ở sao đăng câu hỏi mà ko có bài vậy bạn troll ak
Các câu hỏi tương tự
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)}
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( 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 biếta,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-3right)^4text{}left(4x-3right)^2d,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 xytext{}1200
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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(-1frac{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(35-34frac{8}{9}right).frac{9}{8}1frac{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(4,5-2xright)left(-3frac{2}{3}right)frac{11}{15}backslash34-xbackslashleft(-3right)^4left(4...
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\(3,2.\frac{15}{16}-\left(75\%+\frac{2}{7}\right):\left(-1\frac{1}{28}\right)\)
\(\left(0,25+12,5-\frac{5}{16}\right):\left[12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right]\)
\(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(14,5-\frac{8}{9}:\left(35-34\frac{8}{9}\right).\frac{9}{8}\)
\(1\frac{1}{15}-\left(\frac{1}{15}+\frac{4}{9}:\frac{-2}{3}-\frac{28}{16}.\frac{6}{35}\right)-\frac{3}{10}\)
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(2-1frac{2}{5}right)^2+left|1-frac{12}{5}right|-left(-frac{2}{5}right)^2}{frac{1}{2}-frac{3}{7}:left(-2frac{1}{7}right)+left|frac{1}{5}-frac{1}{2}right|-frac{2}{5}}.text{B}frac{6:frac{3}{5}+1frac{1}{6}cdotfrac{6}{7}+left(frac{1}{3}-frac{1}{4}-frac{1}{12}right)}{frac{4}{11}-frac{2}{22}+5frac{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)
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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)