a,ĐKXĐ:\(x\ne2,x\ne-3\)

\(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x-4}{x-2}\)

c,Để A = - 3/4

thì: \(\frac{x-4}{x-2}=-\frac{3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)

\(4x-16=-3x+6\)

\(4x+3x=6+16\)

\(7x=22\)

\(x=\frac{22}{7}\)

d,\(A=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=\frac{x-2}{x-2}-\frac{2}{x-2}=1-\frac{2}{x-2}\)

Để A nguyên thì: \(x-2\inƯ\left(2\right)\)

Ta có: \(Ư\left(2\right)=\left\{\pm1,\pm2\right\}\)

Xét từng TH:

_ x - 2 = -1 => x = 1

_ x - 2 = 1 => x = 3

_ x - 2 = -2 => x = 0

_ x- 2 = 2 => x= 4

Vậy: \(x\in\left\{0,1,3,4\right\}\)

=.= hok tốt!!

Đề là tìm GTNN?

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Đầu tiên ta chứng minh bổ đề: \(\frac{x}{y}+\frac{y}{x}\ge2\).Áp dụng BĐT AM-GM (Cô si),ta có: \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{yz}}=2^{\left(đpcm\right)}\) (1)

Dấu "=" xảy ra khi x = y

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Đặt \(A=x^2+y^2+z^2-xyz\).Thay giả thiết đề bài,ta có"

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

\(\ge2^2+2^2+2^2-2.2.2=4\)   (BĐT (1) )

Vậy \(A_{min}=4\Leftrightarrow a=b=c\Leftrightarrow x=y=z\)

\(-\left(\frac{a}{b}+\frac{b}{a}\right)\left(\frac{b}{c}+\frac{c}{b}\right)\left(\frac{c}{a}+\frac{a}{c}\right)\le-2.2.2=-8\) ngược dấu nha eiu 

ko đc trừ 2 vế tương ứng của 2 bđt cùng chiều 

aloo 

kho qua

hiok tot

bye >_<

Sửa đề \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(h_a+h_b+h_c\right)\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\) \(\left(1\right)\)

Gọi S là diện tích tam giác \(\Rightarrow\)\(S=\frac{ah_a}{2}=\frac{bh_b}{2}=\frac{ch_c}{2}\)\(\Rightarrow\)\(a=\frac{2S}{h_a};b=\frac{2S}{h_b};c=\frac{2S}{h_c}\)

\(VT=\left(\frac{2S}{h_a}+\frac{2S}{h_b}+\frac{2S}{h_c}\right)\left(\frac{1}{\frac{2S}{h_a}}+\frac{1}{\frac{2S}{h_b}}+\frac{1}{\frac{2S}{h_c}}\right)\) ( thay vào là xong ) 

\(VT=2S\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\left(\frac{h_a+h_b+h_c}{2S}\right)=\left(h_a+h_b+h_c\right)\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\) ( đpcm ) 

Chúc bạn học tốt ~ 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\Leftrightarrow c\left(a+b+c\right)\left(a+b\right)=-ab\left(a+b\right)\)

\(\Leftrightarrow\left(ac+bc+c^2\right)\left(a+b\right)+ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> a=-b hoặc b=-c hoặc c=-a

không mất tính tổng quát ,giả sử a=-b, ta có:

\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{-b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\left(1\right)\)

\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\left(2\right)\)

Từ  (1) và (2) => đpcm

Tương tự với 2 trường hợp còn lại ta cũng có đpcm

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

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

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

\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

Ta có : 

\(\frac{1}{a^2+b^2+c^2}< \frac{1}{a^2};\frac{1}{b^2};\frac{1}{c^2}\)

\(\Rightarrow\)\(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}< 0\)

\(\Rightarrow\)\(x^2=y^2=z^2=0\)\(\Rightarrow\)\(x=y=z=0\) ( đpcm ) 

Chúc bạn học tốt ~ 

a. ĐK: a, b, c khác 0.

 \(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}=1\)

\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)

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

\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)

\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).

b) Không mất tính tổng quả. G/s: a = b + c

Khi đó ta có:

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

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

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

=> Điều phải chứng minh.

Ta có : \(y+z=ax+cz+ax+by=2ax+x\)

\(\Rightarrow\)\(y+z-x=2ax\)\(\Rightarrow\)\(a=\frac{y+z-x}{2x}\)\(\Rightarrow\)\(\frac{1}{a+1}=\frac{2x}{x+y+z}\)

Tương tự, ta cũng có \(\frac{1}{b+1}=\frac{2y}{x+y+z};\frac{1}{c+1}=\frac{2z}{x+y+z}\)

\(\Rightarrow\)\(S=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Chúc bạn học tốt ~