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5 tháng 9 2020

đk: \(x\ge0\)

\(x\sqrt{x}+4\sqrt{x}+12=7x\)

\(\Leftrightarrow\left(x+4\right)\sqrt{x}=7x-12\)

\(\Leftrightarrow\left(x+4\right)^2\cdot x=\left(7x-12\right)^2\)

\(\Leftrightarrow x^3+8x^2+16x=49x^2-168x+144\)

\(\Leftrightarrow x^3-41x^2+184x-144=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(40x^2-40x\right)+\left(144x-144\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-40x+144\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(x-36\right)=0\)

=> \(x\in\left\{1;4;36\right\}\)

16 tháng 8 2016

mình vừa lên lớp 9 , chưa học phương trình bậc 2 

16 tháng 8 2016

hoặc dùng máy nhẩm nghiệm r` chia đa thức 

20 tháng 2 2018

http://k2pi.net.vn/showthread.php?t=24135

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20 tháng 2 2018

gì đây?

7 tháng 11 2015

ĐKXĐ \(x^2-7x+8\ge0\)

\(\Rightarrow x^2-7x+8+\sqrt{x^2-7x+8}=20\)

Đặt a = \(\sqrt{x^2-7x+8}\) (a \(\ge\)0) ta đc:

\(a^2+a=20\)

\(\Rightarrow a^2+a-20=0\)

\(\Rightarrow a=4\) hoặc \(a=-5\) (loại)

Với a = 4

<=> \(\sqrt{x^2-7x+8}=4\)

\(\Leftrightarrow x^2-7x+8=16\)

\(\Leftrightarrow x^2-7x-8=0\)

\(\Rightarrow\left(x-8\right)\left(x+1\right)=0\)

=> x - 8 = 0 => x = 8

hoặc x + 1 = 0 => x = -1

Vậy x = 8 ; x = -1

21 tháng 12 2015

\(\Leftrightarrow\left(x^2-9x+8\right)\left(x^2-6x+8\right)=7x^2\)

Xét x=0: x=0 không là nghiệm của phương trình

Xét x\(\ne\)0

pt \(\Leftrightarrow\left(x-9+\frac{8}{x}\right)\left(x-6+\frac{8}{x}\right)=7\)

Đặt t= x+8/x

Sau đó bạn giải pt tìm t, có t thế vào tìm được x

 

21 tháng 12 2015

ko có đâu bạn ^^ , câu này mình mới phát minh ra mà

3 tháng 4 2020

a)\(\left\{{}\begin{matrix}3x-2y=3\\2x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\3x-2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\3-2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b)\(x^2+7x+12=0\)

\(\Leftrightarrow x^2+3x+4x+12=0\)( chị nghĩ + 12 đúng hơn á )

\(\Leftrightarrow x\left(x+3\right)+4\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)

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

9 tháng 5 2017

NHAN HAI CUM X(X+5) , (X+1)(X+4) TADC (X^2+5X)(X^2+X+4X+4)=12...(X^2+5X)(X^2+5X+4)=12.DAT T=X^2+5X(1)  TADC  T(T+4)-12=0       T^2+4T-12=0 GIAI RA KQUA VA THAY VAO(1)