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120 dm² x 5 + 4m² = 600 dm² +  4 m² = 6 m² + 4m² = 10 m²

3.(⅓x - ¼)² = ⅓ 

=> (\(\dfrac{1}{3x}\)- \(\dfrac{1}{4}\) )² = \(\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{-1}{3}\\\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}=\dfrac{-1}{12}\\\dfrac{1}{3x}=\dfrac{7}{12}\end{matrix}\right.\)        => \(\left[{}\begin{matrix}x=-4\\x=\dfrac{12}{21}=\dfrac{4}{7}\end{matrix}\right.\)

Vậy, tập nghiệm x thỏa mãn là S=\(\left\{-4;\dfrac{4}{7}\right\}\)

\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)

\(\left(\frac{-3}{5}\right)^x=\frac{9}{25}=\left(\frac{-3}{5}\right)^2\)

\(\Rightarrow x=2\)

Ta thấy : \(\left(-\frac{3}{5}\right)^2=\frac{9}{25}\)

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

\(=>x=2\)

Bài này dễ T mik nha

\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)

\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)

\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)

\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)

\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)

3C=1x2x3+2x3x3+3x4x3+...+Xx(X+1)=

=1x2x3+2x3x(4-1)+3x4x(5-2)+...+Xx(X+1)[(X+2)-(X-1)]=

=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-(X-1)xXx(X+1)+Xx(X+1)x(X+2)=

=Xx(X+1)(X+2)

\(M=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(M\ge\left|x-1+3-x\right|=\left|2\right|=2\)

Dấu " = " xảy ra khi \(x-1\ge0;3-x\ge0\)

\(\Rightarrow x\ge1;x\le3\)

\(\Rightarrow1\le x\le3\)

Vậy \(MIN_M=2\) khi \(1\le x\le3\)