mik ko bt

mình ko bt

1.1 Hình vuông có tối đa 4 góc vậy 4 hình vuông có tối đa 20 góc. S

2.1 hình vuông có tối đa 4 góc vậy 4 hình vuông có tối đa 16 góc. Đ

3. 1 hình vuông có tối thiểu 4 góc vậy 4 hình vuông có tối thiểu 16 góc. Đ

4.1 hình vuông có tối thiểu 1 góc vậy 4 hình vuông có tối thiểu 16 góc. S

Nhiêu đó hết tài năng rồi, mình mới lớp 3 thôi.

Ta có \(x1-\frac{1}{9}=x2-\frac{2}{8}=...=x9-\frac{9}{1}\)

\(=\frac{x1-1}{9}=\frac{x2-2}{8}=\frac{x3-3}{7}=...=\frac{x9-9}{1}\)

= \(\frac{x1-1+x2-2+x3-3+...+x9-9}{9+8+7+...+1}\)

\(=\frac{\left(x1+x2+x3+...+x9\right)-\left(1+2+3+...+9\right)}{9+8+7+....+1}\)

=\(\frac{90-45}{45}=\frac{45}{45}=1\)

=> \(\hept{\begin{cases}\begin{cases}x1=10\\x2=10\end{cases}\\.....\\x9=10\end{cases}}\)

Đáp án:ta có :

X1-1/9=X2-2/8=X3-3/7=......X9-9/1

Áp dụn t/c dãy tỉ số bằng nhau

⇒(X1 +X2+X3+........X9)-(1+2+3+...+9)/1=2+3+...+9

=90-45/45=1

⇒X1=X2=X3=X4=..=X9=10

a) \(2011.2013+2012.2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)

b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)

\(=\dfrac{1}{10}\left(9^{64}-1\right)\)

\(=\dfrac{9^{64}-1}{10}\)

Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)

Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)

\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)

c) Ta có:

\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)

Vì x>y>0, ta có:

\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)

Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)

Từ (1), (2) và (3) suy ra:

\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)

a) Ta có:

\(2011.2013+2012.2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

Vậy 2011.2013+2012.2014 = 2012² + 2013² - 2

ta có : \(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\dfrac{\sqrt{x+3}}{\sqrt{x-3}}\)

ta có : \(B=\dfrac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{ 9-x^2}}{x\left(3-x\right)+\left(x+2\right)\sqrt{9-x^2}}\)

\(=\dfrac{\sqrt{x+3}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{x+3}\right)}=\dfrac{\sqrt{x+3}}{\sqrt{3-x}}\)