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\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2-x-2\right)\)
\(=\left(-4\right)\left(x+2\right)\)
a: \(P=\dfrac{a+3}{a}\cdot\dfrac{a^2-9-6a+18}{\left(a-3\right)\left(a+3\right)}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a-3\right)}=\dfrac{a-3}{a}\)
b: Để P=-2 thì -2a=a-3
=>-3a=-3
=>a=1
c: Để P nguyên thì a-3 chia hết cho a
=>-3 chia hết cho a
mà a<>0; a<>3; a<>-3
nên \(a\in\left\{1;-1\right\}\)
\(P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
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\(T=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Rightarrow2T=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^8-1\right)\left(3^8+1\right)=3^{16}-1\)
\(\Rightarrow T=\dfrac{3^{16}-1}{2}=21523360\)
bạn ơi bạn viết rõ bài P dc k o mình đọc chả hiểu j
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab-2ac-2bc-a^2+2ac-c^2-2ab+2bc\)
\(=b^2\)
a) \(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(\Leftrightarrow2b^3\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6ab^2\)
\(\Leftrightarrow2b^3+6a^2b-6ab^2\)