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b, x2 +y2+z2 +2x-4y-6z+14=0
<=> (x2+2x+1)+(y2-4y+4)+(z2-6z+9)=0
<=> (x+1)2+(y-2)2+(z-3)2=0
=>(x+1)2=(y-2)2=(z-3)2=0
=>x+1=y-2=z-3=0
=> x=-1; y=2; z=3
c, 2x2+y2-6x-4y+2xy+5=0
<=> (x2+y2+4+2xy-4x-4y)+(x2-2x+1)=0
<=> (x+y-2)2+(x-1)2=0
=> (x+y-2)2=(x-1)2=0
=>x+y-2=x-1=0
=>x=1; y=1
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
Bài 2 : Phân tích đa thức thành nhân tử
a) \(8x^2-2\)
\(=2\left(4x^2-1\right)\)
\(=2.\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3+y\right)\left(x-3-y\right)\)
1. Tính giá trị biểu thức :
\(Q=x^2-10x+1025\)
\(Q=\left(x^2-2.x.5+25\right)+1000\)
\(Q=\left(x-5\right)^2+1000\)
Thay x=1005 vào biểu thức trên ta có :
\(Q=\left(1005-5\right)^2+1000\)
\(Q=1000000+1000\)
\(Q=1001000\)
Bài 1 :
a) \(A=x^2-6x+11\)
\(A=x^2-2\cdot x\cdot3+3^2+2\)
\(A=\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(B=2\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
c) \(C=5x-x^2\)
\(C=-\left(x^2-5x\right)\)
\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)
\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Bài 2 :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-x^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)
\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
a) \(x+5x^2=0\)
\(=>x\left(1+5x\right)=0\)
\(=>\hept{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
b) \(x^3+x=0\)
\(=>x\left(x^2+1\right)=0\)
\(=>\hept{\begin{cases}x=0\\x^2+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x\in\phi\end{cases}}\)
c) \(5x\left(x-1\right)=x-1\)
\(=>5x\left(x-1\right)-x+1=0\)
\(=>5x\left(x-1\right)-\left(x-1\right)=0\)
\(=>\left(x-1\right)\left(5x-1\right)=0\)
\(=>\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-10x=-25\)
\(=>x^2-10x+25=0\)
\(=>\left(x-5\right)^2=0\)
\(=>x-5=0\)
\(=>x=5\)
\(a,x+5x^2=0\)
\(x.\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
a) \(A=x^2-6x+11\)
\(\Rightarrow A=x^2-6x+9+2\)
\(\Rightarrow A=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = 3
Vậy \(MIN\) \(A=2\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(\Rightarrow B=2\left(x^2+5\right)-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\)
Ta có: \(2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)\ge0\forall x\)
\(\Rightarrow2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\ge-\dfrac{23}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{-5}{2}\)
Vậy \(MIN\) \(B=\dfrac{-23}{2}\Leftrightarrow x=\dfrac{-5}{2}\)
c) \(C=5x-x^2\)
\(\Rightarrow C=-\left(x^2-5x\right)\)
\(\Rightarrow C=-\left(x^2-2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Ta có: \(-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)
Vậy \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
mình thấy là \(\left(x^2+6x-11\right)^2\) sẽ đúng hơn đấy bạn
Ý bạn là sao vậy mk không hiểu