1/4+1/12+1/24+1/40+...+2/x(x+2)=1010/2022
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NT
1
NM
25 tháng 8 2023
\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)
\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)
LN
25 tháng 8 2023
a,50%+127−21=21+127−21=(21−21)+127=127b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12
=25×5−25:3×12=25×(5−13)×12=25×143×12=1400=25×5−25:3×12=25×(5−31)×12=25×314×12=1400
10 tháng 7 2022
1:
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)
\(\Leftrightarrow x^2+5x=0\)
=>x=0 hoặc x=-5
3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
16 tháng 1 2018
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
14 tháng 4 2015
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}
\)
\(=\frac{210}{840}+\frac{70}{840}+\frac{35}{840}+\frac{21}{840}+\frac{14}{840}+\frac{10}{840}\)
\(=\frac{210+70+35+21+14+10}{840}\)
\(=\frac{360}{840}\)
\(=\frac{3}{7}\)
7 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`5 \times 72 \times 10 \times 2`
`= 5 \times 2 \times 10 \times 72`
`= 10 \times 10 \times 72`
`= 100 \times 72`
`= 7200`
`b)`
`40 \times 125`
`= 4 \times 10 \times 25 \times 5`
`= (5 \times 10) \times (4 \times 25)`
`= 50 \times 100`
`= 5000`
`c)`
`4 \times 2021 \times 25`
`= (4 \times 25) \times 2021`
`= 100 \times 2021`
`= 202100`
`d)`
`16 \times 6 \times 25`
`= 4 \times 4 \times 6 \times 25`
`= (4 \times 25) \times 4 \times 6`
`= 100 \times 24`
`= 2400`
`2,`
`a)`
`24 \times 57 + 43 \times 24`
`= 24 \times (57+43)`
`= 24 \times 100`
`= 2400`
`b)`
`12 \times 19 + 80 \times 12 +12`
`= 12 \times (19 + 80 + 1)`
`= 12 \times 100`
`= 1200`
`c)`
`(36 \times 15 \times 169) \div (5 \times 18 \times 13)`
`= 36 \times 15 \times 169 \div 5 \div 18 \div 13`
`= 6 \times 6 \times 3 \times 5 \times 13 \times 13 \div 5 \div 3 \times 6 \div 13`
`= (6 \div 6) \times (3 \div 3) \times (5 \div 5) \times (13 \div 13) \times 6 \times 13`
`= 6 \times 13`
`= 78`
`d)`
`(44 \times 52 \times 60) \div ( 11 \times 13 \times 15)`
`= 44 \times 52 \times 60 \div 11 \div 13 \div 15`
`= 4 \times 11 \times 13 \times 4 \times 15 \times 4 \div 11 \div 13 \div 15`
`= (11 \div 11) \times (13 \div 13) \times (15 \div 15) \times 4 \times 4 \times`
`= 4 \times 4 \times 4`
`= 64`
`3,`
`a)`
`x - 280 \div 35 = 5 \times 54`
`x - 8 = 270`
`x = 270 + 8`
`x = 278`
`b)`
`(x - 280) \div 35 = 54 \div 4`
`(x - 280) \div 35 = 13,5`
`x - 280 = 13,5 \times 35`
`x - 280 = 472,5`
`x = 472,5 + 280`
`x = 752,5`
`c)`
`(x - 128 + 20) \div 192 = 0`
`x - 128 + 20 = 0 \times 192`
`x - 128 + 20 = 0`
`x - 108 = 0`
`x = 0 + 108`
`x = 108`
`d)`
`4 \times (x + 200) = 460 + 85 \times 4`
`4 \times (x+200) = 460 + 340`
`4 \times (x+200) = 800`
`x + 200 = 800 \div 4`
`x + 200 = 200`
`x = 200 - 200`
`x = 0`
`4,`
`a)`
`7/12 - 5/12`
`= (7 - 5)/12`
`= 2/12`
`= 1/6`
`b)`
`8/11 + 19/11`
`= (8+19)/11`
`= 27/11`
`c)`
`3/8 + 5/12`
`= 9/24 + 10/24`
`= 19/24`
`d)`
`3/4 + 7/12`
`= 9/12 + 7/12`
`= 16/12`
`= 4/3`
`5,`
`a)`
`x - 6/7 = 5/2`
`x = 5/2 + 6/7`
`x = 47/14`
`b)`
`12/7 \div x + 2/3 = 7/5`
`12/7 \div x = 7/5 - 2/3`
`12/7 \div x = 11/15`
`x = 12/7 \div 11/15`
`x = 180/77`
`@` `\text {Kaizuu lv uuu}`
7 tháng 7 2023
Bài 1
a) \(5\times72\times10\times2=\left(5\times2\times10\right)\times72=100\times72=7200\)
b) \(40\times125=5\times\left(8\times125\right)=5\times1000=5000\)
