bài dễ lắm bạn tự nghĩ đi

Trả lời giùm đi

Nếu 2a = a+c thì a=c

1/2(1/b+1/d)=1/c => 1/c = 1/4(1/b+1/d)=> b=d

ad=bc(đpcm)

Nhìn BĐT 4 số ngán quá

\(1\ge4\sqrt[4]{\frac{1}{a^2b^2c^2d^2}}\Rightarrow abcd\ge16\)

\(\Rightarrow VT=\frac{abcd}{8}+2\ge4\) (1)

Mà \(VP=\frac{a+c}{\sqrt{ac}}+\frac{b+d}{\sqrt{bd}}\le\frac{2\left(a+c\right)}{a+c}+\frac{2\left(b+d\right)}{b+d}=4\) (2)

(1);(2) \(\Rightarrow\) đpcm

Dấu "=" xảy ra khi \(a=b=c=d=2\)

Akai Haruma

@Nguyễn Việt Lâm anh giúp em vs !!!

18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)

19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được 

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)

Ta có ;

 \(\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=\frac{\left(a+b\right)\left(a-b\right)+\left(b+c\right)\left(b-c\right)+\left(c+a\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

17. Xét vế trái ; 

\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{a}{\left(b-1\right)\left(b^2+b+1\right)}+\frac{b}{\left(a-1\right)\left(a^2+a+1\right)}\)

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

\(=\frac{-\left(a^2+a+1+b^2+b+1\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}=\frac{-\left[\left(a+b\right)^2-2ab+3\right]}{a^2b^2+ab\left(a+b\right)+a^2+b^2+ab+2}\)\(=\frac{2\left(ab-2\right)}{a^2b^2+\left(a^2+2ab+b^2\right)+2}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)

a)

\(\begin{array}{l}f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}{x^2} - \frac{1}{2}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}\left( {{x^2} - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{1}{2}\left( {x + 1} \right) = \frac{1}{2}\left( {1 + 1} \right) = 1\end{array}\)

b) Phương trình đường thẳng \(d\) đi qua điểm \(M\left( {1;\frac{1}{2}} \right)\) và có hệ số góc bằng \(k = f'\left( 1 \right) = 1\) là: \(y - \frac{1}{2} = 1\left( {x - 1} \right) \Leftrightarrow y = x - 1 + \frac{1}{2} \Leftrightarrow y = x - \frac{1}{2}\).

Đường thẳng \(d\) cắt đồ thị hàm số \(\left( C \right)\) tại duy nhất điểm \(M\left( {1;\frac{1}{2}} \right)\).