\(y=\frac{0,1\left(6\right)+0,\left(3\right)}{x\frac{1}{3}+1\frac{15}{9}}:x=10\)
t
Những câu hỏi liên quan
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...
Đọc tiếp
\(3,2.\frac{15}{16}-\left(75\%+\frac{2}{7}\right):\left(-1\frac{1}{28}\right)\)
\(\left(0,25+12,5-\frac{5}{16}\right):\left[12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right]\)
\(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(14,5-\frac{8}{9}:\left(35-34\frac{8}{9}\right).\frac{9}{8}\)
\(1\frac{1}{15}-\left(\frac{1}{15}+\frac{4}{9}:\frac{-2}{3}-\frac{28}{16}.\frac{6}{35}\right)-\frac{3}{10}\)
Tìm x
\(\left(4,5-2x\right)\left(-3\frac{2}{3}\right)=\frac{11}{15}\)
\(\backslash34-x\backslash=\left(-3\right)^4\)
\(\left(4x^2-1\right)\left(\text{\x}\backslash-\frac{2}{3}\right)=0\)
\(\frac{3}{5}x-\frac{1}{2}\)\(x=\frac{-7}{20}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
Đúng 0
Bình luận (0)
nhớ làm giúp mình nhá mai mình phải đi học r :<
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
a) \(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
b) \(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
Mình làm tiếp câu b nha !
b, Bài giải
\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)
\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)
\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)
\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)
\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)
\(x-2016=-2\text{ : }\frac{31}{56}\)
\(x-2016=-\frac{112}{31}\)
\(x=-\frac{112}{31}+2016\)
\(x=\frac{62384}{31}\)
giải hệ phương trình
1 , left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyleft(y-xright)left(y-1right)left(y+xright)left(y-2right)-2xyend{matrix}right.
2, left{{}begin{matrix}2left(frac{1}{x}+frac{1}{2y}right)+3left(frac{1}{x}-frac{1}{2y}right)^29left(frac{1}{x}+frac{1}{2y}right)-6left(frac{1}{x}-frac{1}{2y}right)^2-3end{matrix}right.
3 , left{{}begin{matrix}frac{xy}{x+y}frac{2}{3}frac{yz}{y+z}frac{6}{5}frac{zx}{z+x}frac{3}{4}end{matrix}right.
4 , left{{}begin...
Đọc tiếp
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Bài 1: Tính
a, 22frac{1}{2}.frac{7}{9}+50%-1,25
b,1,4.frac{15}{49}-left(frac{4}{5}+frac{2}{3}right):2frac{1}{5}
c,125%.left(-frac{1}{2}right)^2:left(1frac{5}{6}-1,6right)+2016^0
d,1,4 .frac{15}{49}-left(20%+frac{2}{3}right):2frac{1}{5}
đ,125%.left(frac{1}{2}right)^2:left(1frac{5}{6}-1,5right)+2016^0
Bài 2 Tìm x biết
a,left|frac{3}{4}.x-frac{1}{2}right|frac{1}{4}
b, frac{5}{3}.x-frac{2}{5}.xfrac{19}{10}
c, left|2.x-frac{1}{3}right|frac{2}{9}
d, x-30%.x-1frac{1}{5}
Đọc tiếp
Bài 1: Tính
a, \(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
b,1,4\(.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
c,\(125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
d,1,4 .\(\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
đ,125%.\(\left(\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,5\right)+2016^0\)
Bài 2 Tìm x biết
a,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
b, \(\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
c, \(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
d, x-30%.x=-1\(\frac{1}{5}\)
Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Đúng 0
Bình luận (1)
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
Đúng 0
Bình luận (1)
a)22/1/2 .7/9 + 50% -1,25
=45/2 . 7/9 + 1/2 - 5/4
=35/2 +1/2 -5/4
= 18 - 5/4
=67/4
b)1,4 . 15/49- (4/5+ 2/3):2/1/5
=7/5 . 15/49 - (12/15 + 10/15) : 11/5
=3/7 - 22/15 : 11/5
=3/7 -2/3
=9/21 -14/21
=-5/21
c)125% . (-1/2)^2 : (1/5/6 - 1,6) + 2016^0
=5/4 . (-1/4) : (11/6- 8/5) +1
=-5/16 : (55/30 - 48/30 ) +1
=-5/16 :7/30 +1
=-75/56 + 1
= 131/56
d)1,4 .15/49 -(20% + 2/3) : 2/1/5
=7/5. 15/49 -(1/5 +2/3) :11/5
=3/7 - ( 3/15 +10/15) :11/5
=3/7 - 13/15 :11/5
=3/7 - 13/33
=8/231
đ)125%. (1/2)^2 : (1/5/6 - 1,5) +2016^0
=5/4 . 1/4 : (11/6 - 3/2)+1
=5/16 : 1/3 +1
=15/16 +1
=31/16
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
bài1: tìm x:a)8 2^x 2^9.2^5b)27 81^3:3^x 243c)left(frac{2}{5}right)^xleft(frac{5}{2}right)^{-3}.left(frac{-2}{5}right)^2d)left(5x+1right)^2frac{36}{49}e)left(x-frac{2}{9}right)^3left(frac{2}{3}right)^6 f)left(8x-1right)^{2n+1}5^{2n+1}(n thuộc N)bài 2:tìm x,y biết:a)x^2+left(y-frac{1}{10}right)^40b)left(frac{1}{2}x-5right)^{20}+left(y^2-frac{1}{4}right)^{10} 6c)left(x-7right)^{x+1}-left(x-yright)^{x+11}0bài 3:tìm giá trị nhỏ nhất:Aleft(2x+frac{1}{3}right)^2-1tìm Gía...
