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1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
câu a :
\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)
câu b :
\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)
câu c :
\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)
câu d :
\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)
câu e :
\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)
câu g :
\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)
câu h :
\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)
\(E=1.2.3+2.3.4+3.4.5+...+18.19.20\)
\(4E=1.2.3.4+2.3.4.4+3.4.5.4+...+18.19.20.4\)
\(4E=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+18.19.20.\left(21-17\right)\)
\(4E=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+18.19.20.21-17.18.19.20\)
\(4E=18.19.20.21\)
\(4E=143640\)
\(E=\frac{143640}{4}\)
\(E=35910\)
Chúc bạn học tốt ~
\(G=2.4.6+4.6.8+6.8.10+...+18.20.22\) ( xem lại đề có nhầm dấu ko nha bn )
\(8G=2.4.6.8+4.6.8.8+6.8.10.8+...+18.20.22.8\)
\(8G=2.4.6.8+4.6.8.\left(10-2\right)+6.8.10.\left(12-4\right)+...+18.20.22\left(24-16\right)\)
\(8G=2.4.6.8+4.6.8.10-2.4.6.8+6.8.10.12-4.6.8.10+...+18.20.22.24-16.18.20.22\)
\(8G=18.20.22.24\)
\(8G=190080\)
\(G=\frac{190080}{8}\)
\(G=23760\)
Chúc bạn học tốt ~
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
1+2+3+4+5+6+7+8+9+10=55
11+12+13+14+15+16+17+18+19+20=155
1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362
\(A=\frac{5}{2.4.6}+\frac{5}{4.6.8}+...+\frac{5}{16.18.20}\)
\(4A=5.\left(\frac{4}{2.4.6}+\frac{4}{4.6.8}+...+\frac{4}{16.18.20}\right)\)
\(4A=5.\left(\frac{1}{2.4}-\frac{1}{4.6}+\frac{1}{4.6}-\frac{1}{6.8}+...+\frac{1}{16.18}-\frac{1}{18.20}\right)\)
\(4A=5.\left(\frac{1}{8}-\frac{1}{360}\right)\)
\(4A=5.\left(\frac{45}{360}-\frac{1}{360}\right)\)
\(4A=5.\frac{44}{360}\)
\(A=\frac{5.44}{360.4}\)
\(A=\frac{11}{72}\)
Tham khảo~
\(A=\frac{5}{2\cdot4\cdot6}+\frac{5}{4\cdot6\cdot8}+.....+\frac{5}{16\cdot18\cdot20}\)
\(\frac{4}{5}A=\frac{4}{2\cdot4\cdot6}+\frac{4}{4\cdot6\cdot8}+.........+\frac{4}{16\cdot18\cdot20}\)
\(\frac{4}{5}A=\frac{1}{2\cdot4}-\frac{1}{4\cdot6}+\frac{1}{4\cdot6}-\frac{1}{6\cdot8}+.........+\frac{1}{16\cdot18}-\frac{1}{18\cdot20}\)
\(\frac{4}{5}A=\frac{1}{2\cdot4}-\frac{1}{18\cdot20}\)
\(\frac{4}{5}A=\frac{1}{8}-\frac{1}{360}\)
\(\frac{4}{5}A=\frac{11}{90}\)
\(A=\frac{11}{90}:\frac{4}{5}=\frac{11}{72}\)