\(\left(x+2\right)\left(x+5\right)< 0\)

<=> \(\left\{{}\begin{matrix}x+2>0\\x+5< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+2< 0\\x+5>0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x>-2\\x< -5\end{matrix}\right.\)(vô lý) hoặc \(\left\{{}\begin{matrix}x< -2\\x>-5\end{matrix}\right.\)

=> -5 < x < -2

Vậy .........

a, \(\left(x-3\right)\left(x+7\right)=x^2+7x-3x-21=x^2+4x-21\)

b, \(\left(2x-5\right)\left(x-1\right)=2x^2-2x-5x+5=2x^2-7x+5\)

\(4x^2-36x+56=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot9+9^2-25=0\)

\(\Leftrightarrow\left(2x-9\right)^2-25=0\)

\(\Leftrightarrow\left(2x-9\right)^2=25\Leftrightarrow\left[{}\begin{matrix}2x-9=5\\2x-9=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

Vậy....................

\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)

Áp dụng tccdtsbn ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4\cdot9=36\\y^2=4\cdot16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)

Vậy...........

a, \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}=\dfrac{2x-3y+z}{18-36+15}=\dfrac{6}{-3}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}2x=-2\cdot18=-36\\3y=-2\cdot36=-72\\z=-2\cdot15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-18\\y=-24\\z=-30\end{matrix}\right.\)

Vậy............................

b, \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng t/c của dts = nhau ta có:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

\(\Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)

Vậy.........

\(\sqrt{x-1}=x-3\)

ĐKXĐ: x \(\ge1\)

\(\Leftrightarrow\sqrt{x-1}^2=\left(x-3\right)^2\)

\(\Leftrightarrow x-1=x^2-6x+9\)

\(\Leftrightarrow x-1-x^2+6x-9=0\)

\(\Leftrightarrow-x^2+7x-10=0\)

\(\Leftrightarrow-x^2+2x+5x-10=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy pt có 2 nghiệm là.....

a, \(\left|x^2+2x+1\right|=\left|x^2-x+2\right|\)

\(\Leftrightarrow\left|\left(x+1\right)^2\right|=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)

\(\Leftrightarrow\left(x+1\right)^2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x-\dfrac{1}{2}\right)^2=\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+1+x-\dfrac{1}{2}\right)\left(x+1-x+\dfrac{1}{2}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\left(2x+\dfrac{1}{2}\right)\cdot\dfrac{3}{2}=\dfrac{7}{4}\)

\(\Leftrightarrow2x+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{\dfrac{7}{6}-\dfrac{1}{2}}{2}=\dfrac{1}{3}\)

Vậy......................

Ta có : 

Áp dụng t/c /a/+/b/>(=)/a+b/ (Dấu bằng xảy ra <=>a.b=0) ta có:

y=/2a-1/+/a/>(=)/2a-1+a/=/3a-1/=x

Vậy y>(=)x 

x và y có thể bằng nhau khi:

(2a+1).a=0

Đề? ^~~^