Vì mũ của số trên là 100 và 200, đều là số chẵn

Không số nào trong số trên là số âm

Tổng là số vô âm

Tổng của chúng bằng 0

Các hiệu:  (3x - 5) ; Các tổng:  (2y + 1)

\(\Rightarrow3x-5=0\Rightarrow3x=5\Rightarrow x=\frac{5}{3}\)

\(\Rightarrow2y+1=0\Rightarrow2y=-1\Rightarrow y=-0,5\)

Vậy: \(x=\frac{5}{3};y=-0,5\)

a.

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

TH1:

\(x+\frac{1}{2}=0\)

\(x=-\frac{1}{2}\)

TH2:

\(x-\frac{3}{4}=0\)

\(x=\frac{3}{4}\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)

b.

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

TH1:

\(\frac{1}{2}x-3=0\)

\(\frac{1}{2}x=3\)

\(x=3\div\frac{1}{2}\)

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

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

\(x=-\frac{1}{2}\div\frac{2}{3}\)

\(x=-\frac{1}{2}\times\frac{3}{2}\)

\(x=-\frac{3}{4}\)

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)

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

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

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)

\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)

\(x=5\)

a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)

\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)

b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)

\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)

=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)

=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)

=>\(\frac{2}{3}-\frac{4}{3}x=5\)

=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)

b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)

b)\(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(5-2x\right)=4\\y=5-2x\end{matrix}\right.\)\(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-10+4x=4\\y=5-2x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=14\\y=5-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy nghiệm duy nhất của hpt là: (2;1)

c) \(\left\{{}\begin{matrix}2y-x=2\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\2\left(2y-2\right)-y=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\4y-4-y=-1\end{matrix}\right.\)

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

Vậy nghiệm duy nhất của hpt là: (0;1)

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

Từ (1): \(x=2-2y\) (3)

Thế (3) vào (2), ta được: \(-2\left(2-2y\right)+y=1< =>-4+4y+y=1\)

                                          \(\Leftrightarrow y=1\)\(\Rightarrow\)\(x=2-2.1=0\)

Vậy nghiệm duy nhất của hpt là:  (0;1)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

â)\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(\Leftrightarrow27x-18=2x-3x^2\)

\(\Leftrightarrow27x-18-2x+3x^2=0\)

\(\Leftrightarrow3x^2+25x-18=0\)

\(\Leftrightarrow3x^2+27x-2x-18=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)

       \(\Rightarrow\orbr{\begin{cases}3x-2=0\\x+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)

b)\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)

\(\Leftrightarrow4x^2-4x+1-4x^2+25=18\)

\(\Leftrightarrow26-4x=18\)

 \(\Leftrightarrow4x=8\)

      \(\Rightarrow x=2\)

c)\(5x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow5x^2-10x-2x+10=0\)

\(\Leftrightarrow5x^2-12x+10=0\)

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

\(\Leftrightarrow x^2-6x+9-7=0\)

\(\Leftrightarrow\left(x-3\right)^2=7\)

       \(\Rightarrow\orbr{\begin{cases}x-3=\sqrt{7}\\x-3=-\sqrt{7}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}+3\\x=-\sqrt{7}+3\end{cases}}\)

d)\(x^2-5=0\)

\(\Leftrightarrow x^2=5\)

    \(\Rightarrow x=\sqrt{5};-\sqrt{5}\)

e)\(x^3+5x^2-4x-20=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+10x-20=0\)

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

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

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

\(\Leftrightarrow\left(x^2-4\right)\left(x+5\right)=0\)

      \(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=-2;2\end{cases}}\)