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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
^^
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
=
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
a/ \(3x+3y-4x-4y=3\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(3-4\right)=-1\left(x+y\right)\)
b/ \(7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(7x+1\right)\)
c/ \(5x\left(1-x\right)+\left(x-1\right)=5x\left(1-x\right)-\left(1-x\right)=\left(1-x\right)\left(5x-1\right)\)
d/ \(4x\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left(4x+3x-3y\right)=\left(x-y\right)\left(7x-3y\right)\)
e/ \(4x\left(x-y\right)+3\left(y-x\right)^2=4x\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left(4x+3x-3y\right)=\left(x-y\right)\left(7x-3y\right)\)
g/ \(x^2+8x+7=x^2+x+7x+7=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
h/ \(x^2-6x-16=x^2+2x-8x-16=x\left(x+2\right)-8\left(x+2\right)=\left(x+2\right)\left(x-8\right)\)
i/ \(4x^2-8x+3=4x^2-2x-6x+3=2x\left(2x-1\right)-3\left(2x-1\right)=\left(2x-1\right)\left(2x-3\right)\)
k/ \(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)