<=> \(\frac{x^2-3x+2}{2}=\frac{x^2+2x-3}{2}\)

=> x² - 3x + 2 = x² + 2x - 3

<=> 5x = 5

<=> x = 1

Vậy S = {1}

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

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

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

\(x^2-2x-x+2=x^2+3x-x-3\)

\(x^2-3x+2=x^2+3x-x-3\)

\(x^2+3x+2=2x-3\)

\(-3x+2=2x-3\)

\(2=2x-3+3x\)

\(2=5x-3\)

\(5x=5\Leftrightarrow x=1\)

x^2 - 2x - 15

= x^2 + 3x - 5x - 15

= x(x + 3) - 5(x + 3)

= (x - 5)(x + 3)

\(x^2-2x-15\)\(=\left(x^2-5x\right)+\left(3x-15\right)\)

\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)

(x² - 4x)² + 2(x - 2)² = 43

<=> x²(x - 4)² + 2(x - 2)² = 43

<=> x⁴ - 8x³ + 16x² + 2x² - 8x + 8 = 43

<=> x⁴ - 8x³ + 18x² - 8x + 8 - 43 = 0

<=> x⁴ - 8x³ + 18x² - 8x - 35 = 0

<=> (x³ - 9x² + 27x - 35)(x + 1) = 0

<=> (x² - 4x + 7)(x - 5)(x + 1) = 0

vì x² - 4x + 7 khác 0 nên:

<=> x - 5 = 0 hoặc x + 1 = 5

<=> x = 5 hoặc x = -1

\(\left(x^2-4x\right)^2+2\left(x-2\right)^2=43\)

\(\left[x\left(x-4\right)^2\right]+2\left(x-2\right)^2=43\)

\(x^2\left(x-4\right)^2+2\left(x-2\right)^2=43\)

\(x^4-8x^3+16x^2+2x^2-8x+8=43\)

\(x^4-8x^3+18x^2-8x+8-43=0\)

\(\left(x^3-9x^2+27x-35\right)\left(x+1\right)=0\)

\(x=5;-1\)

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

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

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

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

\(\Rightarrow2x^2+3x^3-3=5x^2+5x\)

\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow-3-5x=0\)

\(\Leftrightarrow-5x=3\)

\(\Leftrightarrow x=\frac{-3}{5}\)

CHÚC EM HỌC TỐT!!!

Đề bài chị ghi thiếu số 2, em bổ sung giúp chị nhé!!! Phần giải thì chị làm đúng rồi đấy, chúc em học tốt!!!

\(\text{a) (5x+2)(x-7)=0}\)

\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=7\end{cases}}\)

Vậy ...

#Thảo Vy#

\(\text{b) (x^2-1)(x+3)=0}\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+3\right)=0\)

\(\hept{\begin{cases}x+1=0\\x-1=0\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\x=1\\x=-3\end{cases}}\)

Vậy...

Ta có : \(\frac{MD}{MA}=\frac{NC}{NB}=\frac{m}{n}\)

\(\Rightarrow\frac{AM}{AD}=\frac{AM}{AM+MD}=\frac{n}{m+n}=\frac{ME}{DC}\)

và  \(\frac{NC}{BC}=\frac{NC}{NC+NB}=\frac{m}{m+n}=\frac{NE}{AB}\)

\(\Rightarrow ME=\frac{nDC}{m+n}\)

và \(NE=\frac{mAB}{m+n}\)

\(\Rightarrow MN=ME+NE=\frac{nDC+mAB}{m+n}\)(ĐPCM)

a,b,c thực dương

dễ thế mà ko làm dc à ngu vậy má