ở sao đăng câu hỏi mà ko có bài vậy bạn troll ak
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
ở sao đăng câu hỏi mà ko có bài vậy bạn troll ak
Tính giá trị biểu thức:
A= \(\frac{\text{(a+1)(a+2)(a+3)....(a+2003)(a+2004) }}{(b+5)(b+6)(b+7)....(b+2006)(b+2007)}\) tại a= 0, b= -4
B= \(\frac{1}{\text{(x−5)(y+7) }}+\frac{1}{(x−4)(y+8)}+....+\frac{1}{(x−1)(y+11)}\)tại x= 6, y= -5
\(\frac{\text{1}}{6}\)x + \(\frac{\text{1}}{\text{1}0}\) x - \(\frac{4}{5}\) x + 1 = 0
( Câu siêu khó dành cho học sinh đội tuyển Toán )
( Giúp với )
Bài 10 :
b) \(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}\)
Thu gọn các đa thức sau:
a) \(3xyz^2+\left(\frac{-4}{8}xyz^5\right)\text{ nhân}\frac{1}{3}xyz\)
b) \(3xyz^5\text{nhân}\left(\frac{-1}{7}xyz^2\right)\text{nhân}\frac{-1}{8}xyz^4\)
\(\frac{\text{1}}{5x8}\) + \(\frac{\text{1}}{8x\text{1}\text{1}}\)+ \(\frac{\text{1}}{\text{1}\text{1}x\text{1}4}\) + ..... + \(\frac{\text{1}}{x\left(x+3\right)}\)= \(\frac{\text{1}0\text{1}}{\text{1}540}\)
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 \(xy\text{=}1200\)
tính:
\(\frac{\text{x+1}}{10}+\frac{\text{x+1}}{11}-\frac{\text{x+1}}{13}+\frac{\text{x+1}1}{12}-\frac{\text{x+1}}{14}=0\)
\(3,2.\frac{15}{16}-\left(75\%+\frac{2}{7}\right):\left(-1\frac{1}{28}\right)\)
\(\left(0,25+12,5-\frac{5}{16}\right):\left[12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right]\)
\(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(14,5-\frac{8}{9}:\left(35-34\frac{8}{9}\right).\frac{9}{8}\)
\(1\frac{1}{15}-\left(\frac{1}{15}+\frac{4}{9}:\frac{-2}{3}-\frac{28}{16}.\frac{6}{35}\right)-\frac{3}{10}\)
Tìm x
\(\left(4,5-2x\right)\left(-3\frac{2}{3}\right)=\frac{11}{15}\)
\(\backslash34-x\backslash=\left(-3\right)^4\)
\(\left(4x^2-1\right)\left(\text{\x}\backslash-\frac{2}{3}\right)=0\)
\(\frac{3}{5}x-\frac{1}{2}\)\(x=\frac{-7}{20}\)
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)