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1.
\(3x+4⋮x-3\)
\(\Rightarrow3x-9+9+4⋮x-3\)
\(\Rightarrow3x-3\cdot3+13⋮x-3\)
\(\Rightarrow3\left(x-3\right)+13⋮x-3\)
\(3\left(x-3\right)⋮x-3\)
\(\Rightarrow13⋮x-3\)
\(\Rightarrow x-3\inƯ\left(13\right)\) ; \(x\in Z\Rightarrow x-3\in Z\)
\(\Rightarrow x-3\in\left\{-1;1;13;-13\right\}\)
\(\Rightarrow x\in\left\{2;4;16;-10\right\}\)
vậy_____
2.
\(x^2+7⋮x+1 \)
\(\Rightarrow x\cdot x+7⋮x+1\)
\(\Rightarrow x\cdot x+x-x+7⋮x+1\)
\(\Rightarrow x\cdot\left(x+1\right)-x+7⋮x+1\)
\(x\cdot\left(x+1\right)⋮x+1\)
\(\Rightarrow x+7⋮x+1\)
\(\Rightarrow x+1+6⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow6⋮x+1\)
\(\Rightarrow x+1\inƯ\left(6\right)\) ; \(x\in Z\Rightarrow x+1\in Z\)
\(\Rightarrow x+1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(\Rightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
vậy______
3x+4 chia hết cho x-3
3x-9+13 chia hết cho x-3
3.(x-3)+13 chia hết cho x-3
ma 3.(x-3) chia hết cho x-3
13 chia hết cho x-3
x-3 thuoc U(13)={1,-1,13,-13}
suy ra x thuộc{2,4,16,-10}
2x-1 chia hết cho x+1
2x+2-3 chia hết cho x+1
2(x+1)-3 chia hết cho x+1
3 chia hết cho x+1
x+1 thuộc Ư(3)={1,-1,3,-3}
suy ra x thuộc {0,2,-2,-4}
CHÚC BẠN HỌC TỐT NHÉ !!!!!!
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{2003\cdot2005}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)
\(2A=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{2005-2003}{2003\cdot2005}\)
\(2A=\left(\frac{3}{1\cdot3}-\frac{1}{1\cdot3}\right)+\left(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}\right)+\left(\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right)+....+\left(\frac{2005}{2003\cdot2005}-\frac{2003}{2003\cdot005}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2003}-\frac{1}{2005}\)
\(2A=1-\frac{1}{2005}\)
\(2A=\frac{2004}{2005}\)
\(A=\frac{2004}{2005}\div2\)
\(A=\frac{2004}{4010}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
( GẠCH BỎ CÁC PHÂN SỐ GIỐNG NHAU )
\(=1-\frac{1}{2005}\)
\(=\frac{2004}{2005}\)
CHÚC BN HỌC TỐT!!!
\(\frac{4}{9}.\left(\frac{5}{7}+\frac{2}{5}\right)-\frac{4}{9}.\left(\frac{-2}{7}+\frac{7}{5}\right)\)
\(=\frac{4}{9}.[\left(\frac{5}{7}+\frac{2}{5}\right)-\left(\frac{-2}{7}+\frac{7}{5}\right)]\)
\(=\frac{4}{9}.\left(\frac{5}{7}+\frac{2}{5}+\frac{2}{7}-\frac{7}{5}\right)\)
\(=\frac{4}{9}.\left(\frac{5}{7}+\frac{2}{7})+(\frac{2}{5}-\frac{7}{5})\right)\)
\(=\frac{4}{9}.\left(1+\left(-1\right)\right)\)
\(=\frac{4}{9}.0\)
\(=0\)
CHÚC BN HỌC TỐT!!!!!!
Đặt A=\(\frac{1}{1\cdot2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(A=1-\frac{1}{2004}\)
\(A=\frac{2003}{2004}\)
\(9^{2007}=\left(9^2\right)^{1003}\cdot9=\left(\overline{......1}\right)^{1003}\cdot9=\overline{.....1}\cdot9=\overline{......9}\)
vậy chữ số tận cùng của 92007 là 9
\(3^{2004}=\left(3^4\right)^{501}=81^{501}=\overline{......1}\)
vậy chữ số tận cùng của 32004 là 1
Vì giá trị tuyệt đối của x luôn \(_{ }\ge\)0 \(\Rightarrow\)-3 <x\(\le\)0 \(\Rightarrow\)x = -2 ,-1 ,0
Vậy ..........
Mik sửa lại đề
\(\frac{1989.1990+3978}{1992\cdot1991-3984}\)
Trả lời
\(\frac{1989.1990+3978}{1992.1991-3984}\)
\(=\frac{1989.1990+1989\cdot2}{1992.1991-1992.2}\)
\(=\frac{1989.\left(1992-2\right)}{1992\left(1991-2\right)}\)
\(=\frac{1989.1990}{1992.1989}\)
\(=\frac{1990}{1992}\)
\(=\frac{995}{996}\)
a) \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}=\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{\left(10+5\right)}{7}+\frac{\left(10+5\right)}{9}-\frac{\left(10+5\right)}{11}}\)
\(=\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\left(\frac{10}{7}+\frac{5}{7}\right)+\left(\frac{10}{9}+\frac{5}{9}\right)-\left(\frac{10}{11}+\frac{5}{11}\right)}=\frac{0}{\frac{10}{7}+\frac{10}{9}-\frac{10}{11}}=0\)
b) \(\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{1}{73}+\frac{4}{73}-\frac{5}{115}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{1}{73}+\frac{4}{73}-\frac{4}{115}+\frac{1}{115}}\)
\(=\frac{4}{5+\frac{1}{73}-\frac{1}{115}}=0\)
tth làm dài quá, làm gọn hơn nè
a, \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}=\frac{5\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}{15\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}=\frac{5}{15}=\frac{1}{3}\)
b, \(\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{5}{115}}=\frac{4\left(1+\frac{1}{73}-\frac{1}{115}\right)}{5\left(1+\frac{1}{73}-\frac{1}{115}\right)}=\frac{4}{5}\)