\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)

\(x^2+\dfrac{9x^2}{\left(x+3\right)^2}=40\)

\(\Leftrightarrow x^2+\dfrac{9x^2}{x^2+6x+9}-40=0\)

\(\Leftrightarrow x^2+9+\dfrac{3}{2}x+x^2-40=0\)

\(\Leftrightarrow2x^2+\dfrac{3}{2}x-31=0\)

\(\Leftrightarrow\dfrac{4x^2}{2}+\dfrac{3x}{2}-\dfrac{62}{2}=0\)

\(\Rightarrow4x^2+3x-62=0\)

\(\Leftrightarrow4\left(x^2+\dfrac{3}{4}x+\dfrac{9}{64}\right)-\dfrac{1001}{16}=0\)

\(\Leftrightarrow4\left(x+\dfrac{3}{8}\right)^2=\dfrac{1001}{16}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{8}=\sqrt{\dfrac{1001}{16}}\\x+\dfrac{3}{8}=-\sqrt{\dfrac{1001}{16}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{1001}{16}}-\dfrac{3}{8}\\x=-\sqrt{\dfrac{1001}{16}}-\dfrac{3}{8}\end{matrix}\right.\)

Bạn vào câu hỏi tương tự nha !!!

2.

\(a,Q=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)Vậy \(Min_Q=\dfrac{-9}{2}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(b,M=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

vậy \(Min_M=\dfrac{3}{4}\)khi \(\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

1,

\(a,H=\left(x-y+z\right)^2+\left(y-z\right)^2+2\left(x-y+z\right)\left(y-x\right)\)\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)\(=\left(x-y+z-y+z\right)=x-2y+2z\)

\(b,P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^{32}-1\right)\)

\(x-y=4\Rightarrow\left(x-y\right)^2=16\)

\(\Leftrightarrow x^2-2xy+y^2=16\)

\(\Leftrightarrow x^2+y^2=16+2xy=16+5.2=26\)

\(\left(x+y\right)^2=x^2+y^2+2xy=26+2.5=36\)

\(\Rightarrow\left[{}\begin{matrix}x+y=\sqrt{36}\\x+y=-\sqrt{36}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x+y=4\\x+y=-4\end{matrix}\right.\)

Vì x , y < 0 \(\Rightarrow x+y=-4\)

\(a,\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2=x^2-9-x^2-2x-1=-10-2x\) \(b,\left(4x-3\right)\left(4x+3\right)-16x^2=16x^2-9-16x^2=-9\)\(c,\left(x+4\right)\left(x^2-4x+16\right)-x^3=x^3+64-x^3=64\)

Tên: Trương Thị Thùy Ninh

Lớp 8a

\(8,2NaOH+H_2SO_4\rightarrow NaSO_4+2H_2O\)

\(15,N_2O_5+H_2O\rightarrow2HNO_3\)

\(PTK_{X_2O}=2.X+O=62\Rightarrow2X=62-O=46\Rightarrow X=23\)Vậy X là nguyên tố Natri