Lấy \(2.\left(2\right)-\left(1\right)\) ta được:

\(2b+4a+6-\left(a-1-2b\right)=0\)

\(\Leftrightarrow4b+3a+7=0\Rightarrow b=\dfrac{-3a-7}{4}\)

Thế vào (2):

\(\sqrt{a^2+\left(\dfrac{-3a-7}{4}\right)^2}=\dfrac{-3a-7}{4}+2a+3\)

\(\Leftrightarrow\sqrt{25a^2+42a+49}=5a+5\) (\(a\ge-1\))

\(\Leftrightarrow25a^2+42a+49=25a^2+50a+25\)

\(\Rightarrow a=...\Rightarrow b=...\)

\(\left\{{}\begin{matrix}x+y=5\\\dfrac{3}{5}+\dfrac{2}{x-y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\\dfrac{2}{x-y}=\dfrac{12}{5}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y=5\\6\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\x-y=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{35}{6}\\y=x-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{35}{12}\\y=\dfrac{25}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^3-3\left|y+1\right|=18\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x^3=40\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3=8\\2x^3+3\left|y+1\right|=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3=8\\\left|y+1\right|=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x-2y=-2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=-2\\4x+2y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=0\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2.0+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\0+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy...

\(\left\{{}\begin{matrix}3x-2y=-2\\2x+y=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3x-2y=-2\\4x+2y=2\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7x=0\\2x+y=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=0\\2.0+y=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

vậy...

Ta có

\(x\left(-2x+1\right)=-1\)

\(\Rightarrow x=1,-1;\left(-2x+1\right)=1;-1\\ \)
Mà \(\left(-2x+1\right)\le0\\ \left(-2x+1\right)=-1\Leftrightarrow x=1\)

Mik mới lớp 6 nên không chắc lắm

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

\(\Leftrightarrow-2x+2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=0\)

\(\Leftrightarrow\left(-2x-\frac{1}{2}\right)^2+\frac{3}{4}=0\)

Ta có : \(\left(-2x-\frac{1}{2}\right)^2>0\)

\(\Rightarrow\left(-2x-\frac{1}{2}\right)^2+\frac{3}{4}>0_{ }\)

Kết luận vậy pt VÔ NGHIỆM 

[125-20]+[75-5]=105+70=175

125-(-75)+(-4)

200+(-4)

196

\(x^2+42=4320\)

\(\Rightarrow x^2+90x-48x-4320=0\)

\(\Rightarrow x\left(x+90\right)-48\left(x-90\right)\)

\(\Rightarrow\left(x+90\right)\left(x-48\right)\)

\(\Rightarrow\orbr{\begin{cases}x+90=0\\x-48=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-90\\x=48\end{cases}}}\)

Ta có x^2+42x=4320

=>x^2+90x-48x-4320=0

=>x(x+90)-48(x+90)=0

=>(x-48)(x+90)=0

=>x=48 hoặc x=-90