<=>x^2-y^2+x^2-xy=8

<=>(x-y)(2x+y)=8

2x+y>x-y

tự xét tiếp

lớp 9 kém thế

\(4-2xy+\left(2-x-y\right)^2=0\)

\(\Leftrightarrow y^2+x^2-4x-4y+8=0\)

\(\Leftrightarrow\left(y-2\right)^2+\left(x-2\right)^2=0\)

TA có \(\frac{2}{b}=\frac{1}{a}+\frac{1}{b}\)

=>\(\frac{2}{b}-\frac{1}{b}=\frac{1}{a}\)

=>\(\frac{1}{b}=\frac{1}{a}\)

=>\(a=b\)thay vào P:

\(P=\frac{a+b}{2a-b}+\frac{c+d}{2c-b}\)

=>\(P=\frac{2a}{a}+\frac{2c}{c}\)

=>\(P=4\)

\(x^2-8\sqrt{3x+7}=3x-32\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(3x+7-8\sqrt{3x+7}+16\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(4-\sqrt{3x+7}\right)^2=0\)

\(\Leftrightarrow x=3\)

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

\(\Leftrightarrow2x+1=\sqrt{2}\)

\(\Leftrightarrow4x^2+4x-1=0\)
Giờ thế vô A đi

bạn ơi cho hỏi sao chỗ kia từ dòng số 2 sao xuống dòng số 3 được vậy 

Ta có :

\(2\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\right)\)

\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\frac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\frac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\frac{9}{2}\)

\(\Rightarrow\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}>\frac{9}{4}\left(đpcm\right)\)

Chúc bạn học tốt !!!

Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{3}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Rightarrow2A=\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{1}\right)^2}{\sqrt{1}+\sqrt{3}}+...+\frac{\left(\sqrt{99}\right)^2-\left(\sqrt{97}\right)^2}{\sqrt{97}+\sqrt{99}}\)

\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{99}-\sqrt{97}\)

\(=\sqrt{99}-1\)

\(\Rightarrow A=\frac{\sqrt{99}-1}{2}=\frac{2\sqrt{99}-2}{4}>\frac{9}{4}\left(đpcm\right)\)