Violympic toán 8

NC

Tìm x:

a) 2x (x-5) -(x2-10x +25)=0

b) x2 - 9 +3x(x+3) = 0

c) x3 - 16x = 0

d) (2x+3)(x-2) - (x2 -4x+4) = 0

e) 9x2 -(x2 -2x +1)=0

f)x3-4x2 -9x +36 = 0

g) 3x - 6 = (x-1).(x-2)

i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5

k) x2 -1 = (x-1).(2x+3)

l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6

H24
12 tháng 10 2020 lúc 10:44

a) 2x (x-5) -(x2-10x +25)=0

\(\Leftrightarrow\)2x(x-5)-(x-5)2=0

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

\(\Leftrightarrow\)(x-5)(x+5)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

b) x2 - 9 +3x(x+3) = 0

\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0

\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0

\(\Leftrightarrow\)(x+3)(x-3+3x)=0

\(\Leftrightarrow\)(x+3)(4x-3)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)

c) x3 - 16x = 0

\(\Leftrightarrow\)x(x2-16)=0

\(\Leftrightarrow\)x(x-4)(x+4)=0

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

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

d) (2x+3)(x-2) - (x2 -4x+4) = 0

\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0

\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0

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

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

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

e) 9x2 -(x2 -2x +1)=0

\(\Leftrightarrow\)(3x)2-(x-1)2=0

\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0

\(\Leftrightarrow\)(2x+1)(4x-1)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

f)x3-4x2 -9x +36 = 0

\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0

\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0

\(\Leftrightarrow\)(x-4)(x2-9)=0

\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)

g) 3x - 6 = (x-1).(x-2)

\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)

\(\Leftrightarrow\)x-1=3

\(\Leftrightarrow\)x=4

i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)

k) x2 -1 = (x-1).(2x+3)

\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)

\(\Leftrightarrow\)x+1=2x+3

\(\Leftrightarrow\)x-2x=3-1

\(\Leftrightarrow\)-x=2

\(\Leftrightarrow\)x=-2

l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6

\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6

\(\Leftrightarrow\)6x-8=6

\(\Leftrightarrow\)6x=14

\(\Leftrightarrow\)x=\(\frac{7}{3}\)

Bình luận (0)
 Khách vãng lai đã xóa

Các câu hỏi tương tự
Xem chi tiết
DD
Xem chi tiết
TB
Xem chi tiết
HG
Xem chi tiết
PP
Xem chi tiết
MO
Xem chi tiết
QD
Xem chi tiết
NN
Xem chi tiết
TD
Xem chi tiết