\(4x^2-12x=-9\)
b) \(\left(5-2x\right)\left...">
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23 tháng 10 2016

a) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

\(\Leftrightarrow\left(2x-3\right)^2=0\)

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)

d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)

8 tháng 6 2016

a) 4x2 - 12x + 9 = 0 <=> (2x - 3)2 = 0 <=> 2x - 3 = 0 <=> x = 3/2.KL

b) ( 5 - 2x )( 2x + 7 ) + ( 25 - 4x2 ) = 0 <=> ( 5 - 2x )( 2x + 7 ) + ( 5 + 2x )( 5 - 2x ) = 0 <=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0. KL

<=> ( 5 - 2x )( 4x + 12 ) = 0 <=>\(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=2\frac{1}{2}\\x=-3\end{cases}}\)KL.

c) ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 3 ) = 0 <=> ( x + 3 )( x2 - 3x + 9 + x - 3 ) = 0 <=> ( x + 3 )( x2 -2x + 6 ) = 0 <=> x + 3 = 0 (vi x2 - 2x + 6 = ( x + 1 )2 + 5 > 0 voi moi x) KL

<=>x=-3.KL

d) [ 2 ( 2x + 7 ) ]2 - [ 3 ( x + 3 ) ]2 = 0 <=> ( 4x + 14 )2 - ( 3x + 9 )2 = 0 <=> ( 4x + 14 + 3x + 9 )( 4x + 14 - 3x -9 ) = 0

<=> ( 7x + 23 )( x + 5 ) = 0 <=> 7x + 23 = 0 hoac x + 5 = 0 <=> x = -23/7 hoac x = -5.KL

14 tháng 10 2018

\(x^3-8-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\)

Lại có : \(x^2+x+16=\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}>0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy...

b/ \(A=x^2-2x+9=\left(x-1\right)^2+8\ge8\)

Dấu "=" xảy ra khi : \(x=1\)

Vậy...

c/ \(B=x^2+6x-3=\left(x^2+6x+9\right)-12=\left(x+3\right)^2-12\ge-12\)

Dấu "=" xảy ra khi \(x=-3\)

Vậy...

d/ \(C=\left(x-1\right)\left(x-3\right)+9=\left(x^2-4x+3\right)+9-\left(x^2-4x+4\right)+8=\left(x-2\right)^2+8\ge8\)

Dấu "=" xảy ra khi : \(x=2\)

Vậy ...

e/ \(D=-x^2-4x+7=-\left(x^2+4x+4\right)+3=-\left(x+2\right)^2+3\le3\)

Dấu "=" xảy ra khi \(x=-2\)

Vậy...

g/ \(E=5-4x^2+4x=-\left(4x^2-4x+4\right)+9=-\left(x-2\right)^2+9\le9\)

Dấu "=" xảy ra khi \(x=2\)

Vậy...

https://i.imgur.com/u6zkAVa.jpg
14 tháng 2 2020

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

\(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

31 tháng 8 2017

a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)

b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)

c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)

d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)

e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)

f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)

i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)

30 tháng 8 2017

A),(-2)5:(-2)3=(-2)2=4

B) (-y)7 :(-y)3=y4

3 tháng 3 2020

\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)

\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)

\(\Leftrightarrow4x+4x>-1\)

\(\Leftrightarrow8x>-1\)

\(\Leftrightarrow x>-\frac{1}{8}\)

\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)

\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)

\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)

\(\Leftrightarrow4x^2-6x^2< 1+3\)

\(\Leftrightarrow-2x^2< 4\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow x>\pm\sqrt{2}\)

3 tháng 7 2017

1. \(f\left(x\right)=25x^2-20x+\dfrac{9}{2}\)

=>\(f\left(x\right)=25x^2-20x+4+\dfrac{1}{2}\)

=> \(f\left(x\right)=(25x^2-20x+4)+\dfrac{1}{2}\)

=> \(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\)

Ta thấy: \((5x-2)^2\ge0\)

=>\(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)(đpcm)

2. \(f\left(x\right)=4x^2-28x+50\)

=> \(f\left(x\right)=(4x^2-28x+49)+1\)

=> \(f\left(x\right)=(2x-7)^2+1\)

Ta thấy: \((2x-7)^2\ge0\)

=> \(f\left(x\right)=(2x-7)^2+1\ge1>0\) (đpcm)

3. \(f\left(x\right)=-16x^2+72x-82\)

=> \(f\left(x\right)=-(16x^2-72x+82)\)

=> \(f\left(x\right)=-(16x^2-72x+81+1)\)

=> \(f\left(x\right)=-[(4x-9)^2+1]\)

Ta thấy: \((4x-9)^2\ge0\)

=> \((4x-9)^2+1\ge1>0\)

=> \(f\left(x\right)=-[(4x-9)^2+1]< 0\)

5. \(f\left(x;y\right)=4x^2+9y^2-12x+6y+11\)

=> \(f\left(x;y\right)=4x^2+9y^2-12x+6y+9+1+1\)

=> \(f\left(x;y\right)=(4x^2-12x+9)+(9y^2+6y+1)+1\)

=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\)

Ta thấy: \((2x-3)^2\ge0\)

\((3y+1)^2\ge0\)

=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\) \(\ge1>0\) (đpcm)

24 tháng 9 2016

a) Áp dụng hằng đẳng thức số 3 bạn nhé

b) (2x + 3)(4x^2 - 6x +9) = 8x^3 + 9 

Thay x= 120:2 = 60 vào biểu thức.

8* 60^3 + 9 = 1728009 

c) = (2x + 1)^3 

Thay x= -0,5 vào biểu thức

[2*(-0,5)+1]^3 = 0

d) = x^2 - 49 - x^2 - 2x - 1 = -50 - 2x 

Thay x=49 vào biểu thức.

-50 - 2* 49 = -148 

  

24 tháng 9 2016

\(2010^2-2009^2=\left(2010+2009\right)\left(2010-2009\right)=4019\)