm là \(x=1\)
Giải pt
a) 2x(x+5) -(x-3) ²=x²+6
b) 6(x-3) +(x-1) ²-(x+1) ²=2x
c) 2x-3(x-2) =x+6+(x+1)
d) (x+4) ²-(x+8) (x-8) =96
e) 5x-2/6+3-4x/2=2-x+7/3
f) 2x-1/2=2x+1/4-1-2x/8
Tìm x, biết
a)3x³-27x=0
b) 2x³-12x²+18x=0
Giải pt :
a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)
\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)
\(\Leftrightarrow16x-15=0\)
\(\Leftrightarrow x=\frac{15}{16}\)
b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)
\(\Leftrightarrow2x-18=2x\)
\(\Leftrightarrow-18=0\)( vô lí )
=> x thuộc rỗng
c)d) tương tự
e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)
\(\Leftrightarrow5x-2+9-12x=12-2x-14\)
\(\Leftrightarrow-5x+9=0\)
\(\Leftrightarrow x=\frac{9}{5}\)
f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)
\(\Leftrightarrow8x-4=4x+2-1+2x\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
Tìm x :
a) \(3x^3-27x=0\)
\(\Leftrightarrow3x\left(x^2-9\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(2x^3-12x^2+18x=0\)
\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
giải pt:
a) x^5 + 2x^4 + 3x^3 + 3x^2 + 2x +1=0
b) x^4 + 3x^3 - 2x^2 + x - 3 = 0
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy....
b) \(x^4+3x^3-2x^2+x-3=0\)
\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)
\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)
...
\(\Leftrightarrow x=1\)
p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
d) 5/-x2+5x-6 + x+3/2-x = 0
e) x/2x+2 - 2x/x2-2x-3 = x/6-2x
f) 1/x-1 - 3x2/x3-1 = 2x/x2+x-1
giải pt: x^5 + 2x^4 +3x^3 + 3x^2 + 2x +1=0
giải pt: x^4 + 3x^3 - 2x^2 +x - 3=0
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
Giải pt sau
a.(2x+3)(x-5)=4x2+6x
b.x/2x-6 - x/2x+2 = 2x/(x+1)(x-3)
c.giải bpt sau : 12x+1/12 ≤ 9x+1/3 - 8x+1/4
Bài 1: giải các pt sau:
1,(x-1)^2-(x+1)^2=2(x+3)
2,(2x-1)^2-(2x+1)^2=4(x-3)
3,(2x+3)^2-(2x+3).(2x-4)=-(x-2)^2
4,8x^3-(x+1)^3=3x-3
5,(3x-2).(9x^2+6x+4)-(3x+1).(9x^2-3x+1)=(2x+1).(2x-1)-4x(x-3)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)