Nhầm.

A B C M D N

a) Xét \(\Delta MBN\) và \(\Delta ABC\)có:

\(\widehat{NBC}\)chung.

\(\widehat{BMN}=\widehat{BAC}\left(=90^0\right)\).

\(\Rightarrow\Delta MBN~\Delta ABC\left(g.g\right)\)(điều phải chứng minh).

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

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

\(\Leftrightarrow\left(-x-2\right)\left(5x-2\right)=0\Leftrightarrow x=-2;x=\frac{2}{5}\)

Vậy tập nghiệm của phương trình là S = { -2 ; 2/5 } 

4( x - 1 )² = 9x²

<=> ( 2x - 2 )² - (3x)² = 0

<=> ( -x - 2 )( 5x - 2 ) = 0

<=> x = -2 hoặc x = 2/5

lồn cạc địt nhau ko

\(\frac{x-3}{x+5}+\frac{x+5}{x-3}< 2\)

\(\Leftrightarrow\frac{x-3}{x+5}+\frac{x+5}{x-3}-2< 0\)

\(\Leftrightarrow\frac{\left(x-3\right)^2+\left(x+5\right)^2-2\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}< 0\)

\(\Leftrightarrow\frac{x^2-6x+9+x^2+10x+25-2\left(x^2+2x+2\right)}{\left(x-3\right)\left(x+5\right)}< 0\)

\(\Leftrightarrow\frac{2x^2+4x+34-2x^2-4x-4}{\left(x-3\right)\left(x+5\right)}< 0\)

\(\Leftrightarrow\frac{30}{\left(x-3\right)\left(x+5\right)}< 0\Rightarrow\left(x-3\right)\left(x+5\right)< 0\)do 30 > 0 

mà \(x+5>x-3\)suy ra : \(\hept{\begin{cases}x-3< 0\\x+5>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 3\\x>-5\end{cases}\Leftrightarrow-5< x< 3}\)

\(\frac{x-15}{173}+\frac{x-13}{171}\le\frac{x-11}{169}+\frac{x-9}{167}\)

\(\Leftrightarrow\frac{x-15}{173}+1+\frac{x-13}{171}+1\le\frac{x-11}{169}+1+\frac{x-9}{167}+1\)

\(\Leftrightarrow\frac{x-15+173}{173}+\frac{x-13+171}{171}\le\frac{x-11+169}{169}+\frac{x-9+167}{167}\)

\(\Leftrightarrow\frac{x+158}{173}+\frac{x+158}{171}-\frac{x+158}{169}-\frac{x+158}{167}\le0\)

\(\Leftrightarrow\left(x+158\right)\left(\frac{1}{173}+\frac{1}{171}-\frac{1}{169}-\frac{1}{167}\right)\le0\Rightarrow x+158\le0\)

\(\Leftrightarrow x\le-158\)Vậy tập nghiệm bfp là \(\left\{x|x\le-158\right\}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne5\end{cases}}\)

\(A=\frac{x-5}{\left(x-2\right)\left(x-5\right)}+\frac{x^2-x-2}{\left(x-2\right)\left(x-5\right)}-\frac{2\left(x-2\right)^2}{\left(x-2\right)\left(x-5\right)}\)

\(=\frac{x-5+x^2-x-2-2x^2+8x-8}{\left(x-2\right)\left(x-5\right)}\)

\(=\frac{-x^2+8x-15}{\left(x-2\right)\left(x-5\right)}=\frac{\left(3-x\right)\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3-x}{x-2}\)

\(A=\frac{1}{x-2}+\frac{x^2-x-2}{x^2-7x+10}-\frac{2x-4}{x-5}\)ĐK : \(x\ne2;5\)

\(=\frac{x-5+x^2-x-2-\left(2x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\frac{x^2-7-2x^2+8x-8}{\left(x-2\right)\left(x-5\right)}\)

\(=\frac{-x^2+8x-15}{\left(x-2\right)\left(x-5\right)}=\frac{-\left(x-5\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5-x}{x-2}\)

có công thức trong sgk mà bn

rút gọn\(\frac{1}{x-2}+\frac{x^2-x-2}{x^2-7x+10}-\frac{2x-4}{x-5}\)

Áp dụng bất đẳng thức Cauchy ta có:

\(m^2+n^2+p^2+q^2+1\)

\(=\left(\frac{1}{4}m^2+n^2\right)+\left(\frac{1}{4}m^2+p^2\right)+\left(\frac{1}{4}m^2+q^2\right)+\left(\frac{1}{4}m^2+1\right)\)

\(\ge2\sqrt{\frac{1}{4}m^2\cdot n^2}+2\sqrt{\frac{1}{4}m^2\cdot p^2}+2\sqrt{\frac{1}{4}m^2\cdot q^2}+2\sqrt{\frac{1}{4}m^2\cdot1}\)

\(=2\cdot\frac{1}{2}mn+2\cdot\frac{1}{2}mp+2\cdot\frac{1}{2}mq+2\cdot\frac{1}{2}m\)

\(=mn+mp+mq+m\)

\(=m\left(n+p+q+1\right)\)

Dấu "=" xảy ra khi: \(\frac{1}{4}m^2=n^2=p^2=q^2=1\)\(\Rightarrow\hept{\begin{cases}m=2\\n=p=q=1\end{cases}}\)