\(\frac{a}{3}=\frac{2a+2}{5}\)

\(\Rightarrow5a=3.\left(2a+2\right)\)

\(\Rightarrow5a=6a+6\)

\(\Rightarrow5a-6a=6\)

\(\Rightarrow a=-6\)

\(\frac{a}{3}=\frac{2a+2}{5}\Rightarrow5a=3\left(2a+2\right)\Rightarrow5a=6a+10\Rightarrow a=6\)

sory mk giải nhầm

\(\frac{a}{3}=\frac{2a+2}{5}\Rightarrow\frac{5a}{15}=\frac{6a+6}{15}\Rightarrow5a=6a+6\Rightarrow6=5a-6a\Rightarrow a=-6\)

Vậy a=-6

a/b=-4/5

nên a/-4=b/5

Đặt a/-4=b/5=k

=>a=-4k; b=5k

\(a^2+2b^2=16.5\)

\(\Leftrightarrow16k^2+50k^2=16.5\)

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

Trường hợp 1: k=1/2

=>a=-2; b=5/2

=>a+b=1/2

Trường hợp 2: k=-1/2

=>a=2; b=-5/2

=>a+b=-1/2

Vậy: Giá trị lớn nhất của a+b là 1/2

\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)

\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)=x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)

Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{2016}>\frac{1}{2017}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}>\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\)

=> x = 0

Vậy x = 0

\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)

\(\Rightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)

\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)=0\)

\(\Rightarrow x=0\).Do \(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\ne0\)

Vậy x=0

Theeo mình là x = 0 

= 6 cặp 

mk làm trong violympic rùi tin mk đi

a³+b³+c³=3abc

<=>(a+b)³-3ab(a+b)-3abc+c³=0

<=>(a+b+c)[(a+b)²-(a+b)c+c²]-3ab.(a+b+c)=0

<=>(a+b+c)(a²+b²+c²-ab-bc-ac)=0

<=>(a+b+c)(2a²+2b²+2c²-2ab-2bc-2ac)=0

<=>(a+b+c)[(a-b)²+(b-c)²+(c-a)²]=0

<=>a+b+c=0 [(a-b)²+(b-c)²+(c-a)² khác 0]

=>a²+b²-c²=-2ab;b²+c²-a²=-2bc;c²+a²-b²=-2ac

Suy ra : P=\(-\left(\dfrac{1}{2ab}+\dfrac{1}{2bc}+\dfrac{1}{2ac}\right)=-\dfrac{a+b+c}{2abc}=0\)

\(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow B=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\left(\frac{\left(1-x\right)\left(1+x\right)}{\left(x-1\right)^2\left(x+1\right)}\right)\)

\(\Leftrightarrow B=\left(1+x+x^2-x\right):\left(\frac{-1}{x-1}\right)\)

\(\Leftrightarrow B=-\left(x^2+1\right).\left(x-1\right)\)

\(\Leftrightarrow B=-x^3+x^2-x+1\)

b) Để B < 0

\(\Leftrightarrow-x^3+x^2-x+1< 0\)

\(\Leftrightarrow-\left(x^2+1\right)\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)>0\)

TH1 : \(\hept{\begin{cases}x^2+1>0\left(tm\right)\\x-1>0\end{cases}\Leftrightarrow x>1}\)

TH2 : \(\hept{\begin{cases}x^2+1< 0\left(ktm\right)\\x-1< 0\end{cases}}\Leftrightarrow x\in\varnothing\)

Vậy để \(B< 0\Leftrightarrow x>1\)

c) Khi \(x-4=5\)

\(\Leftrightarrow x=9\)

\(\Leftrightarrow B=-\left(9^3\right)+9^2-9+1\)

\(\Leftrightarrow B=-729+81-9+1\)

\(\Leftrightarrow B=-656\)

Vậy khi \(x-4=5\Leftrightarrow B=-656\)

Lớp 7 gì mà dễ ẹc :))

\(\frac{2a-b}{a+b}=\frac{2}{3}\)

\(\Leftrightarrow6a-3b=2a+2b\)

\(\Rightarrow4a=5b\)

\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)

\(\Leftrightarrow4a-2b=3b-3c+3a\)

\(\Leftrightarrow a=5b-3c\)

\(\Leftrightarrow a-5b=-3c\)

\(\Leftrightarrow a-4a=-3c\)

\(\Leftrightarrow-3a=-3c\)

\(\Rightarrow a=c\)

Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)

8 nhé bn !