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Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Câu a nhé: 2x . x^2 - 2x . 7x - 2x . 3 = 2x^3 - 14x^2 - 6x
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
1
a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)
=\(6x^2+2x+3x+1-\left(6x^2+6x-x-1\right)\)
\(=6x^2+5x+1-6x^2-6x+x+1\)
\(=2\)
c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)
\(=\left(a^3+1\right)+\left(a^2-1\right)\)
\(=a^3+1+a^2-1\)
\(=a^3+a^2\)
2,
a,\(4ab+a^2-3a-12b\)
\(=\left(4ab-12b\right)+\left(a^2-3a\right)\)
\(=4b\left(a-3\right)+a\left(a-3\right)\)
\(=\left(4b+a\right)\left(a-3\right)\)
b,\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-\left(3y\right)^3\)
\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)
\(=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)
4
a,\(2004^2-16\)
\(=2004^2-4^2\)
\(=\left(2004-4\right)\left(2004+4\right)\)
\(=2000.2008\)
\(=4016000\)
b,\(892^2+892.216+108^2\)
\(=\left(892+108\right)^2\)
\(=1000^2=1000000\)
c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)
\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)
\(=9,8.10+10,2.10\)
\(=98+102\)
\(=200\)
d,\(36^2+26^2-52.36\)
=\(\left(36-26\right)^2\)
\(=10^2=100\)
3)\(A=-x^2+2x-3\)
\(\Leftrightarrow A=-x^2+2x-1-2\)
\(\Leftrightarrow A=-\left(x^2-2x+1\right)-2\)
\(\Leftrightarrow A=-\left(x-1\right)^2-2\)
Vậy GTLN của A=-2 khi x=1
Bài 1 :
a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)
TH1 : \(x^2-2x+3=0\)
\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm
TH2 : \(x-4=0\Leftrightarrow x=4\)
b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)
TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)
\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)
TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)
c, đưa về hệ đc ko ?
d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)
TH1 : \(x=0,74...\) ( bấm máy cx ra )
TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm
KL : vô nghiệm
Bài 2 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)
\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)
Vậy biểu thức ko phụ thuộc vào biến
b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-yx^3-y^2x^2-y^3x-y^4-x^4y^4\)
\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến
6.
\(A=x^2-2x+2\\ =x^2-2x+1+1\\ =\left(x-1\right)^2+1\\ \left(x-1\right)^2\ge0\forall x\\ \left(x-1\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
Vậy \(Min_A=1\) khi \(x=1\)
5,
a,
\(x\left(x-2\right)+x-2=0\\ \left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b,
\(5x\left(x-3\right)-x+3=0\\ 5x\left(x-3\right)-\left(x-3\right)=0\\ \left(5x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
c,
\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\\ 3x^2-15x-\left(2x-2+3x^2-3x\right)\\ 3x^2-15x-\left(3x^2-x-2\right)=30\\ 3x^2-15x-3x^2+x+2=30\\ -14x+2=30\\ -14x=28\\ x=-2\)
Vậy \(x=-2\)
d,
\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\\ x^2+5x+6-\left(x^2+3x-10\right)=0\\ x^2+5x+6-x^2-3x+10\\ 8x+16=0\\ 8x=-16\\ x=-2\)
Vậy \(x=-2\)