\(^2\) + 8y - 20= 0

b, x\(^2\)...">

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12 tháng 10 2018

a)\(y^2+8y-20=0\)

\(\Leftrightarrow y^2+2\cdot y\cdot4+16-16-20=0\)

\(\Leftrightarrow\left(y+4\right)^2-36=0\)

\(\Leftrightarrow\left(y+4\right)^2=36\)

\(\Leftrightarrow y+4=\pm6\)

\(\Leftrightarrow y=2\)hoặc \(y=-10\)

Vậy.....

b)\(x^2+7x=0\)

\(\Leftrightarrow x\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

Vậy .....

c)\(2y^2-5y=0\)

\(\Leftrightarrow y\left(2y-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=0\\2y-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{5}{2}\end{matrix}\right.\)

Vậy ......

d)\(y^2-5y^2+4=0\)

\(\Leftrightarrow-4y^2+4=0\)

\(\Leftrightarrow-4\left(y^2-4\right)=0\)

\(\Leftrightarrow-4\left(y+4\right)\left(y-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}y=-4\\y=4\end{matrix}\right.\)

Vậy....

2) Bạn thực hiện phép chia đi

Cuối cùng có:

Để (x2+3x+a)\(⋮\)(x+1) thì a-2=0=>a=2

Chúc bạn học tốt

13 tháng 10 2018

thank you

7 tháng 11 2017

1)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)

d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)

e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)

f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)

\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)

2)

a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)

b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)

c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )

d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)

e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)

5 tháng 4 2017

tớ không biết

5 tháng 4 2017

cj lậy chú

nhây vừa thoi

Câu 1: 

a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)

b: \(D=x^3+y^3+3xy\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)

\(=1-3xy+3xy=1\)

5 tháng 11 2017

1.

a. x2 - 2x + 1 = 0

x2 - 2x*1 + 12 = 0

(x-1)2 = 0

............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)

1, Tìm x biết:

a, x2 - 2x +1 = 0

(x-1)2 = 0

x-1 = 0

x = 1. Vậy ...

b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30

25x2 +10x + 1 - (25x2 -9) = 30

25x2 +10x + 1 - 25x2 +9 = 30

10x + 10 =30

10(x+1) = 30

x+1 =3

x = 2. vậy ...

c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5

(x3 - 1) - x(x2 -4) = 5

x3 - 1 - x3 + 4x = 5

4x - 1 = 5

4x = 6

x = \(\dfrac{3}{2}\) .vậy ...

d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15

x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15

x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15

24x + 25 = 15

24x = -10

x = \(\dfrac{-5}{12}\) vậy ...

22 tháng 10 2021

a) \(\left(2a+b\right)^2-\left(2b+a\right)^2\)

\(=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)

\(=3\left(a+b\right)\left(a-b\right)\)

22 tháng 10 2021

b) \(x^4+2x^2y+y^2\)

\(=\left(x^2+y\right)^2\)

24 tháng 7 2017

Bài 1:

\(x^2+x-6=x^2+3x-2x+6\)

\(=x\left(x+3\right)-2\left(x+3\right)\)

\(=\left(x-2\right)\left(x+3\right)\)

\(b,x^4+2x^3+x^2=\left(x^2+x\right)^2\)

\(e,x^2+5x-6=x^2+6x-x-6\)

\(=x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)

\(f,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)\(g,7x-6x^2-2=-6x^2+3x+4x-2\)

\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2-3x\right)\left(2x-1\right)\)\(i,2x^2+3x-5=2x^2-2x+5x-5\)

\(=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)

\(j,16x-5x^2-3=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)=\left(5x-1\right)\left(x+3\right)\)

Bài 2,

\(a,5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)

\(b,2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

24 tháng 7 2017

được chừng nào bạn đăng hết chẳng chịu suy nghĩ gì cả

7 tháng 9 2019

a)10x22+10xy+5x+5y

= 10x( x + y ) + 5( x + y )= ( 10x + 5 )( x + y )= 5( 2x + 1 )( x + y )

b)5ay-3bx+ax-15by

= ( 5ay + ax ) - ( 3bx + 15by ) = a( 5y + x ) - 3b( x + 5y ) = ( a - 3b ) ( x + 5y )

10 tháng 9 2017

Bài 3 :

a ) \(x\left(x-1\right)+x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy...........

b ) \(3\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\) \(\left(x-3\right)=0\Rightarrow x=3\)

Vậy............

Các câu sau tương tự

10 tháng 9 2017

Đăng từ từ thôi

Bài 1: Phân tích đa thức thành nhân tử: a) \(2x\left(x+1\right)+2\left(x+1\right)\) b) \(y^2\left(x^2+y\right)-zx^2-zy\) c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\) d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\) e) \(x^2-6xy+9y^2\) f) \(x^3+6x^2y+12xy^2+8y^3\) g) \(x^3-64\) h) \(125x^3+y^6\) k) \(0,125\left(a+1\right)^3-1\) t) \(x^2-2xy+y^2-xz+yz\) q) \(x^2-y^2-x+y\) p) \(a^3x-ab+b-x\) đ)...
Đọc tiếp

Bài 1: Phân tích đa thức thành nhân tử:

a) \(2x\left(x+1\right)+2\left(x+1\right)\)

b) \(y^2\left(x^2+y\right)-zx^2-zy\)

c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\)

d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

e) \(x^2-6xy+9y^2\)

f) \(x^3+6x^2y+12xy^2+8y^3\)

g) \(x^3-64\)

h) \(125x^3+y^6\)

k) \(0,125\left(a+1\right)^3-1\)

t) \(x^2-2xy+y^2-xz+yz\)

q) \(x^2-y^2-x+y\)

p) \(a^3x-ab+b-x\)

đ) \(3x^2\left(a+b+c\right)+36xy\left(a+b+c\right)+108y^2\left(a+b+c\right)\)

l) \(x^2-x-6\)

i) \(x^4+4x^2-5\)

m) \(x^3-19x-30\)

j) \(x^4+x+1\)

y) \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

o) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

ê) \(4a^2b^2-\left(a^2+b^2+c^2\right)^2\)

w) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)

z) \(\left(x^2-8\right)^2+36\)

u) \(81x^4+4\)

Bài 2 : Tìm x

a)\(\left(2x-1\right)^2-25=0\)

b) \(8x^3-50x=0\)

c) \(\left(x-2\right)\left(x^2+2+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

d) \(3x\left(x-1\right)+x-1=0\)

e) \(2\left(x+3\right)-x^2-3x\) =0

f) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

5
12 tháng 10 2017

Bài 1 :

a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)

b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)

c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)

d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)

e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)

12 tháng 10 2017

Bài 1 :

f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)

g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)