câu d, xem lại đề bài nha!

\(\left(x-4\right)\left(x-5\right)\left(x-8\right)\left(x-10\right)=72x^2\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)\left(x-8\right)\left(x-10\right)-72x^2=0\)

\(\Leftrightarrow\left(x^2-14x+40\right)\left(x^2-13x+40\right)-72x^2=0\)

\(\Leftrightarrow\left(x^2-13,5x+40-0,5x\right)\left(x^2-13,5x+40+0,5x\right)-72x^2=0\)

\(\Leftrightarrow\left(x^2-13,5x+40\right)^2-\left(0,5x\right)^2-72x^2=0\)

\(\Leftrightarrow\left(x^2-13,5x+40\right)^2-72,25x^2=0\)

\(\Leftrightarrow\left(x^2-13,5x+40+8,5x\right)\left(x^2-13,5x+40-8,5x\right)=0\)

\(\Leftrightarrow\left(x^2-5x+40\right)\left(x^2-22x+40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+40=0\left(VN\right)\\x^2-22x+40=0\Leftrightarrow\left[{}\begin{matrix}x=20\\x=2\end{matrix}\right.\end{matrix}\right.\)

Câu a,c xem lại đề, cách làm giống câu b, còn câu e giống câu d

b) \(2x^4+5x^3+x^2+5x+2=0\)

Ta nhận thấy x=0 không phải là 1 nghiệm của phương trình, chia cả 2 vế của phương trình cho \(x^2\ne0\), ta được:

\(2x^2+5x+1+\dfrac{5}{x}+\dfrac{2}{x^2}=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)+1=0\)

Đặt \(y=x+\dfrac{1}{x}\Rightarrow x^2+\dfrac{1}{x^2}=y^2-2\)

\(\Leftrightarrow2\left(y^2-2\right)+5y+1=0\)

\(\Leftrightarrow2y^2+5y-3=0\)

PT đơn giản, tự giải nha, ta được nghiệm y=1/2 và y=-3

Với y=1/2 thì không tìm được x

Với y=-3 thì tìm được 2 nghiệm, tự giải

a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)

b. \(\Leftrightarrow x^3+x+3x^2+3=0\)

\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+1=0\left(vn\right)\end{matrix}\right.\)

c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)

\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)

d.

\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)

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

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

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

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

e.

\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)

\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)

c/

\(x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt \(x^2+3x=t\)

\(t\left(t+2\right)-24=0\Leftrightarrow t^2+2t-24=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3x=4\\x^2+3x=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\)

d/

\(\Leftrightarrow x^4-2x^3+x^2+3x^2-3x-10=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+3\left(x^2-x\right)-10=0\)

Đặt \(x^2-x=t\)

\(t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\end{matrix}\right.\)

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

a/ ĐKXĐ: ...

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

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

\(\Leftrightarrow2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=2\\x+\frac{1}{x}=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x=1=0\\2x^2-x+2=0\end{matrix}\right.\)

b/ Với \(x=0\) ko phải nghiệm

Với \(x\ne0\) chia 2 vế của pt cho \(x^2\)

\(x^2+\frac{1}{x^2}-5x+\frac{5}{x}-8=0\)

\(\Leftrightarrow x^2+\frac{1}{x^2}-2-5\left(x-\frac{1}{x}\right)-6=0\)

Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2+\frac{1}{x^2}-2\)

\(t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=-1\\x-\frac{1}{x}=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x-1=0\\x^2-6x-1=0\end{matrix}\right.\)

a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

Mk năm nay lên lớp 9 nên chỉ làm bài 1 đc thôi

Câu 1:

a)\(\left(2x+3\right)^2-\left(x+1\right)^2=0\)

    \(\left(2x+3+x+1\right)\left(2x+3-x-1\right)=0\)

       \(\left(3x+4\right)\left(x+2\right)=0\)

             \(\Rightarrow\orbr{\begin{cases}3x+4=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)

b)\(x^2-6x+5=0\)

   \(x^2-5x-x+5=0\)

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

           \(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

c)\(3x^2-5x+2=0\)

    \(3x^2-3x-2x+2=0\)

     \(\left(3x-2\right)\left(x-1\right)=0\)

               \(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=1\end{cases}}\)

\(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-5=9\\2x-3=9\\x-1=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=6\\x=10\end{matrix}\right.\)

Vậy \(x=\left\{3,5;6;10\right\}\)

d: Sửa đề: \(\left(4x-5\right)^2\cdot\left(2x-3\right)\left(x-1\right)=9\)

a: \(\Leftrightarrow\left(2x^2+x\right)^2-3\left(2x^2+x\right)-\left(2x^2+x\right)+3=0\)

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

\(\Leftrightarrow\left(2x^2+x-3\right)\left(2x^2+x-1\right)=0\)

\(\Leftrightarrow\left(2x^2+3x-2x-3\right)\left(2x^2+2x-x-1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)

hay \(x\in\left\{-\dfrac{3}{2};1;-1;\dfrac{1}{2}\right\}\)