2

1+1=2

1 + 1 = 2

\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=\dfrac{1}{2}\)

\(\Leftrightarrow1-2a^2b^2=\dfrac{1}{2}\)

\(\Leftrightarrow a^2b^2=\dfrac{1}{4}\)

mà \(a^2+b^2=1\)

Nên \(a^2;b^2\) là nghiệm của phương trình:

\(X^2-X+\dfrac{1}{4}=0\Leftrightarrow\left(X-\dfrac{1}{2}\right)^2=0\Leftrightarrow X=\dfrac{1}{2}\)

\(\Rightarrow a^2=b^2=\dfrac{1}{2}\)

\(P=a^{2024}+b^{2024}=\left(a^2\right)^{1012}+\left(b^2\right)^{1012}=\left(\dfrac{1}{2}\right)^{1012}+\left(\dfrac{1}{2}\right)^{1012}\)

\(\Rightarrow P=2.\left(\dfrac{1}{2}\right)^{1012}=\dfrac{1}{2^{1011}}\)

k cs nghĩa

\(1+1=2\)

=2

=2

đề bài đâu bn

sao đăng trống màu trắng ko thấy bài đâu hết vậy

bằng 2 nhé bn
1+1=2

2

\(1+1=2\)

`a+a = 10`

`2 xx a = 10`

`a = 10 : 2`

`a = 5`

`b + b = 12`

`2 xx b = 12`

`b = 12 :2`

`b=6`

`c + c =20`

`2 xx c = 20`

`c = 20:2`

`c = 10`

`-> a+b+c=5+6+10=21`

a + a = 10

=> a = 5

b + b = 12

=> b = 6

c + c = 20

=> c = 10

a + b + c = 5 + 6 + 10

= 21

Chà sao lại đặt trong lớp 1 vậy em

1.35

\(x=\dfrac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{9-4\sqrt{5}}-2}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{5-2.2.\sqrt{5}+4}-2}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{\left(\sqrt{5}-2\right)^2}-2}=\dfrac{\left|\sqrt{3}+1\right|-\sqrt{3}}{\left(\sqrt{5}+2\right)\left|\sqrt{5}-2\right|-2}\)

\(=\dfrac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\dfrac{1}{5-4-2}=\dfrac{1}{-1}=-1\)

\(\Rightarrow P=\left(\left(-1\right)^2-1+1\right)^{2017}=1^{2017}=1\)

1.36.

\(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\right)\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}\right)\)

\(=\left(\dfrac{x+2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{1}{2\sqrt{x}+3}\right)\)

\(=\dfrac{\left(x+2\sqrt{x}+1\right)}{\sqrt{x}+1}.\dfrac{1}{2\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}\)

b.

\(\dfrac{x}{4}=\sqrt{\dfrac{1009+\sqrt{2017}}{2}}-\sqrt{\dfrac{1009-\sqrt{2017}}{2}}\)

\(x=4\sqrt{\dfrac{1009+\sqrt{2017}}{2}}-4\sqrt{\dfrac{1009-\sqrt{2017}}{2}}\)

\(x=2.\sqrt{2018+2\sqrt{2017}}-2\sqrt{2018-2\sqrt{2017}}\)

\(x=2\sqrt{2017+2\sqrt{2017}+1}-2\sqrt{2017-2\sqrt{2017}+1}\)

\(x=2\sqrt{\left(\sqrt{2017}+1\right)^2}-2\sqrt{\left(\sqrt{2017}-1\right)^2}\)

\(x=2\left|\sqrt{2017}+1\right|-2\left|\sqrt{2017}-1\right|\)

\(x=2\left(\sqrt{2017}+1\right)-2\left(\sqrt{2017}-1\right)=4\)

\(\Rightarrow A=\dfrac{\sqrt{4}+1}{2\sqrt{4}+3}=\dfrac{3}{7}\)

1.37

\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\left(3x-6\sqrt{x}\right)+\left(x+3\sqrt{x}+2\right)-\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}+2}\)

b.

\(x=\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\)

\(\Rightarrow x^3=\left(\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\right)^3\)

Áp dụng HĐT: \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Rightarrow x^3=2+3\sqrt[3]{\left(1+\dfrac{\sqrt[]{84}}{9}\right)\left(1-\dfrac{\sqrt[]{84}}{9}\right)}.\left(\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\right)\)

\(\Rightarrow x^3=2+3.\sqrt[3]{1-\dfrac{84}{81}}.x\)

\(\Rightarrow x^3=2-x\)

\(\Rightarrow x^3+x-2=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=1\)

\(\Rightarrow P=\dfrac{4\sqrt[]{1}}{\sqrt[]{1}+2}=\dfrac{4}{3}\)

\(a,\sqrt{4-\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}\\ =\sqrt{\left(\sqrt{4-\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}\right)^2}\\ =\sqrt{4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4-\sqrt{10+2\sqrt{5}}\right)\left(4+\sqrt{10+2\sqrt{5}}\right)}+4+\sqrt{10+2\sqrt{5}}}\\ =\sqrt{8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}}\\ =\sqrt{8+2\sqrt{16-10-2\sqrt{5}}}\\ =\sqrt{8+2\sqrt{6-2\sqrt{5}}}\\ =\sqrt{8+2\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}}\\ =\sqrt{8+2\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{8+2\sqrt{5}-2}\\ =\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}\\ =\sqrt{5}+1\)

\(b,\left(2+\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2+\sqrt{4+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2+\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2+\sqrt{4+\sqrt{5}-1}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{2}+\sqrt{8+2\sqrt{5}-2}\right)\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{2}+\sqrt{6+2\sqrt{5}}\right)\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}\right)\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{2}+\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\sqrt{10}-2\sqrt{2}+5-\sqrt{5}+\sqrt{5}-1\\ =2\sqrt{10}-2\sqrt{2}+4\)