Haizzz lâu r mới trở lại cái này ko bt có nhớ ko nx:

Ta có: x và y là 2 đại lượng tỉ lệ thuận

\(\Rightarrow y_1=k.x_1;y_2=k.x_2\)

Mà \(y_1+y_1=k.x_1+k.x_2=k\left(x_1+x_2\right)=4\)

Hay \(-1.k=4\Rightarrow k=-4\)\(\)

a, Biểu diễn: \(y=-4x\)

b,*) x = 1

Suy ra: \(y=-4x=-4.1=-4\)

*) x = 0,5 suy ra \(y=-4x=-4.0,5=-2\)

c, Cx tương tự câu b áp dụng y = -4x là ra ngay

*) y = -12 suy ra x = 3

*) y = \(\dfrac{4}{3}\) suy ra x = \(-\dfrac{1}{3}\)

có 60 em .

tích cho tớ nhé !

\(\frac{a}{b}=\frac{b}{3c}=\frac{c}{9a}=k\Leftrightarrow\left(\frac{a}{b}\right)^3=\frac{a.b.c}{b.3c.9a}=\frac{1}{27}=k^3\Leftrightarrow k=\frac{1}{3}\)

\(\frac{b}{3c}=\frac{1}{3}\Leftrightarrow b=c\)

\(3\sqrt{\dfrac{9}{8}}-\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{25}{18}}=3.\dfrac{3}{\sqrt{8}}-\dfrac{7}{\sqrt{2}}+\dfrac{5}{\sqrt{18}}=3.\dfrac{3}{2\sqrt{2}}-\dfrac{7}{\sqrt{2}}+\dfrac{5}{3\sqrt{2}}\)\(=\dfrac{27-42+10}{6\sqrt{2}}=\dfrac{-5}{6\sqrt{2}}\)

Ta có: \(L=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\)

\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ca+c+abc}\) ( Do abc = 1)

\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{ab+a+1}+\dfrac{1}{ab+a+1}=\dfrac{ab+a+1}{ab+a+1}=1\)

\(\dfrac{x^2-16}{4x-x^2}=\dfrac{x^2-4^2}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=-\dfrac{x+4}{x}\)

\(x^{10}+x^8+x^6+x^4+x^2+1=x^8\left(x^2+1\right)+x^4\left(x^2+1\right)+\left(x^2+1\right)\)\(=\left(x^2+1\right)\left(x^8+x^4+1\right)=\left(x^2+1\right)\left(x^8-x^2+x^4+x^2+1\right)\)

\(=\left(x^2+1\right)[x^2\left(x-1\right)\left(x^3+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\left(x^2-x+1\right)]\)

\(=\left(x^2+1\right)\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\dfrac{-1}{c^3}\) hay \(\dfrac{1}{a^3}+\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{b^3}=\dfrac{-1}{c^3}\)

\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{3}{abc}\)

\(a^2b^2c^2.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3}{abc}.a^2b^2c^2\)

\(\Leftrightarrow\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}=3abc\) hay\(M=3abc\left(đpcm\right)\)

a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)

b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

hay \(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)Ta có: \(a^2+b^2+c^2\ge0\) .Dấu "=" xảy ra \(\Leftrightarrow a=b=c=0\)

Suy ra \(ab+bc+ca=-\dfrac{a^2+b^2+c^2}{2}\le-\dfrac{0}{2}=0\)

Dấu "=" xảy ra \(\Leftrightarrow a^2=b^2=c^2=0\Leftrightarrow a=b=c=0\)