Tính giá trị biểu thức sau theo cách hợp lí:
\(A=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
Tính giá trị cua biểu thức sau
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(\sqrt[7]{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
Rút gọn biểu thức
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
1) Rút gọn biểu thức theo là cách hợp lý:
A = \(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
2) Tính hợp lý:
M = \(1-\frac{5}{\sqrt{196}}-\frac{5}{\left(2\sqrt{21}\right)^2}-\frac{\sqrt{25}}{204}-\frac{\left(\sqrt{5}\right)^2}{374}\)
3) Có hay không giá trị của x thỏa mãn điều kiện sau:
\(2002.\sqrt{\left(1+x\right)^2}+2003.\sqrt{\left(1-x\right)^2}=0\)
4) Tìm các số x, y, z thỏa mãn đẳng thức:
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
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
tính:
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(\sqrt[7]{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
rút gọn phân số sau:
A = \(\frac{1-\sqrt[1]{49}+\sqrt[1]{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\sqrt{\frac{64}{2}}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
b1: rút gọn biểu thức:
\(A=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7.\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
b2: tìm x, y, z thỏa mãn:
\(\sqrt{\left(x-\sqrt{2}\right)^2}_{ }\)+ \(\sqrt{\left(y+\sqrt{2}\right)^2}^{ }\)+ |x+y+z| = 0
nhanh nhé, ai đúng mk t*** cho !!!
) Tính giá trị của biểu thức sau bằng các hợp lý : A=\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
b) Tính: B=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2017}\right)\)
c) Giả sử x+y+z=2017 và \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{672}\)
TÍNH tổng C=\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
d) Cho ba sô x,y,z thỏa mãn xyz=2017
Tính tổng: D= \(\frac{2017x}{xy+2017x+2017}+\frac{y}{yz+y+2017}+\frac{z}{zx+z+1}\)
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
b
Tổng quát:\(1-\frac{1}{1+2+3+....+n}=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n^2+2n\right)-\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{n\left(n+2\right)-\left(n+2\right)}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Thay số vào,ta được:
\(\frac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\cdot\frac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}\cdot.....\cdot\frac{\left(2017-1\right)\left(2017+2\right)}{2017\left(2017+1\right)}\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{2016\cdot2019}{2017\cdot2018}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2019}{3\cdot4\cdot5\cdot...\cdot2018}\)
\(=\frac{1}{2017}\cdot\frac{2019}{3}=\frac{2019}{6051}\)
1/ Tính giá trị biểu thức
A=\(\frac{7}{35}+\left(-1\frac{3}{4}+\frac{12}{7}\right)-\left(\frac{1}{4}-\frac{2}{7}-\frac{13}{35}\right)-\frac{3}{7}\)\(\frac{3}{7}\)
B=\(\sqrt{\frac{25}{16}-\frac{\left(-3\right)^2}{\left|-4\right|}}+\frac{\sqrt{\left(-7\right)^2}}{4}-3.\left(\frac{5}{2}\right)^2\)
Tính
a) \(2\sqrt{\frac{25}{16}}-3\sqrt{\frac{49}{36}}+4\sqrt{\frac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\frac{1}{2}}\right)^2+\frac{1}{16}.\left(\sqrt{\frac{3}{4}}\right)^2\)
c) \(\frac{2}{3}\sqrt{\frac{81}{16}}-\frac{3}{4}\sqrt{\frac{64}{9}}+\frac{7}{5}.\sqrt{\frac{25}{196}}\)
a) = \(\frac{7}{2}\)
b) = \(\frac{643}{64}\)
c) = 0