Tìm x :
a ) \(\left(x+3\right)^2=\left(x-2\right)\left(x+4\right)\)
b ) \(\left(x+4\right)^2=2x^2+16\)
Câu 1: Rút gọn các biểu thức sau:
1. \(\left(x+y-z\right)^2+\left(y-z\right)^2+2z\left(z-y\right)\)
2. \(\left(3x+4\right)^2+\left(x-4\right)^2+2\left(3x+4\right)\left(x-4\right)\)
3.\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
4. \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)\)
5. \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
Câu 2: Tìm x
1. \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=1\)
2. \(\left(3x+1\right)^2+\left(5x-2\right)^2=34\left(x+2\right)\left(x-2\right)\)
3. \(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\)
4. \(4x^2-9-x\left(2x-3\right)=0\)
5. \(4x^2-12x+9=0\)
Câu 3: Tìm GTNN
D = \(\left(2x-1\right)^2+\left(x+2\right)^2\)
Câu 4: Cho \(a^2+b^2+c^2=ab+bc+ac\) . Chứng minh rằng a=b=c
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
Tìm x :
a) \(\left(x^2-2x\right)^2+3\left(x^2-2x\right)+2=0\)
b) \(\left(x^2+x\right)\left(x^2+x-4\right)+4=0\)
Lời giải:
a. Đặt $x^2-2x=a$ thì pt trở thành:
$a^2+3a+2=0$
$\Leftrightarrow (a+1)(a+2)=0$
$\Leftrightarrow a+1=0$ hoặc $a+2=0$
$\Leftrightarrow x^2-2x+1=0$ hoặc $x^2-2x+2=0$
Nếu $x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1$
Nếu $x^2-2x+2=0\Leftrightarrow (x-1)^2=-1<0$ (vô lý)
Vậy pt có nghiệm duy nhất $x=1$
b.
Đặt $x^2+x=a$ thì pt trở thành:
$a(a-4)+4=0$
$\Leftrightarrow a^2-4a+4=0$
$\Leftrightarrow (a-2)^2=0$
$\Leftrightarrow a-2=0$
$\Leftrightarrow x^2+x-2=0$
$\Leftrihgtarrow (x-1)(x+2)=0$
$\Rightarrow x=1$ hoặc $x=-2$
tìm x biết:
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
b, \(\left(2x-1\right)^2-\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
a,\((x+4)^2-(x+1)(x-1)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)
b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)
\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)
\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)
\(\Rightarrow -x^2-10x-244=0\)
\(\Rightarrow -(x^2-10x+25)-219=0\)
\(\Rightarrow -(x-5)^2-219=0\)
\(\Rightarrow (x-5)^2+219=0\)
Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm
tìm GTLN
a)\(A=x^2+5y^2+2xy-4x-8y+2015\)
b)\(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
c)\(C=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)
d)\(D=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)
\(B=2x^2+2x+2012^2+2013^2\)
\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)
\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)
\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)
Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
a) ( x + 4 )2 - ( x + 1 )( x - 1 ) = 16
<=> x2 + 8x + 16 - ( x2 - 1 ) = 16
<=> x2 + 8x + 16 - x2 + 1 = 16
<=> 8x + 17 = 16
<=> 8x = -1
<=> x = -1/8
b) ( 2x - 1 )2 + ( x + 3 )2 - 5( x + 7 )( x - 7 ) = 0
<=> 4x2 - 4x + 1 + x2 + 6x + 9 - 5( x2 - 49 ) = 0
<=> 5x2 + 2x + 10 - 5x2 + 245 = 0
<=> 2x + 255 = 0
<=> 2x = -255
<=> 2x = -255/2
c) ( x + 2 )( x2 - 2x + 4 ) - x( x2 + 2 ) = 15
<=> x3 + 23 - x3 - 2x = 15
<=> 8 - 2x = 15
<=> 2x = -7
<=> x = -7/2
d) ( x + 3 )3 - x( 3x + 1 )2 + ( 2x + 1 )( 4x2 - 2x + 1 ) = 28
<=> x3 + 9x2 + 27x + 27 - x( 9x2 + 6x + 1 ) + [ ( 2x )3 + 13 ] = 28
<=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 + 1 = 28
<=> 3x2 + 26x + 28 = 28
<=> 3x2 + 26x = 0
<=> x( 3x + 26 ) = 0
<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)
Tìm x:
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
c, \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
d, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\) \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)\(\Leftrightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
\(c,\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6-49=0\)\(\Leftrightarrow24x=-13\Rightarrow x=-\dfrac{13}{24}\)
\(d,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=23\Rightarrow x=-\dfrac{23}{2}\)
Tìm x biết :
\(a)\)\(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
\(b)\)\(\left|x^2+\left|6x-2\right|\right|=x^2+4\)
Chứng minh rằng biểu thức sau không phụ thuộc vào biến x:
a/A= \(\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
b/B=\(\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)
a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)