Tìm Min mà mấy chế nói cái huần hòe gì thế

\(x\left(x+1\right)=x^2+x=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{2}^2-\dfrac{1}{2}^2=\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

Vậy: Min=\(-\dfrac{1}{4}\) khi x= \(-\dfrac{1}{2}\)

a) \(35^6-35^5=35^5\left(35-1\right)=35^5.34\)

\(\Rightarrow35^6-35^5⋮34\)

b) \(43^4+43^5=43^4\left(1+43\right)=43^4.44\)

\(\Rightarrow43^4+44^5⋮44\)

3) \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)

\(\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)

\(\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{a+c}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)

\(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c\)

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)

Vậy: \(P=0\)

\(g\left(x\right)=x^2-6x+10\)

\(=x^2-6x+9+1\)

\(=\left(x-3\right)^2+1\ge1>0\)

Vậy ta có đpcm

\(\left\{{}\begin{matrix}\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\)

Đặt \(u=\sqrt{x};v=\sqrt{y}\) cho đơn giản (\(u;v\ge0\))

\(\left\{{}\begin{matrix}u-2v=-2\\2u-3v=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2u-4v=-4\\2u-3v=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-v=-1\\u-2v=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}v=1\\u-2.1=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}v=1\\u=0\end{matrix}\right.\left(TMĐK\right)\)

\(\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)

Đề thiếu \(0\le a,b,c\le\dfrac{4}{3}\)

\(b+c=2-a\)

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

Áp dụng BĐT \(\left(x+y\right)^2\ge4xy\), ta có

\(\left(2-a\right)^2\ge4\left(1-2a+a^2\right)\)

\(4-4a+a^2\ge4-8a+4a^2\)

\(4-4a+a^2-4+8a-4a^2\ge0\)

\(-3a^2+4a\ge0\)

\(3a^2-4a\le0\)

\(a\left(3a-4\right)\le0\)

\(\Rightarrow0\le a\le\dfrac{4}{3}\)

Tương tự với b,c

\(\dfrac{1}{1+a+ab}+\dfrac{a}{a+ab+abc}+\dfrac{1}{abc+ac+c}\)

\(=\dfrac{1}{1+a+ab}+\dfrac{a}{1+a+ab}+\dfrac{1}{c\left(1+a+ab\right)}\)

\(=\dfrac{ac+c+1}{c\left(1+a+ab\right)}=\dfrac{c\left(a+1+ab\right)}{c\left(1+a+ab\right)}=1\)

là hình 

Bài 6,8 có gì đó sai sai

4) 2 vế đều không âm nên bình phương

\(\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2\ge64x^2y^2z^2\)(2)

Ta có: \(\left(x+y\right)^2\ge4xy\)

\(\left(y+z\right)^2\ge4yz\)

\(\left(z+x\right)^2\ge4xz\)

Nhân từng vế ta được (2) đúng

VT và VP của 2 không âm nên khai phương ta được đpcm