b/ x²-2x=24

=> x²-2x-24=0

=> (x-6)(x+4)=0

=>x=6 hoặc x =-4

a/ (x-3)² - 4 = 0

=> (x-3-2)(x-3+2)=0

=> (x-5)(x-1)=0

=> x = 5 hoặc x=1

ý c đâu ạ

#)Giải :

a)\(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}\)

\(\Leftrightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]\)

\(\Leftrightarrow5x-3\left(4x-8x+30x-12\right)\)

\(\Leftrightarrow5x-12x+24x-90x+36\)

\(\Leftrightarrow-73x+36=182\)

\(\Leftrightarrow-73x=146\)

\(\Leftrightarrow x=-2\)

\(=>2^3-x^3+\left(x^2-3x\right)\left(x+4\right)-x^2+24=0\)

\(=>8-x^3+x^3+x^2-12x-x^2+24=0\)

\(=>-12x=-16=>x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

\(\left(x^2+2x+4\right)\left(2-x\right)+x\left(x-3\right)\left(x+4\right)-x^2+24=0\)

\(\Leftrightarrow2x^2-x^3+4x-2x^2+8-4x+\left(x^2-3x\right)\left(x+4\right)-x^2+24=0\)

\(\Leftrightarrow2x^2-x^3+4x-2x^2+8-4x+x^3+4x^2-3x^2-12x-x^2+24=0\)

\(\Leftrightarrow-12x+8+24=0\)

\(\Leftrightarrow-12x+32=0\)

\(\Leftrightarrow-12x=-32\)

\(\Leftrightarrow x=\frac{8}{3}\)

á , mình sai rồi -12x=-32 nha

bạn đăng tách ra nhé

a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)

\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)

b, sửa đề :  \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)

\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)

c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)

d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)

\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)

e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)

\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)

f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)

\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)

g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)

\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)

vô nghiệm 

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-7\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

b) \(x^2-2x=24\)

\(x^2-2x-24=0\)

\(\left(x-6\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(5x^2+10x+10-5x^2+245=0\)

\(10x+255=0\)

\(x=-25.5\)

A) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)

th1\(\left(x-3\right)^2=2^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

th2: \(\left(x-3\right)^2=\left(-2\right)^2\)

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

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

\(\Rightarrow x=1\)

\(\Leftrightarrow x\in\left\{1;5\right\}\)

a) \(\left(x-3\right)^2-4=0\)

Ta xét 2TH:

TH1: \(x-3=2\)

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

\(x=5\)

TH2: \(x-3=-2\)

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

\(x=1\)

=> \(x\in\left\{1;5\right\}\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

\(\left(x-5\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)

tíc mình nha

dễ mà bạn

giúp mình giải bài này với ^_​​

3^{39 và 11 mũ 21}

1) \(\left(x-3\right)^2-4=0\)

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

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop