Ta có :

A=\(\left(-3x^5y^3\right)^4\ge0\forall x,y\)

B=\(\left(2x^2z^4\right)^5=\left(2xz^2\right)^{10}\ge0\forall x,z\)

Mà A+B = 0

\(\Rightarrow\left\{{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x^5y^3\\2xz^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\z=0\end{matrix}\right.\end{matrix}\right.\)

Vậy x =0 ; y = 0 ; z = 0 là các giá trị cần tìm

\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+1\)

\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)+15xy\left(y-x\right)+1\)

\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1\)

\(=\left(x-y\right)\left(2+13x^3y^2-15xy\right)+1\)

\(1-\left(3\dfrac{3}{8}+x-5\dfrac{9}{24}\right):7\dfrac{2}{3}=0\)

\(\Leftrightarrow1-\left(\dfrac{27}{8}+x-\dfrac{43}{8}\right):\dfrac{23}{3}=0\)

\(\Leftrightarrow1-\left(x-2\right)\cdot\dfrac{3}{23}=0\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{3}{23}=1\\ \Leftrightarrow x-2=1:\dfrac{3}{23}\)

\(\Leftrightarrow x-2=\dfrac{23}{3}\\ \Leftrightarrow x=\dfrac{23}{3}+2\\ \Leftrightarrow x=\dfrac{29}{3}\)

Vậy........................

Ta có \(\dfrac{x+5}{7}=\dfrac{40}{140}\)

\(\Leftrightarrow\dfrac{x+5}{7}=\dfrac{2}{7}\\ \Leftrightarrow x+5=2\\ \Leftrightarrow x=-3\)

Tương tự : \(\dfrac{-30}{5y+5}=\dfrac{40}{140}\)

\(\Leftrightarrow\dfrac{-6}{y+1}=\dfrac{2}{7}\\ \Leftrightarrow\left(y+1\right)\cdot2=-6\cdot7\\ \Leftrightarrow2y+2=-42\)

\(\Leftrightarrow2y=-44\\ \Leftrightarrow y=-22\)

Vậy..................................

1a,

Đổi 75cm = \(\dfrac{3}{4}\) m \(\Rightarrow\left(\dfrac{2}{3}:\dfrac{3}{4}\right)\cdot100\%\approx88.89\%\)

Đổi \(\dfrac{3}{10}\)h = 18 phút \(\Rightarrow\left(18:20\right)\cdot100\%=90\%\)

b).

\(\left(2\dfrac{3}{7}:1\dfrac{13}{21}\right)\cdot100\%=\left(\dfrac{17}{7}:\dfrac{34}{21}\right)\cdot100\%=150\%\)

Đổi 0,3 tạ = 30 kg

\(\Rightarrow\left(30:50\right)\cdot100\%=60\%\)

Ta có \(A=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\)

\(\Leftrightarrow2A=\dfrac{2}{99\cdot97}-\dfrac{2}{97\cdot95}-...-\dfrac{2}{3\cdot1}\)

\(=-\dfrac{1}{99}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{95}-...-\dfrac{1}{3}+1\)

\(=-\dfrac{1}{99}+1=\dfrac{98}{99}\)

\(\Rightarrow A=\dfrac{49}{99}\)

1) . \(\dfrac{1}{2}-\left|\dfrac{1}{5}-\dfrac{1}{4}\right|+\left(-\dfrac{1}{3}\right)^2\\ =\dfrac{1}{2}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\dfrac{1}{9}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{9}\)

\(=\dfrac{61}{180}\)

2) . \(\dfrac{1}{3}+\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{-2}{3}\right)^2\\ =\dfrac{1}{3}+\dfrac{4}{3}\cdot\dfrac{1}{6}+\dfrac{4}{9}\\ =\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{9}\\ =1\)

a, 10m

b, 8m

c, 80\(m^2\)

A>B

Số dưới gấp 3 số trên \(\Rightarrow\)(?)=6