Nhãngiải giúp bài 6 7 8 9 nha
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NA
1
14 tháng 8 2020
Ta có:1-2+3-4+5-6+7-8+9-10+11
=(-1)+(-1)+(-1)+(-1)+(-1)+11
=(-5)+11
=6
Vây:1-2+3-4+4-5+5-6+7-8+9-10+11=6
ks nhé!Học tốt!:))
4 tháng 2 2017
\(\frac{2}{9}\)Và \(\frac{1}{6}\)
\(\frac{2}{9}=\frac{2\times2}{9\times2}=\frac{4}{18}\)
\(\frac{1}{6}=\frac{1\times3}{6\times3}=\frac{3}{18}\)
\(\frac{7}{6}\)Và \(\frac{3}{8}\)
\(\frac{7}{6}=\frac{7\times4}{6\times4}=\frac{28}{24}\)
\(\frac{3}{8}=\frac{3\times3}{8\times3}=\frac{9}{24}\)
\(\frac{1}{12}\)Và \(\frac{1}{9}\)
\(\frac{1}{12}=\frac{1\times3}{12\times3}=\frac{3}{36}\)
\(\frac{1}{9}=\frac{1\times4}{9\times4}=\frac{4}{36}\)
NH
13
HA
0
MN
6
PT
15 tháng 9 2018
1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1028=(1+2+3+4+5+6+7+8+9)x3+1028=45x3+1028=135+1028=1163
VT
20 tháng 8 2017
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
C
20 tháng 4 2022
\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)
\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)
mấy câu kia áp dụng là dc!
6, \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)ĐK : \(x\ge0;x\ne4\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)
7, \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)ĐK : \(x\ge0;x\ne4;9\)
\(=\frac{x-4-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{1}{\sqrt{x}-3}\)
9, \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x-7\sqrt{x}-4}\)ĐK : \(x\ge0;x\ne\frac{1}{4}\)
\(=\frac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}+4\right)+\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{4x+17\sqrt{x}+4}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{\left(\sqrt{x}+4\right)\left(4\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{4\sqrt{x}+1}{2\sqrt{x}-1}\)
8, bạn tự làm nhé