Giúp đc sẽ giúp.Mấy bác đi qua nhường em câu này.Lăng nhăng là em chém :"D

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

\(=x^2-9-2x+3\)

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

\(\left(a+b\right)^2+\left(a-b\right)^2-2\left(b-a\right)\left(b+a\right)\)

\(=a^2+2ab+b^2+a^2-2ab+b^2-2\left(b^2-a^2\right)\)

\(=2a^2+2b^2-2a^2-2b^2\)

\(=0\)

2) \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(đpcm\right)\)

\(x-x^2-2=-x^2+x-2=-\left(x^2-x+2\right)=-\left(x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\right)=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}< 0\left(đpcm\right)\)

\(A=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

\(A=\left(a+b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2-\left(b-c\right)^2\)

\(A=\left[\left(a+b-c\right)^2-\left(b-c\right)^2\right]+\left[\left(a-b+c\right)^2-\left(b-c^2\right)\right]\)

\(A=\left[\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)\right]+\left[\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\right]\)

\(A=\left[a\left(a+2b\right)\right]+\left[a\left(a-2b+2c\right)\right]\)

\(A=a\left(a+2b+a-2b+2c\right)\)

\(A=a\left(2a+2c\right)\)

ngu thì im

\(3.6=18\)

\(4.8=32\)

\(5.10=50\)

\(6.12=72\)

\(7.14=98\)

Như vậy \(10.20=200\)

1) - ko di chuyển đc

-tự tổng hợp đc cá chất hữu cơ

-phản ứng chậm vs các chất kích thích từ bên ngoài

\(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(A=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{5^{32}-1}{2}\)

Gọi 3 phần đó lần lượt là \(a;b;c\)

Theo đề bài ta có:

\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{c}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{c}{6}=\dfrac{a+b+c}{4+5+6}=\dfrac{30}{15}=2\)

Suy ra: \(\left\{{}\begin{matrix}a=2.4=8\\b=2.5=10\\c=2.6=12\end{matrix}\right.\)

Theo đề bài ta có:

\(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}\\\dfrac{b}{6}=\dfrac{c}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{4}=\dfrac{b}{6}\\\dfrac{b}{6}=\dfrac{c}{5}\end{matrix}\right.\)\(\Leftrightarrow\dfrac{a}{4}=\dfrac{b}{6}=\dfrac{c}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{4}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+b+c}{4+6+5}=\dfrac{30}{15}=2\)

Suy ra:

\(\left\{{}\begin{matrix}a=2.4=8\\b=2.6=12\\c=2.5=10\end{matrix}\right.\)

\(A=4.4^1.4^3.4^5.....4^{57}=4^{1+3+5+...+57}=4^{\left[\left(\dfrac{57-1}{2}\right)+1\right]:2\left(57+1\right)}=4^{841}\)\(B=3+3^2+3^3+3^4+...+3^{100}\)

\(3B=3\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(3B=3^2+3^3+3^4+3^5+...+3^{101}\)

\(3B-B=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4....+...+3^{100}\right)\)

\(2B=3^{101}-3\Leftrightarrow B=\dfrac{3^{101}-3}{2}\)

2)

Từ \(1\rightarrow9\) có: \(\left(9-1\right):1+1=9\)(chữ số)

Từ \(10\rightarrow99\) có:\(2\left[\left(99-10\right):1+1\right]=180\)(chữ số)

Từ \(100\rightarrow386\) có:\(3\left[\left(386-100\right):1+1\right]=816\)(chữ số)

Như vậy,Để đánh số trang từ \(1\rightarrow386\) thì cần:

\(9+180+816=1005\)(chữ số)