Đọc tiếp
bài1: tìm x:
a)\(8< 2^x< =2^9.2^5\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\frac{2}{5}\right)^x\left(\frac{5}{2}\right)^{-3}.\left(\frac{-2}{5}\right)^2\)
d)\(\left(5x+1\right)^2=\frac{36}{49}\)
e)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\) f)\(\left(8x-1\right)^{2n+1}=5^{2n+1}\)(n thuộc N)
bài 2:tìm x,y biết:
a)\(x^2+\left(y-\frac{1}{10}\right)^4=0\)
b)\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}< =6\)
c)\(\left(x-7\right)^{x+1}-\left(x-y\right)^{x+11}=0\)
bài 3:tìm giá trị nhỏ nhất:
\(A=\left(2x+\frac{1}{3}\right)^2-1\)
tìm Gía trị lớn nhất :\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)
baif4: tìm x,y:
\(x.\left(x-y\right)=\frac{1}{10}\) \(\)
giúp mình với nhé
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
Đọc tiếp
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
3 tháng 7 2019 lúc 7:56
Tìm x , biết :
a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{-1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
Đúng 0
Bình luận (0)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
Đúng 0
Bình luận (0)
a) (1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3) = 0
3 trường hợp:
TH1: 1/7x - 2/7 = 0 <=> 1/7x = 0 + 2/7 <=> 1/7x = 2/7 <=> x = 2.7/7 = 2
=> x = 2
TH2: -1/5x + 3/5 = 0 <=> -1/5x = 0 - 3/5 <=> -1/5x = -3/5 <=> x = (-3/5).(-5) = 3
=> x = 3
TH3: 1/3x + 4/3 = 0 <=> 1/3x = 0 - 4/3 <=> 1/3x = -4/3 <=> x = x = 3.(-4/3) = -4
=> x = -4
Vậy: x = 2, 3, -4
b) 1/6x + 1/10x - 4/15x + 1 = 0
<=> 1/6x + 1/10x - 4/15x = 0 - 1
<=> 1/6x + 1/10x - 4/15x = -1
<=> 1/6x.30 + 1/10x.30 - 4/15x.30 = -1.30
<=> 5x + 3x - 8x = -30
<=> 0 = -30
=> không có x thỏa mãn
Đúng 0
Bình luận (0)
Giải hệ phương trình :
1, left{{}begin{matrix}frac{2}{x}+frac{3}{y-2}4frac{4}{x}+frac{1}{y-2}1end{matrix}right.
2 , left{{}begin{matrix}frac{2}{2x-y}-frac{1}{x+y}0frac{3}{2x-y}-frac{6}{x+y}-1end{matrix}right.
3, left{{}begin{matrix}5left(x+2yright)3x-12x+43left(x-2yright)-15end{matrix}right.
4, left{{}begin{matrix}2x+y7-x+4y10end{matrix}right.
Đọc tiếp
Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
2 , \(\left\{{}\begin{matrix}\frac{2}{2x-y}-\frac{1}{x+y}=0\\\frac{3}{2x-y}-\frac{6}{x+y}=-1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-2y\right)-15\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\)
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
Đúng 0
Bình luận (0)