ngáo à chó

\(M=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(M=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(M=\left[x\left(x+5\right)+2\left(x+5\right)\right]\left[x\left(x+4\right)+3\left(x+4\right)\right]-24\)

\(M=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)

\(M=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(M=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)

\(M=\left(x^2+7x+11\right)^2-1-24\)

\(M=\left(x^2+7x+11\right)^2-25\)

\(M=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)

\(M=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

Biểu thức đâu bạn ? :)))

Sau khi ib với Đinh Lan Anh  thì \(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)

\(a,ĐKXĐ:\hept{\begin{cases}a+1\ne0\\a-1\ne0\end{cases}\Leftrightarrow a\ne\pm1}\)

\(b,P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)

       \(=\frac{2a^2+a\left(a-1\right)-a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)

       \(=\frac{2a^2+a^2-a-a^2-q}{\left(a-1\right)\left(a+1\right)}\)

       \(=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}\)

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

      \(=\frac{2a}{a+1}\)

\(c,P=\frac{2a}{a+1}=\frac{2a+2}{a+1}-\frac{2}{a+1}=2-\frac{2}{a+1}\)

Để \(P\inℤ\)thì \(2-\frac{2}{a+1}\inℤ\)

                    \(\Leftrightarrow\frac{2}{a+1}\inℤ\)

Mà \(a\inℤ\Rightarrow a+1\inℤ\)

Ta có bảng

Kết hợp ĐKXĐ \(a\ne\pm1\)ta  được \(a\in\left\{-3;-2;0\right\}\)

Vậy //////

Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2a+2b+2c}{a+b+c}=2\)

\(\Rightarrow\) a + b = 2c; b + c = 2a; c + a = 2b

\(\Rightarrow\) M = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

= \(\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{a+c}{a}\right)\)

= \(\frac{2c}{b}\times\frac{2a}{c}\times\frac{2b}{a}\)

= 8

Vậy: M = 8.

M=8

\(\left\{{}\begin{matrix}x-y-z=0\\x+2y-10z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=3z\\x=y+z=4z\\x+2y=10z\end{matrix}\right.\)

\(B=\dfrac{2x^2+4xy}{y^2+z^2}=\dfrac{2x\left(x+2y\right)}{9z^2+z^2}=\dfrac{2.4z.10z}{10.z^2}=8\)

ĐK \(a\ne\left\{-1;1\right\}\)

a. Ta có \(Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)^2}{a+1}\)

b. Khi \(\left|x\right|=5\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Với \(x=5\Rightarrow Q=\frac{\left(5-1\right)^2}{5+1}=\frac{16}{6}=\frac{8}{3}\)

Với \(x=-5\Rightarrow Q=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-9}=-4\)

ĐK \(x\ne\left\{-2;2\right\}\)

a. Ta có \(A=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\frac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}:\frac{x^2-4+10-x^2}{x+2}=-\frac{6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=-\frac{1}{x-2}\)

b. Ta có \(\left|x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

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

Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-1}{-\frac{1}{2}-2}=\frac{2}{5}\)

c. Để \(A< 0\Rightarrow-\frac{1}{x-2}< 0\Rightarrow x-2>0\Rightarrow x>2\)

Vậy với \(x>2\)thì \(A< 0\)

Biểu thức đâu bạn :V