Ta có : \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)

\(\Leftrightarrow\left(x^2+x-1\right)^2-1=24\)

\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)

<=> 2 trường hợp sảy ra là bằng 5 hoặc -5 nhé 

bạn lam được cả câu a thì mk k

a,\(\left(x+y\right)^2-y^2=\left(x+y-y\right)\left(x+y+y\right)=x\left(x+2y\right)\left(đpcm\right)\)

b,\(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2+2xy+y^2\right)\left(x^2-2xy+y^2\right)=\left(x+y\right)^2\left(x-y\right)^2\left(đpcm\right)\)

c, áp dụng quy tắc dấu ngoặc

Nhanh Nha

5x² - 5y² - x - y

<=> 5(x² - y²) - (x + y

<=> 5(x - y)(x + y) - (x + y)

<=> [5( x - y) - 1](x + y)

\(5x^2-5y^2-x-y\)

\(\left[\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\frac{1}{2.\sqrt{5}}+\left(\frac{1}{2.\sqrt{5}}\right)^2\right]-\left[\left(\sqrt{5}y\right)^2+2.\sqrt{5}y.\frac{1}{2.\sqrt{5}}+\left(\frac{1}{2.\sqrt{5}}\right)^2\right]\)

\(=\left(\sqrt{5}x-\frac{1}{2.\sqrt{5}}\right)^2-\left(\sqrt{5}y+\frac{1}{2.\sqrt{5}}\right)^2\)

\(=\left(\sqrt{5}x-2.\frac{1}{2.\sqrt{5}}-\sqrt{5}y\right)\left(\sqrt{5}x-\sqrt{5}y\right)\)

tham khảo nhé

\(A=x^2+5x+7\)

\(A=\left(x^2+5x+\frac{25}{4}\right)+\frac{3}{4}\)

\(A=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x+\frac{5}{2}\right)^2=0\)

\(\Leftrightarrow\)\(x+\frac{5}{2}=0\)

\(\Leftrightarrow\)\(x=\frac{-5}{2}\)

Vậy GTNN của \(A\) là \(\frac{3}{4}\) khi \(x=\frac{-5}{2}\)

Chúc bạn học tốt ~ 

\(B=6x-x^2-5\)

\(-B=x^2-6x+5\)

\(-B=\left(x^2-6x+9\right)-4\)

\(-B=\left(x-3\right)^2-4\ge-4\)

\(B=-\left(x-3\right)^2+4\le4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x-3\right)^2=0\)

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

\(\Leftrightarrow\)\(x=3\)

Vậy GTLN của \(B\) là \(4\) khi \(x=3\)

Chúc bạn học tốt ~ 

a/ \(25x^2-9=0\)

<=> \(\left(5x-3\right)\left(5x+3\right)=0\)

<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)

<=> \(x^2+8x+16-x^2+8x-9=16\)

<=> \(16x+7=16\)

<=> \(16x=9\)

<=> \(x=\frac{9}{16}\)

a) \(25x^2-9=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)

Vậy S = {3/5 ; -3/5}

b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)

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

\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)

\(\Leftrightarrow9=0\left(vl\right)\)

Vậy S = \(\varnothing\)

xy + 1 - x - y

<=> xy - x + 1 - y 

<=> x(y - 1) - (y - 1)

<=> (x - 1)(y - 1)

Nếu bạn tìm nghiệm thì:

<=> (x - 1)(y - 1) = 0

<=> \(\orbr{\begin{cases}x-1=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\y=1\end{cases}}}\)

Vậy (x,y) = (1,1)

cảm ơn bn nhìu nhé mk biết ơn lắm lun:D