Các bạn nào mà làm bài này thì ghi chữ ❝ trả lời ❞ và các cố gắng ghi từng bước ra nhé !
Câu 1)
1) \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)=
2)\(12\div\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)=
3)\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(0,8-\dfrac{3}{4}\right)^2\)=
4)\(16\dfrac{2}{7}\div\left(\dfrac{-3}{5}\right)+28\dfrac{2}{7}\div\dfrac{3}{5}\)
5)\(\left(2^2\div\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
6)\(\left(\dfrac{1}{3}\right)^{50}\times-9^{25}-\dfrac{2}{3}\div4\)
7)\(10\times\sqrt{0,01}\times\sqrt{\dfrac{16}{9}}+3\sqrt{49}-\dfrac{1}{6}\sqrt{4}\)
Bài 2) Tìm x biết
1)\(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
2)\(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\)
3)\(\dfrac{\left(-3\right)^x}{81}=-27\)
4)\(\left|x+0,237\right|=0\)
5)\(\left(x-1\right)^2=25\)
6)\(\left|2x-1\right|=5\)
7)\(\left(x-1\right)^3=\dfrac{-8}{27}\)
8)\(1\dfrac{2}{3}\div\dfrac{x}{4}=6\div0,3\)
9)\(2\dfrac{2}{3}\div x=1\dfrac{7}{9}\div2\dfrac{2}{3}\)
Bài 3)Tìm các số x,y,z biết
1) \(\dfrac{x}{7}=\dfrac{y}{3}\) và \(x-24=y\)
2) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\) và x - y = 48
3) \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\) và x - y = 4009
4) \(\dfrac{x}{2}=\dfrac{y}{3};=\) và x - y - z = 28
5) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\) và 2x +3y -z = -14
6) 3x = y ; 5y = 4z và 6x +7y +8z =456
2.
\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)
Vậy \(A< 3\)
b,
\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)
Vậy \(B< 2\)
\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)
\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)
\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)
\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)
Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)
Vậy P < Q