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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

4 tháng 2 2018

              \(\left(x^2+7x+12\right)\left(x^2-15x+56\right)=180\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x+4\right)\left(x-7\right)\left(x-8\right)-180=0\)

\(\Leftrightarrow\)\(\left(x^2-4x-21\right)\left(x^2-4x-32\right)-180=0\)

Đặt     \(x^2-4x-21=t\)  ta có:

                         \(t\left(t-11\right)-180=0\)

           \(\Leftrightarrow\)\(t^2-11t-180=0\)

           \(\Leftrightarrow\)\(t^2-20t+9t-180=0\)

           \(\Leftrightarrow\)\(\left(t-20\right)\left(t+9\right)=0\)

           \(\Leftrightarrow\)\(\orbr{\begin{cases}t-20=0\\t+9=0\end{cases}}\)

  P/S:đến đây bn thay trở lại rồi tìm   x   nhé! chúc bn hok tốt

6 tháng 1 2022

⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)

⇔ \(x^2+8x+12=32\)

⇔ \(x^2+8x-20=0\)

⇔ \(\left(x-2\right)\left(x+10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

6 tháng 1 2022

Sửa lại đề nha:

 \(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)

25 tháng 5 2017

Đk:\(x\ne2;x\ne3;x\ne4;x\ne5;x\ne6\)

\(pt\Leftrightarrow\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+...+\frac{1}{\left(x-3\right)\left(x-2\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}+...+\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)\(\Leftrightarrow\frac{x-2}{\left(x-6\right)\left(x-2\right)}-\frac{x-6}{\left(x-2\right)\left(x-6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\Leftrightarrow\left(x-2\right)\left(x-6\right)=32\)

\(\Leftrightarrow x^2-8x+12=32\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

25 tháng 5 2017

! là gì vậy bn sao ghi vào?

24 tháng 4 2022

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x- 2x - x + 2 - x- 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !