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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
^^
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
1)
a) => 16x2 - 8x + 1 - 8(2x2 + 3x - 4x - 6) = 15
=> 16x2 - 8x + 1 - 8(2x2 - x - 6) = 15
=> 16x2 - 8x + 1 - 16x2 + 8x + 48 = 15
=> 49 = 15 (?) (vô lí)
=> Không tìm được x thoả mãn
b) (5x - 2)(x - 2) - 4(x - 3) = x2 + 3
=> 5x2 - 10x - 2x + 4 - 4x + 12 = x2 + 3
=> 5x2 - 16x + 16 = x2 + 3
=> 4x2 - 16x + 16 = 3
=> (2x)2 - 2.2x.4 + 42 = 3
=> (2x - 4)2 = 3
=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)
Mong bạn xem lại đề bài!
2)
a) 5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x - y)2 - (2z)2]
= 5(x - y - 2z)(x - y + 2z)
b) a3 - ay - a2x + xy
= a(a2 - y) - x(a2 - y)
= (a - x)(a2 - y)
c) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x - y)2 - (2z)2]
= 3(x - y - 2z)(x - y + 2z)
d) x2 - 2xy + tx - 2ty
= x(x - 2y) + t(x - 2y)
= (x + t)(x - 2y)
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)=-\left[\left(6x^2-3x\right)-\left(4x+2\right)\right]=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left[\left(3x-2\right)\left(2x-1\right)\right]\)
d) \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)
a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3)
b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
f.2(x^5)-x^2-5x ( mik ko bik làm)
g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)
\(=\left(x^2-2\right)-x^2=\left(x^2-2+x\right)\left(x^2-2-x\right)\)
a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)
b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)
d,x2 - 4x + 3 = x2 - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
e,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
f,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2
\(\text{1) }\)
\(\text{a) }5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
\(\text{b) }5x^2+5xy-x-y\)
\(=\left(5x^2-x\right)+\left(5xy-y\right)\)
\(=x\left(5x-1\right)+y\left(5x-1\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
\(\text{c) }2\left(x+4\right)-x^2+16\)
\(=2\left(x+4\right)-\left(x^2-16\right)\)
\(=2\left(x+4\right)-\left(x+4\right)\left(x-4\right)\)
\(=\left(x+4\right)\left(2-x+4\right)\)
\(=\left(x+4\right)\left(6-x\right)\)
\(\text{d) }x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=\left(x^2+3x\right)+\left(x+3\right)\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x+1\right)\)
\(\text{e) }x^2+5x-6\)
\(=x^2+6x-x-6\)
\(=\left(x^2+6x\right)-\left(x+6\right)\)
\(=x\left(x+6\right)-\left(x+6\right)\)
\(=\left(x+6\right)\left(x-1\right)\)