Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
A=5,19+2,43/8,16.1,13=7,62/9,72=0,78
B=(56/24+2/3).(64/10-4/7)=3.204/35=612/35
Mk chỉ ghi kết quả thôi,cái này nhóm trưởng giảng đấy
Đặt \(A=\left[\left(\frac{-1}{3}\right)^2.\frac{27}{7}+\sqrt{\frac{4}{49}}-3\right]:\frac{4}{7}\)
Ta có : \(\sqrt{\frac{4}{49}}=\pm\frac{2}{7}\)
\(\Rightarrow\orbr{\begin{cases}A=\left(\frac{1}{9}.\frac{27}{7}+\frac{2}{7}-3\right).\frac{7}{4}\\A=\left(\frac{1}{9}.\frac{27}{7}+\frac{-2}{7}-3\right).\frac{7}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=\left(\frac{3}{7}+\frac{2}{7}-3\right).\frac{7}{4}\\A=\left(\frac{3}{7}-\frac{2}{7}-3\right).\frac{7}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}A=\left(\frac{5}{7}-3\right).\frac{7}{4}\\A=\left(\frac{1}{7}-3\right).\frac{7}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=\frac{-16}{7}.\frac{7}{4}\\A=\frac{20}{7}.\frac{7}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=-4\\A=5\end{cases}}\)
TL:
A=[( −1 3 )2. 27 7 +√ 4 49 −3]: 4 7 Ta có : √ 4 49 =± 2 7 ⇒[ A=( 1 9 . 27 7 + 2 7 −3). 7 4 A=( 1 9 . 27 7 + −2 7 −3). 7 4 ⇒[ A=( 3 7 + 2 7 −3). 7 4 A=( 3 7 − 2 7 −3). 7 4 ⇒[ A=( 5 7 −3). 7 4 A=( 1 7 −3). 7 4 ⇒[ A= −16 7 . 7 4 A= 20 7 . 7 4.
^HT^