Giúp mk với mọi người

a/ \(x^2-2x=-1\)

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

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy..............

b/ \(x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\Rightarrow x=-1\)

Vậy.......

c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)

\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

Vậy...................

d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3-2016x^2-2017x=0\)

\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)

\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)

Vậy pt có 3 nghiệm là.....(tự ghi ra)

\(a,x^2-2x=-1\)

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

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

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(b,x^2+2x+1=0\)

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

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3+x^2-2017x-2017=0\)

\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)

a)  \(5\left(x+3\right)-6x-2x^2=0\)   \(\Leftrightarrow5.\left(x+3\right)-2x.\left(x+3\right)=0\) 

\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\Leftrightarrow\hept{\begin{cases}x+3=0\\5-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\2x=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\x=\frac{5}{2}\end{cases}}}\)

b)  \(6x.\left(x^2-2\right)-\left(2-x^2\right)=0\)  \(\Leftrightarrow6x.\left(x^2-2\right)+\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\Leftrightarrow\hept{\begin{cases}x^2-2=0\\6x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=2\\6x=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\sqrt{2}\\x=\frac{-1}{6}\end{cases}}}\)

c)  \(4x.\left(x-2017\right)-x+2017=0\) \(\Leftrightarrow4x.\left(x-2017\right)-\left(x-2017\right)=0\)

\(\Leftrightarrow\left(x-2017\right).\left(4x-1\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x-2017=0\\4x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2017\\4x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}}\)

d)  \(12x=x^2+36\) \(\Leftrightarrow x^2-12x+36=0\) \(\Leftrightarrow\left(x-6\right)^2=0\) \(\Rightarrow x-6=0\) \(\Leftrightarrow x=6\)

em xét dấu trị tuyệt đối với mũ 2 nhé

ã) x=-3

Bài 2 : 

a) \(A=3,7+\left|4,3-x\right|\ge3,7\)

Min A = 3,7 \(\Leftrightarrow x=4,3\)

b) \(B=\left|3x+8,4\right|-14\ge-14\)

Min B = -14 \(\Leftrightarrow x=\frac{-14}{5}\)

c) \(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Min C = 17,5 \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{-3}{2}\end{cases}}\)

d) \(D=\left|x-2018\right|+\left|x-2017\right|\)

\(D=\left|2018-x\right|+\left|x-2017\right|\ge\left|2018-x+x-2017\right|=1\)

Min D =1 \(\Leftrightarrow\left(2018-x\right)\left(x-2017\right)\ge0\)

\(\Leftrightarrow2017\le x\le2018\)

\(A=3,7+\left|4,3-x\right|\)

Ta có \(\left|4,3-x\right|\ge0\Leftrightarrow A=3,7+\left|4,3-x\right|\ge3,7\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|4,3-x\right|=0\Leftrightarrow4,3-x=0\Leftrightarrow x=4,3\)

\(B=\left|3x+8,4\right|-14\)

Ta có \(\left|3x+8,4\right|\ge0\Leftrightarrow B=\left|3x+8,4\right|-14\ge-14\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|3x+8,4\right|=0\Leftrightarrow3x=-8,4\Leftrightarrow x=2,8\)

\(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Ta có \(\hept{\begin{cases}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{cases}}\Leftrightarrow C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu '' = '' xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

\(D=\left|x-2018\right|+\left|x-2017\right|\)

\(\Leftrightarrow D=\left|x-2018\right|+\left|2017-x\right|\)

Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)ta có

\(D\ge\left|x-2018+2017-x\right|=\left|-1\right|=1\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left(2017-x\right)\left(x-2018\right)\ge0\Leftrightarrow2018\ge x\ge2017\)

Trần văn ổi ()

đù khó thế

tl j z mấy chế , k câu dc đâu :))