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

    \(3x^2-3x-2x+2=0\)

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

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

ta có \(x-\sqrt{x}+1=\left(x-1\right)^2+\sqrt{x}\)

mà \(\left(\sqrt{x}-1\right)^2>=0>-1\Leftrightarrow\sqrt{x}-1< \sqrt{x}+\left(\sqrt{x}-1\right)^2\)-1)^2

 hay\(\sqrt{x}-1< x-\sqrt{x}+1\)

vậy đpcm

xin lỗi mk mới hc lp 7 ko thể giúp bn đc !

tui làm bên học24 r` mà, muốn đưa link mà lỗi, thôi làm lại :(

\(pt\Leftrightarrow x^9-12x^6+48x^3-64=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+8\sqrt[3]{\left(x^2+4\right)^2}+16\)

\(\Leftrightarrow x^9-12x^6+48x^3-128=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2-16+8\sqrt[3]{\left(x^2+4\right)^2}-32\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\frac{\left(x^2+4\right)^4-4096}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x^2+4\right)^2-32768}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\frac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\frac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\frac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\right]=0\)

Dễ thấy: pt trong ngoặc vuông vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

Đ/S :2