a) \(\left(x-5\right)^2+\left(x^2-25\right)=0\)

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

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

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

b) \(\frac{x-2}{4}+\frac{2x-3}{3}=\frac{x-18}{6}\)

\(\Rightarrow\frac{3x-6}{12}+\frac{8x-12}{12}=\frac{2x-36}{12}\)

\(\Rightarrow\frac{11x-18}{12}=\frac{2x-36}{12}\)

\(\Rightarrow11x-18=2x-36\)

\(\Rightarrow11x-2x=18-36\)

\(\Rightarrow9x=-18\Rightarrow x=-2\)

c) \(\frac{1}{x-3}+\frac{x-3}{x+3}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-6x+9}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x^2-5x+12}{x^2-9}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow x^2-5x+12=5x-6\)

\(\Rightarrow x^2-10x+18=0\)

Giải biệt thức sẽ ra 2 nghiệm \(5+\sqrt{7}\)và \(5-\sqrt{7}\)

Gửi Cool: Lần sau đừng quên tìm điều kiện nhé. Câu c. ĐK: x khác 3 và x khác -3

khó quá 

\(a.\frac{-5}{9}+\frac{5}{9}:A=\left(-\frac{5}{9}+\frac{5}{9}\right):A=0:A=0\)

\(b.\frac{7}{25}.\frac{11}{13}-\frac{7}{25}.\frac{2}{13}-\frac{8}{25}=\frac{7}{25}.\left(\frac{11}{13}-\frac{2}{13}\right)-\frac{8}{25}=\frac{7}{25}.\frac{9}{13}-\frac{8}{25}=\frac{63}{325}-\frac{8}{25}=\frac{-41}{325}\)

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ad biy sai câu b

Thay \(x=-4\) vào pt elip ta được:

\(\frac{y^2}{9}=1-\frac{16}{25}=\frac{9}{25}\Rightarrow\left[{}\begin{matrix}y=\frac{9}{5}\\y=-\frac{9}{5}\end{matrix}\right.\)

\(\Rightarrow MN=2.\frac{9}{5}=\frac{18}{5}\)

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)