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

\(a,\) Điều kiện xác định: \(x^2+x\ne0\Leftrightarrow x\left(x+1\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

\(b,A=\dfrac{x+1}{x^2+x}=\dfrac{x+1}{x\left(x+1\right)}=\dfrac{1}{x}\)

Cảm ơn bạn. Bạn sửa lại cho mình chỗ \(\left(x-\dfrac{9}{8}\right)^2=-\dfrac{143}{64}\left(\text{vô lí}\right)\) và bước thử nghiệm thì cả hai nghiệm đều thỏa hết nha.

\(3\sqrt{x^3+8}=2x^2-3x+10\)

\(\Leftrightarrow3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=2x^2-3x+10\left(1\right)\)

\(\Leftrightarrow9\left(x+2\right)\left(x^2-2x+4\right)=\left(2x^2-3x+10\right)^2\)

\(\Leftrightarrow9\left(x^3-2x^2+4x+2x^2-4x+8\right)=4x^4-6x^3+9x^2-30x+20x^2-30x+100\)

\(\Leftrightarrow9x^3-18x^2+36x+18x^2-36x+72-4x^4+6x^3-20x^2+6x^3-9x^2+30x-20x^2+30x-100=0\)

\(\Leftrightarrow-4x^4+21x^3-49x^2+60x-28=0\left(2\right)\)

Nhận thấy, \(x=1\) và \(x=2\) là nghiệm của phương trình \(\left(2\right)\)

\(\left(2\right)\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(-4x^2+9x-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\-4x^2+9x-14=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\\left(x-\dfrac{9}{8}\right)^2=-\dfrac{143}{16}\left(\text{vô lí}\right)\end{matrix}\right.\)

Thử lại nghiệm \(x=1;x=2\) vào phương trình \(\left(1\right)\) thấy nghiệm \(x=2\) thỏa mãn.

\(2,\)

\(y=\dfrac{x}{\left(x^2-1\right)\sqrt{x-1}}\)

Hàm số xác định \(\Leftrightarrow\left\{{}\begin{matrix}x^2-1\ne0\\\sqrt{x-1}\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\\sqrt{x-1}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x>1\end{matrix}\right.\)

\(\Rightarrow D=\left(1;\text{+∞}\right)\)

\(\Rightarrow\) Chọn đáp án \(B\)

\(3\overrightarrow{AP}-2\overrightarrow{AC}=\overrightarrow{0}\)

\(VT=3\left(\overrightarrow{AD}+\overrightarrow{DP}\right)-2\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\)

\(=3\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{AD}-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\left(\overrightarrow{DC}+\overrightarrow{CP}\right)-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\overrightarrow{DC}+3\overrightarrow{CP}-2\overrightarrow{DC}\)

\(=\widehat{AD}+\overrightarrow{DC}+3.\dfrac{2}{3}\overrightarrow{CO}\)

\(=\overrightarrow{AD}+\overrightarrow{DC}+2.\dfrac{1}{2}\overrightarrow{CA}\)

\(=\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CA}\)

\(=\overrightarrow{AC}+\overrightarrow{CA}\)

\(=\overrightarrow{AA}=\overrightarrow{0}=VP\) (điều phải chứng minh)

\(\Delta AHE\left(\widehat{AHE}=90^o\right):\widehat{HAE}+\widehat{HEA}=90^o\) (2 góc phụ nhau)

\(\Rightarrow\widehat{HEA}=90^o-55^o=35^o=\widehat{KEB}\)

\(\Delta KBE\left(\widehat{BKE}=90^o\right):\widehat{KBE}+\widehat{KEB}=90^o\) (2 góc phụ nhau)

\(\Rightarrow\widehat{KBE}=90^o-35^o=55^o\)

\(\widehat{KBH}+\widehat{KBE}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{KBH}=180^o-55^o=125^o\)

\(c,\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13y=26\\6x-3y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\6x-3.2=-24\end{matrix}\right.\)

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

\(d,\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\left(I\right)\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)

\(\left(I\right)\Rightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\3a+4.\dfrac{2}{7}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{7}\Leftrightarrow x=\dfrac{7}{2}\\\dfrac{1}{y}=\dfrac{9}{7}\Leftrightarrow y=\dfrac{7}{9}\end{matrix}\right.\)

\(x-\sqrt{x+3+4\sqrt{x-1}}=1\) (Điều kiện xác định: \(x\ge1\))

\(\Leftrightarrow x-1-\sqrt{x-1+4+4\sqrt{x-1}}=0\left(1\right)\)

Đặt \(\sqrt{x-1}=t\left(t\ge0\right)\)

\(\Rightarrow x-1=t^2\)

\(\left(1\right)\Rightarrow t^2-\sqrt{t^2+4+4t}=0\)

\(\Leftrightarrow t^2=\sqrt{\left(t+2\right)^2}\)

\(\Leftrightarrow t^2=\left|t+2\right|\left(2\right)\)

Vì \(t^2\ge0\forall x\in R\Rightarrow\left|t+2\right|\ge0\forall x\in R\) (luôn đúng)

\(\left(2\right)\Leftrightarrow t^2=t+2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(\text{thỏa mãn điều kiện}\right)\\t=-1\left(\text{không thỏa mãn điều kiện}\right)\end{matrix}\right.\)

Với \(t=2\Rightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=5\) (thỏa mãn điều kiện)

\(A=\left(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\right).\dfrac{2x+6}{8x}\)

\(a,\) Điều kiện xác định: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\8x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\\x\ne0\end{matrix}\right.\)

\(b,A=\left(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\right).\dfrac{2x+6}{8x}\)

\(=\left[\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{2\left(x+3\right)}{8x}\)

\(=\dfrac{\left(x-3-x-3\right)\left(x-3+x+3\right)}{\left(x+3\right)\left(x-3\right)}.\dfrac{x+3}{4x}\)

\(=\dfrac{-6.2x}{\left(x-3\right)}.\dfrac{1}{4x}\)

\(=\dfrac{-12x}{4x\left(x-3\right)}\)

\(=\dfrac{-3}{x-3}\)

\(c,A=\dfrac{1}{2}\Rightarrow\dfrac{-3}{x-3}=\dfrac{1}{2}\Leftrightarrow x=-3\)

\(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=35\\12x-15y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\12x-15.5=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)