c) \(16\times6\times25=4\times4\times6\times25=\left(4\times6\right)\times\left(4\times25\right)=24\times100=2400\) Bài 2:
a) \(24\times57+43\times24=24\times\left(57+43\right)=24\times100=2400\)
b) \(12\times19+80\times12+12=12\times\left(19+80+1\right)=12\times100=1200\)
c) \(\left(36\times15\times169\right)\div\left(5\times18\times13\right)\)
\(=\left(18\times2\times3\times5\times13\times13\right)\div\left(5\times18\times13\right)\)
\(=\left(2\times3\times13\right)\times\left(18\times5\times13\right)\div\left(5\times18\times13\right)\)
\(=2\times3\times13\)
\(=78\)
d) \(\left(44\times52\times60\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times11\times4\times13\times4\times15\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times4\times4\right)\times\left(11\times13\times15\right)\div\left(11\times13\times15\right)\)
\(=4\times4\times4\)
\(=64\)
Bài 3:
a) \(x-280\div35=5\times54\)
\(x-8=270\)
\(x=270+8\)
\(x=278\)
b) \(\left(x-280\right)\div35=54\div4\)
\(\left(x-280\right)\div35=\dfrac{27}{2}\)
\(x-280=\dfrac{27}{2}\times35\)
\(x-280=\dfrac{945}{2}\)
\(x=\dfrac{945}{2}+280\)
\(x=\dfrac{1505}{2}\)
c) \(\left(x-128+20\right)\div192=0\)
\(x-128+20=0\times192\)
\(x-128+20=0\)
\(x-128=0-20\)
\(x-128=-20\)
\(x=-20+128\)
\(x=108\)
d) \(4\times\left(x+200\right)=460+85\times4\)
\(4\times\left(x+200\right)=460+340\)
\(4\times\left(x+200\right)=800\)
\(x+200=800\div4\)
\(x+200=200\)
\(x=200-200\)
\(x=0\)
Bài 4:
a) \(\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{8}{11}+\dfrac{19}{11}=\dfrac{27}{11}\)
c) \(\dfrac{3}{8}+\dfrac{5}{12}=\dfrac{9}{24}+\dfrac{10}{24}=\dfrac{19}{24}\)
d) \(\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)
Bài 5:
a) \(x-\dfrac{6}{7}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{6}{7}\)
\(x=\dfrac{47}{14}\)
b) \(\dfrac{12}{7}\div x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\dfrac{12}{7}\div x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\dfrac{12}{7}\div x=\dfrac{11}{15}\)
\(x=\dfrac{12}{7}\div\dfrac{11}{15}\)
\(x=\dfrac{180}{77}\)
27 tháng 3 2021
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2021
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
9 tháng 2 2017
a)đặt x^2-5x=y
<=> y^2+10y+24=0
<=>(y^2+2.5y+25)=1
<=>(y+5)^2=1
\(\left[\begin{matrix}y+5=1\\y+5=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}y=-4\\y=-6\end{matrix}\right.\)
với y=-4 <=> x^2-5x=-4<=> x(x-4)-(x-4)=0
<=> (x-4)(x-1)=0=>\(\left[\begin{matrix}x=1\\x=4\end{matrix}\right.\)
với y=-6<=> x^2-5x=-6<=> x(x-2)-3(x-2)=(x-2)(x-3)=>\(\left[\begin{matrix}x=2\\x=3\end{matrix}\right.\)
d) trôi hết đề bạn đăng quá nhiều
(x+2)(x+3)(x+4)(x+5)-24=0
<=>[(x+2)(x+5)][(x+3)(x+4)]-24=0
<=>(x^2+7x+10)(x^2+7x+12)-24=0
đặt x^2+7x+11=t
<=> (t-1)(t+1)-24=0
<=>t^2-1-25=0
<=>t^2=25=> t=+-5
với t=5
x^2+7x+11=5<=> x^2+7x+6=0
{a-b+c=0}=> x=-1 hoặc -6
với t=-5
x^2+7x+11=-5<=> x^2+7x+17=0=> vô nghiệm
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{2}{x\left(x+2\right)}\)
\(=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{x\left(x+2\right)}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}\)
\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{x+2-x}{x\left(x+2\right)}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\)
\(=\frac{1}{2}-\frac{1}{x+2}=\frac{x}{2\left(x+2\right)}=\frac{1010}{2022}\)
\(\Leftrightarrow2022x=2020\left(x+2\right)\)
\(\Leftrightarrow x=2020\)