Ta có :\(b=\left(a+b+c+d\right)-\left(a+c+d\right)\)

\(\Rightarrow b=1-2=-1\)

\(c=\left(a+b+c+d\right)-\left(a+b+d\right)\)

\(\Rightarrow c=1-3=-2\)

\(d=\left(a+b+c+d\right)-\left(a+b+c\right)\)

\(\Rightarrow d=1-4=-3\)

Ta có :\(a+b+c+d=1\)

\(\Rightarrow a-1-2-3=1\)

\(\Rightarrow a-6=1\)

\(\Rightarrow a=7\)

Vậy \(a=7;b=-1;c=-2;d=-3\)

a/ Đặt: \(x+\frac{1}{x}=a\)

Ta có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)

\(x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)^2-2\)

\(=\left(a^3-3a\right)^2-2\)

\(\Rightarrow M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)

\(=\frac{a^6-\left(a^3-3a\right)^2+2-2}{a^3+a^3-3a}\)

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

\(=3.\left(x+\frac{1}{x}\right)=\frac{3x^2+3}{x}\)

b/ \(\frac{3x^2+3}{x}=3x+\frac{3}{x}\ge2.3=6\)

Đấu =  xảy ra khi \(x=\frac{1}{x}\Leftrightarrow x=1\)

x + 300 = 500  +100

x + 300 = 600

x = 600 - 300

x =  300 

Mk tk bạn rồi tk mk nha 

Nhầm tí S=\(\frac{1}{2}\)a.h

\(\Rightarrow a=5,3\)

x + 90 = 190

        x = 190 - 90

        x = 100

tk mk nha

1/ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow a=b=c=1\)

2/ \(\left(a+b+c\right)^2=3\left(a+b+c\right)\Leftrightarrow\left(a+b+c\right)\left(a+b+c-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a+b+c=3\end{cases}}\)

Bạn xem lại đề ! 

102 : 2 = 51

thua . là gì bạn biết hông chỉ mình , mình lên lớp đố mấy đứa chơi

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