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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
^^
nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu
a) x2 +5x -6 = x2 -x +x + 5x -6
= x2 -x +6x -6
= x( x-1) + 6(x-1) = (x-1)(x+6)
b) 5x2 +5xy -x-y = 5x(x+y) -(x+y)
= (x+y)(5x-1)
c) 7x -6x2 -2 = 6( x+2/3)(x+1/2)
d) x2 +4x +3 = x2 +x +3x +3
= x(x+1) + 3(x+1)
= (x+1)(x+3)
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
=
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)=-\left[\left(6x^2-3x\right)-\left(4x+2\right)\right]=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left[\left(3x-2\right)\left(2x-1\right)\right]\)
d) \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)
\(x^2+4x+3\)
\(=\left(x+1\right)\left(x+3\right)\)
\(2x^2+3x-5\)
\(\left(x-1\right)\left(x+\frac{5}{2}\right)\)
Tìm x:
\(5x\left(x-1\right)=x-1\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy x=\(\dfrac{1}{5}\)hoặc x=1
\(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
A)\(x^2+5x-6=x^2-x+6x-6\\ =\left(x-1\right)\left(x+6\right)\)
B)\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(5x-1\right)\)
C)\(7x-6x^2-2=-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2x-1\right)\left(2-3x\right)\)
D)\(x^2+4x+3=x^2+x+3x+3=\left(x+1\right)\left(x+3\right)\)
E)\(2x+3x-5=5x-5=5\left(x-1\right)\)
F)\(16x-5x^3=x\left(16-5x^2\right)\)