\(\frac{11}{40}+\frac{4}{5}-\text{|}\frac{3}{4}-1\frac{5}{12}\text{|}=\frac{3}{20}-x\)
\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}.\text{{}\text{[}5.\left(8-\frac{13}{15}\right)\text{]}-1\frac{1}{3}\)
Các bạn giải giúp mình
Tìm x biết
a,\(\left(\frac{4}{5}\right)^{2x+7}\text{=}\frac{625}{256}\)
b,\(\frac{7^{x+2}+7^{x+1}+7^x}{57}\text{=}\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
c,\(\left(4x-3\right)^4\text{=}\left(4x-3\right)^2\)
d,\(\frac{2x+3}{5x+2}\text{=}\frac{4x+5}{10x+2}\)
e,\(\frac{3x-1}{40-5x}\text{=}\frac{2x-3x}{5x-34}\)
f,\(\frac{15}{x-9}\text{=}\frac{20}{y-12}\text{=}\frac{40}{z-2x}\) và \(xy\text{=}1200\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
Bài 4: Tính hợp lý
A=\(\frac{4}{\text{1⋅2}}+\frac{4}{\text{3⋅5}}+......+\frac{4}{\text{20⋅11⋅2013}}\)
Bài 5: So sánh với 1:
A=\(\frac{1}{\text{1⋅2}}+\frac{1}{\text{2⋅3}}+\frac{1}{\text{3⋅4}}+......+\frac{1}{\text{49⋅50}}\)
Bài 5 :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{59}\)
\(A=1-\frac{1}{50}\)
từ trên ta có : \(1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)
Cho \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{\text{n}}{5^{\text{n}-2}}+...+\frac{11}{5^{12}}\) với \(\text{n}\in\text{N }\).CMR:\(A< \frac{1}{16}\)
tinh nhanh
\(\frac{4}{3\text{x}7}\)+ \(\frac{5}{7\text{x}12}\)+ \(\frac{1}{12\text{x}13}\)+ \(\frac{7}{13\text{x}20}\)+ \(\frac{3}{20\text{x}23}\)+ \(\frac{6}{23\text{x}29}\)
Chó biểu thức :
\(\text{A}=\frac{\left(2-1\frac{2}{5}\right)^2+\left|1-\frac{12}{5}\right|-\left(-\frac{2}{5}\right)^2}{\frac{1}{2}-\frac{3}{7}:\left(-2\frac{1}{7}\right)+\left|\frac{1}{5}-\frac{1}{2}\right|-\frac{2}{5}}\).
\(\text{B}=\frac{6:\frac{3}{5}+1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)}{\frac{4}{11}-\frac{2}{22}+5\frac{2}{11}-\left(\frac{6}{-11}\right)}\).
Tính : \(\frac{\text{A}}{\text{B}}\).
(Trích đề thi HKI 7_Cô giáo dạy toán Phạm Thị Hồng Duyên-THCS Nghiêm Xuyên--Huyện Thường Tín--Tp. Hà Nội)
\(\text{Tính nhanh}:P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
P=(2/3-1/4+5/11).12.11/(5/12+1-7/11).12.11
P=88-33+60/55+132-84
P=115/103
\(\frac{2017}{1\times2\text{×}3}+\frac{2017}{2\text{×}3\text{×}4}+\frac{2017}{3\text{×}4\text{×}5}+..+\frac{2017}{19\text{×}20\text{×}21}\)
\(\frac{2017}{1.2.3}+\frac{2017}{2.3.4}+\frac{2017}{3.4.5}+...+\frac{2017}{19.20.21}\)
\(=2017\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\right)\)
\(=2017.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right)\)
\(=2017.\left(1-\frac{1}{2}-\frac{1}{3}-\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\right)\right)\)
\(=2017.\left(1+\frac{1}{21}\right)\)phá ngoặc trước dấu trừ đổi dấu,rút gọn:
\(=2017.\frac{20}{21}=\frac{40340}{21}\)
a)\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
b )\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
c)\(\text{|}x\text{|}-\frac{3}{4}=\frac{5}{3}\)
d) \(\frac{17}{2}-\text{|}2x-\frac{3}{4}\text{|}=-\frac{7}{4}\)
a) -2 /3 x + 1/5 = 3/10
-2/3x =1/10
x = -3/20
vậy x = -3/20
b) 25/9 - 12/13x = 7/
12/13x = 2
x = 13/6
c) (x) - 3/4 =5/3
(x) = 29/12
x = 29/12 ; -29/-12
d) x = 11/2