\(x\left(x-\dfrac{1}{3}\right)< 0\)

Ta có : \(x>x-\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{1}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{1}{3}\end{matrix}\right.\Rightarrow0< x< \dfrac{1}{3}\)

Vậy \(0< x< \dfrac{1}{3}\)

Em làm vậy có dài lắm không Nguyễn Huy Tú

\(\text{a) }\dfrac{4^5\cdot4^2}{16^4}=\dfrac{4^7}{\left(4^2\right)^4}=\dfrac{4^7}{4^8}=\dfrac{1}{4}\)

\(\text{b) }\dfrac{2^8\cdot9^4}{6^6\cdot8^3}\\ =\dfrac{2^8\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\\ =\dfrac{2^8\cdot3^8}{2^6\cdot3^6\cdot2^9}\\ =\dfrac{2^8\cdot3^8}{2^{15}\cdot3^6}\\ =\dfrac{3^2}{2^7}\\ =\dfrac{9}{128}\)

\(\text{c) }\dfrac{6^3+3\cdot6^2+3^3}{-13}\\ =\dfrac{6^3+3\cdot6^2+3^3}{-13}\\ =\\ \dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}\\ =\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}\\ =\dfrac{8\cdot3^3+4\cdot3^3+3^3}{-13}\\ =\dfrac{3^3\cdot\left(8+4+1\right)}{-13}\\ =\dfrac{3^3\cdot13}{-13}\\ =-3^3\\ =-27\)

\(\text{d) }E=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\\ 3E=3\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\right]\\ 3E=1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^{99}\\ 3E-E=\left[1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^{99}\right]-\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\right]\\ 2E=1-\left(\dfrac{1}{3}\right)^{100}\\ 2E=1-\dfrac{1}{3^{100}}\\ E=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

\(\text{e) }\dfrac{\left(0.9\right)^7}{\left(0.3\right)^8}\\ =\dfrac{\left(0.3\cdot3\right)^7}{\left(0.3\right)^8}\\ =\dfrac{\left(0.3\right)^7\cdot3^7}{\left(0.3\right)^8}\\ =\dfrac{3^7}{0.3}=\dfrac{3^6}{0.1}\\ =\dfrac{729}{\dfrac{1}{10}}\\ =729\cdot10\\ =7290\)

\(A=100^2+200^2+300^2+...+1000^2\)

\(\Rightarrow A=100\left(1^2+2^2+3^2+...+10^2\right)\)

Mà \(1^2+2^2+3^2+...+10^2=385\)

\(\Rightarrow A=100\cdot385\)

\(\Rightarrow A=38500\)

Vậy \(A=38500\)

\(\text{Câu 1 : }\\ \sqrt{\dfrac{4}{9}-\sqrt{\dfrac{25}{36}}}=\sqrt{\dfrac{4}{9}-\dfrac{5}{6}}=\sqrt{\dfrac{-7}{18}}\\ \text{Mà }-\dfrac{7}{18}< 0\\ \Rightarrow\text{ Biểu thức không có giá trị}\)

\(\text{Câu 2 :}\)

\(\text{a) }\left(x-1\right)^2=\dfrac{9}{16}\\ \Leftrightarrow\left(x-1\right)=\left(\dfrac{3}{4}\right)^2\\ \Leftrightarrow x-1=\dfrac{3}{4}\\ \Leftrightarrow x=\dfrac{7}{4}\\ \text{Vậy }x=\dfrac{7}{4}\)

\(\text{b) }x-2\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=\sqrt{2}\)

\(\text{c) }x=\sqrt{x}\\ \Leftrightarrow x-\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=1\)

\(71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\\ =71\dfrac{38}{45}-43\dfrac{8}{45}+1\dfrac{17}{57}\\ =\left(71\dfrac{38}{45}-43\dfrac{8}{45}\right)+1\dfrac{17}{57}\\ =28\dfrac{2}{3}+1\dfrac{17}{57}\\ =\dfrac{86}{3}+\dfrac{74}{57}\\ =\dfrac{1708}{57}\)

\(\text{Ta có : }\dfrac{a}{b}=\dfrac{-2.5}{4.5}\Rightarrow\dfrac{a}{b}=\dfrac{-5}{9}\Rightarrow\dfrac{a}{-5}=\dfrac{b}{9}\\ a+b=1.44\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{a}{-5}=\dfrac{b}{9}=\dfrac{a+b}{-5+9}=\dfrac{1.44}{4}=\dfrac{9}{25}=0.36\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra : \(\dfrac{a}{-5}=0.36\Rightarrow a=-1.8\)

Vậy \(a=-1.8\)

\(\text{Câu 1 :}\)

\(A=\dfrac{5}{17}+\dfrac{-4}{9}-\dfrac{20}{31}+\dfrac{12}{17}-\dfrac{11}{31}\\ A=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{20}{31}+\dfrac{11}{31}\right)+\dfrac{-4}{9}\\ A=1-1+-\dfrac{4}{9}\\ A=-\dfrac{4}{9}\)

\(B=\dfrac{-3}{7}+\dfrac{7}{15}+\dfrac{-4}{7}+\dfrac{8}{15}-\dfrac{-2}{3}\\ B=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{7}{15}+\dfrac{8}{18}\right)-\dfrac{-2}{3}\\ B=\left(-1\right)+1+\dfrac{2}{3}\\ B=\dfrac{2}{3}\)

\(\text{Câu 2 : }\)

\(A< \dfrac{x}{9}\le B\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{2}{3}\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{6}{9}\\ \Rightarrow-4< x\le6\\ \Rightarrow x\in\left\{\pm4;\pm3;\pm2;\pm1;0;5;6\right\}\)

Mình đặt tên cho từng đường thẳng và điểm nhé Trần Nguyễn Hoài Thư

Ta có : \(x\) \(\perp\) \(z\)

\(y\) \(\perp\) \(z\)

\(\Rightarrow x\) \(\backslash\backslash\) \(y\)

\(\Rightarrow xCD+yDC=180^o\left(\text{2 góc trong cùng phía}\right)\)

\(hay:155^o+x=180^o\)

\(\Rightarrow x=180^o-155^o\)

\(\Rightarrow x=35^o\)

Vậy \(x=35^o\)

Ta có nghiệm của đa thức thỏa mãn :

\(x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy nghiệm của đa thức là \(x=0\) hoặc \(x=-1\)