giải các phương trình:
a) 2x2 + 2xy + y2 + 9 = 6x - /y+3\
b)(2x2 + x - 2013)2 + 4(x2-5x-2012)2 = 4(2x2+ x -2013)(x2 -5x -2012)
Giải các phương trình :
a)2x2+2xy+y2+9=6x-/y+3/
b)(2x2+x-213)2+4(x2-5x-2012)2=4(2x2+x-213)(x2-5x-2012)
Đặt \(2x^2+x-213=a;x^2-5x-2012=b\)
Khi đó phương trình tương đương với:
\(a^2+4b^2=4ab\)
\(\Leftrightarrow\left(a-2b\right)^2=0\)
\(\Leftrightarrow a-2b=0\)
\(\Leftrightarrow a=2b\)
Đến đây dễ rồi
giải các phương trình sau
a) 2x2 + 2xy + y2 + 9 = 6x - /y+3\
b) (2x2 + x -2013)2 + 4(x2-5x-2012)2 = 4(2x2+x-2013)(x2-5x-2012)
1) Tìm GTLN:
A = x2 + 5y2 + 2xy -4x -8y +2015
B= (x- 2012)2 + (x+2013)2
C= (x-1) (2x-1) (2x2- 3x -1) +2017
D= (x-1) (x-3) (x-4) (x-6) +10
C=(2x-1)(x-1)(2x^2-3x-1)+2017
=(2x^2-3x+1)(2x^2-3x-1)+2017
=(2x^2-3x)^2-1+2017
=(2x^2-3x)^2+2016>=2016
Dấu = xảy ra khi 2x^2-3x=0
=>x=0 hoặc x=3/2
D=(x-1)(x-6)(x-3)(x-4)+10
=(x^2-7x+6)(x^2-7x+12)+10
=(x^2-7x)^2+18*(x^2-7x)+72+10
=(x^2-7x+9)^2+1>=1
Dấu = xảy ra khi x^2-7x+9=0
=>\(x=\dfrac{7\pm\sqrt{13}}{2}\)
a,Giải phương trình:
(2x2+x-2013)2+4(x2-5x-2012)2=(2x2+x-2013)(x2-5x-2012)
b,Tìm x,y € Z : x3+2x2+3x+2=y3
Giúp nha mn.Thanks nhiều!
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
Giải phương trình:
a)x2-4x+4=0
b)2x2-x=0
c)x2-5x+6=0
d)x2+y2=0
e)x2+6x+10=0
\(a.x^2-4x+4=0\)
\(\left(x-2\right)^2=0\)
=>x=2
b) \(2x^2-x=0\)
\(x\left(2x-1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(x^2-5x+6=0\)
\(x^2-2x-3x+6=0\)
\(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
d) \(x^2+y^2=0\)
Vì \(x^2,y^2\ge0\forall x,y\)
=>x=y=0
e) \(x^2+6x+10=0\)
\(\left(x+3\right)^2+1=0\)
Vì \(\left(x+3\right)^2\ge0\forall x\)
=> VT>0 \(\forall x\)
=> phương trình vô nghiệm
a) \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
b) \(2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(x^2-5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) \(\left(a+b+c=0\right)\)
d) \(x^2+y^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
e) \(x^2+6x+10=0\)
\(\Leftrightarrow x^2+6x+9+1=0\)
\(\Leftrightarrow\left(x+3\right)^2+1=0\left(1\right)\)
mà \(\left(x+3\right)^2+1\ge1>0,\forall x\in R\)
Nên phương trình (1) vô nghiệm
bài 1 giải các bất phương trình sau
a, -x2 +5x-6 ≥ 0
b, x2-12x +36≤0
c, -2x2 +4x-2≤0
d, x2 -2|x-3| +3x ≥ 0
e, x-|x+3| -10 ≤0
bài 2 xét dấu các biểu thức sau
a,<-x2+x-1> <6x2 -5x+1>
b, x2-x-2/ -x2+3x+4
c, x2-5x +2
d, x-< x2-x+6 /-x2 +3x+4 >
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
\(m,x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
\(n,2x^2+5x-3=0\)
\(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(k,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-4x-7x+14=0\)
\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
Giải phương trình :
1) √x2+x+2 + 1/x= 13-7x/2
2) x2 + 3x = √1-x + 1/4
3) ( x+3)√48-x2-8x= 28-x/ x+3
4) √-x2-2x +48= 28-x/x+3
5) 3x2 + 2(x-1)√2x2-3x +1= 5x + 2
6) 4x2 +(8x - 4)√x -1 = 3x+2√2x2 +5x-3
7) x3/ √16-x2 + x2 -16 = 0